{"year": "1972", "tier": "T1", "problem_label": "1", "problem_type": null, "exam": "USAMO", "problem": "Let $(\\mathrm{a}, \\mathrm{b}, \\ldots, \\mathrm{k})$ denote the greatest common divisor of the integers $\\mathrm{a}, \\mathrm{b}, \\ldots \\mathrm{k}$ and $[\\mathrm{a}, \\mathrm{b}, \\ldots, \\mathrm{k}]$ denote their least common multiple. Show that for any positive integers $\\mathrm{a}, \\mathrm{b}, \\mathrm{c}$ we have $(\\mathrm{a}, \\mathrm{b}, \\mathrm{c})^{2}[\\mathrm{a}, \\mathrm{b}][\\mathrm{b}, \\mathrm{c}][\\mathrm{c}, \\mathrm{a}]=[\\mathrm{a}, \\mathrm{b}$, $c]^{2}(a, b)(b, c)(c, a)$.", "solution": "If we express $\\mathrm{a}, \\mathrm{b}, \\mathrm{c}$ as a product of primes then the $\\mathrm{gcd}$ has each prime to the smallest power and the $1 \\mathrm{~cm}$ has each prime to the largest power. So the equation given is equivalent to showing that $2 \\min (\\mathrm{r}, \\mathrm{s}, \\mathrm{t})+\\max (\\mathrm{r}, \\mathrm{s})$ $+\\max (\\mathrm{s}, \\mathrm{t})+\\max (\\mathrm{t}, \\mathrm{r})=2 \\max (\\mathrm{r}, \\mathrm{s}, \\mathrm{t})+\\min (\\mathrm{r}, \\mathrm{s})+\\min (\\mathrm{s}, \\mathrm{t})+\\min (\\mathrm{t}, \\mathrm{r})$ for non-negative integers $\\mathrm{r}, \\mathrm{s}, \\mathrm{t}$. Assume $\\mathrm{r} \\leq \\mathrm{s} \\leq \\mathrm{t}$. Then each side is $2 \\mathrm{r}+\\mathrm{s}+2 \\mathrm{t}$.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl", "problem_match": "# Problem 1", "solution_match": "# Solution"}}
{"year": "1972", "tier": "T1", "problem_label": "2", "problem_type": null, "exam": "USAMO", "problem": "A tetrahedron has opposite sides equal. Show that all faces are acute-angled.", "solution": "Let the tetrahedron be $A B C D$. Let $M$ be the midpoint of $B C$. We have $A M+M D>A D(*)$. Now the triangles $\\mathrm{ABC}$ and $\\mathrm{DCB}$ are congruent because $\\mathrm{AB}=\\mathrm{DC}, \\mathrm{BC}=\\mathrm{CB}$ and $\\mathrm{AC}=\\mathrm{DB}$. Hence $\\mathrm{AM}=\\mathrm{DM}$. Also $\\mathrm{AD}=\\mathrm{BC}=2 \\mathrm{MC}$. So $(*)$ implies that $\\mathrm{AM}>\\mathrm{MC}$. But that implies that angle $\\mathrm{BAC}$ is acute. Similarly for all the other angles.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl", "problem_match": "# Problem 2", "solution_match": "# Solution"}}
{"year": "1972", "tier": "T1", "problem_label": "3", "problem_type": null, "exam": "USAMO", "problem": "$\\mathrm{n}$ digits, none of them 0 , are randomly (and independently) generated, find the probability that their product is divisible by 10 .", "solution": "Answer: $1-(8 / 9)^{n}-(2 / 9)^{n}+(4 / 9)^{n}$.\n\nA number is divisible by 10 iff it has an even number and a 5 amongst its digits. The probability of no 5 is $(8 / 9)^{\\mathrm{n}}$. The probability of no even number is $(5 / 9)^{\\mathrm{n}}$. The probability of no 5 and no even number is $(4 / 9)^{\\mathrm{n}}$. Hence result.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl", "problem_match": "# Problem 3", "solution_match": "# Solution"}}
{"year": "1972", "tier": "T1", "problem_label": "4", "problem_type": null, "exam": "USAMO", "problem": "Let $k$ be the real cube root of 2. Find integers A, B, C, a, b, c such that $\\left|\\left(A x^{2}+B x+C\\right) /\\left(a x^{2}+b x+c\\right)-k\\right|<\\mid$ $\\mathrm{x}-\\mathrm{k} \\mid$ for all non-negative rational $\\mathrm{x}$.", "solution": "Taking the limit, we must have $\\left(\\mathrm{Ak}^{2}+\\mathrm{Bk}+\\mathrm{C}\\right)=\\mathrm{k}\\left(\\mathrm{ak}^{2}+\\mathrm{bk}+\\mathrm{c}\\right)$, so $\\mathrm{A}=\\mathrm{b}, \\mathrm{B}=\\mathrm{c}, \\mathrm{C}=2 \\mathrm{a}$. Now notice that $\\left(b x^{2}+c x+2 a\\right)-k\\left(a x^{2}+b x+c\\right)=(b-a k) x^{2}+(c-b k) x+(2 a-c k)=(x-k)\\left((b-a k) x+c-a k^{2}\\right)$. So we require $\\left|(b-a k) x+c-\\mathrm{ak}^{2}\\right|<\\left|a x^{2}+b x+c\\right|$ for all $x \\geq 0$.\n\nThere are many ways to satisfy this. For example, take $\\mathrm{a}=1, \\mathrm{~b}=\\mathrm{c}=2$. Then ( $\\mathrm{b}-\\mathrm{ak}) \\mathrm{x}$ is always positive and less than $\\mathrm{bx}$ for positive $\\mathrm{x}$, and $\\mathrm{c}-\\mathrm{ak}^{2}$ is positive and less than $\\mathrm{c}$.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl", "problem_match": "# Problem 4", "solution_match": "# Solution"}}
{"year": "1972", "tier": "T1", "problem_label": "5", "problem_type": null, "exam": "USAMO", "problem": "A pentagon is such that each triangle formed by three adjacent vertices has area 1. Find its area, but show that there are infinitely many incongruent pentagons with this property.", "solution": "Let the pentagon be ABCDE. Triangles BCD and ECD have the same area, so $\\mathrm{B}$ and $\\mathrm{E}$ are the same perpendicular distance from $\\mathrm{CD}$, so $\\mathrm{BE}$ is parallel to $\\mathrm{CD}$. The same applies to the other diagonals (each is parallel to the side with which it has no endpoints in common). Let BD and CE meet at X. Then ABXE is a parallelogram, so area $\\mathrm{BXE}=$ area $\\mathrm{EAB}=1$. Also area $\\mathrm{CDX}+$ area $\\mathrm{EDX}=$ area $\\mathrm{CDX}+$ area $\\mathrm{BCX}=1$. Put\narea $\\mathrm{EDX}=\\mathrm{x}$. Then $\\mathrm{DX} / \\mathrm{XB}=$ area $E D X /$ area $\\mathrm{BXE}=\\mathrm{x} / 1$ and also $=$ area $\\mathrm{CDX} / \\operatorname{area} \\mathrm{BCX}=(1-\\mathrm{x}) / \\mathrm{x}$. So $\\mathrm{x}^{2}+$ $x-1=0, x=\\sqrt{5}-1) / 2$ (we know $x<0$, so it cannot be the other root). Hence area $\\operatorname{ABCDE}=3+x=(\\sqrt{ } 5+$ $5) / 2$.\n\nTake any triangle XCD of area $(3-\\sqrt{5}) / 2$ and extend DX to $B$, so that $B C D$ has area 1 , and extend CX to E so that CDE has area 1. Then take BA parallel to CE and EA parallel to BD. It is easy to check that the pentagon has the required property.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl", "problem_match": "# Problem 5", "solution_match": "# Solution"}}
{"year": "1973", "tier": "T1", "problem_label": "1", "problem_type": null, "exam": "USAMO", "problem": "Show that if two points lie inside a regular tetrahedron the angle they subtend at a vertex is less than $\\pi / 3$.", "solution": "Let the tetrahedron be $\\mathrm{ABCD}$ and the points be $\\mathrm{P}$ and $\\mathrm{Q}$. Note that we are asked to prove the result for any vertex, not just some. So consider angle PAQ. Let the rays AP, AQ meet the plane BCD at $\\mathrm{P}^{\\prime}, \\mathrm{Q}^{\\prime}$ respectively. So we have to show that angle $\\mathrm{P}^{\\prime} \\mathrm{AQ}^{\\prime}<60^{\\circ}$ for $\\mathrm{P}^{\\prime}$ and $\\mathrm{Q}^{\\prime}$ interior points of the triangle $\\mathrm{BCD}$. Extend $\\mathrm{P}^{\\prime} \\mathrm{Q}^{\\prime}$ to meet the sides of the triangle at $\\mathrm{X}$ and $\\mathrm{Y}$. Without loss of generality, $\\mathrm{X}$ lies $\\mathrm{BC}$ and $\\mathrm{Y}$ lies on CD. Obviously if is sufficient to show that angle XAY $<60^{\\circ}$.\n\n$\\mathrm{X}$ and $\\mathrm{Y}$ cannot both be vertices (or $\\mathrm{P}^{\\prime}, \\mathrm{Q}^{\\prime}$ would not have been interior points of the triangle and hence $\\mathrm{P}$ and $\\mathrm{Q}$ would not have been strictly inside the tetrahedron). So suppose $\\mathrm{X}$ is not a vertex. We show that $\\mathrm{XY}<=$ $\\mathrm{XD}$. Consider triangle XYD. $\\angle \\mathrm{XDY}<\\angle \\mathrm{BDC}=60^{\\circ}$, but $\\angle \\mathrm{XYD}=\\angle \\mathrm{XCD}+\\angle \\mathrm{CXY} \\geq 60^{\\circ}$, so $\\angle \\mathrm{XDY}<$ $\\angle \\mathrm{XYD}$. Hence $\\mathrm{XY} \\leq \\mathrm{XD}$. But XD = AX (consider, for example, the congruent triangles AXB and DXB). Hence $X Y<$ AX. Similarly, XY $\\leq$ AY. Hence angle XAY $<60^{\\circ}$.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl", "problem_match": "# Problem 1", "solution_match": "# Solution"}}
{"year": "1973", "tier": "T1", "problem_label": "2", "problem_type": null, "exam": "USAMO", "problem": "The sequence $a_{n}$ is defined by $a_{1}=a_{2}=1, a_{n+2}=a_{n+1}+2 a_{n}$. The sequence $b_{n}$ is defined by $b_{1}=1, b_{2}=7$, $b_{n+2}=2 b_{n+1}+3 b_{n}$. Show that the only integer in both sequences is 1 .", "solution": "We can solve the first recurrence relation to give $a_{n}=A(-1)^{n}+B 2^{n}$. Using $a_{1}$ and $a_{2}$, we get $a_{n}=\\left(2^{n}+(-\\right.$ $\\left.1)^{n+1}\\right) / 3$. Similarly, for the second recurrence relation we get $b_{n}=2 \\cdot 3^{n-1}+(-1)^{n}$. So if $a_{m}=b_{n}$ then $2 \\cdot 3^{n}+3(-$ $1)^{\\mathrm{n}}=2^{\\mathrm{m}}+(-1)^{\\mathrm{m}+1}$ or $2^{\\mathrm{m}}=2.3^{\\mathrm{n}}+3(-1)^{\\mathrm{n}}+(-1)^{\\mathrm{m}}$.\n\nIf $m=1$ or 2 , then we find $n=1$ is the only solution, corresponding to the fact that the term 1 is in both sequences. If $\\mathrm{m}>2$, then $2^{\\mathrm{m}}=0 \\bmod 8$. But $3^{\\mathrm{n}}=(-1)^{\\mathrm{n}} \\bmod 4$, so $2.3^{\\mathrm{n}}+3(-1)^{\\mathrm{n}}+(-1)^{\\mathrm{m}}=5(-1)^{\\mathrm{n}}+(-1)^{\\mathrm{m}} \\bmod 8$ which cannot be $0 \\bmod 8$. So there are no solutions for $\\mathrm{m}>2$.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl", "problem_match": "# Problem 2", "solution_match": "# Solution"}}
{"year": "1973", "tier": "T1", "problem_label": "3", "problem_type": null, "exam": "USAMO", "problem": "Three vertices of a regular $2 n+1$ sided polygon are chosen at random. Find the probability that the center of the polygon lies inside the resulting triangle.", "solution": "Answer: $(n+1) /(4 n-2)$.\n\nLabel the first vertex picked as 1 and the others as $2,3, \\ldots, 2 n+1$ (in order). There are $2 n(2 n-1) / 2$ ways of choosing the next two vertices. If the second vertex is 2 (or $2 n+1$ ), then there is just one way of picking the third vertex so that the center lies in the triangle (vertex $n+2$ ). If the second vertex is 3 (or $2 n$ ), then there are two $(n+2, n+3)$ and so on. So the total number of favourable triangles is $2(1+2+\\ldots+n)=n(n+1)$. Thus the required probability is $(n+1) /(4 n-2)$.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl", "problem_match": "# Problem 3", "solution_match": "# Solution"}}
{"year": "1973", "tier": "T1", "problem_label": "4", "problem_type": null, "exam": "USAMO", "problem": "Find all complex numbers $x, y, z$ which satisfy $x+y+z=x^{2}+y^{2}+z^{2}=x^{3}+y^{3}+z^{3}=3$.", "solution": "Answer: $1,1,1$.\n\nWe have $(x+y+z)^{2}=\\left(x^{2}+y^{2}+z^{2}\\right)+2(x y+y z+z x)$, so $x y+y z+z x=3$.\n\n$(x+y+z)^{3}=\\left(x^{3}+y^{3}+z^{3}\\right)+3\\left(x^{2} y+x y^{2}+y^{2} z+y z^{2}+z^{2} x+z x^{2}\\right)+6 x y z$, so $8=\\left(x^{2} y+x y^{2}+y^{2} z+y z^{2}+z^{2} x\\right.$ $\\left.+z x^{2}\\right)+2 x y z$. But $(x+y+z)(x y+y z+z x)=\\left(x^{2} y+x y^{2}+y^{2} z+y z^{2}+z^{2} x+z x^{2}\\right)+3 x y z$, so $x y z=-1$. Hence $x, y, z$ are the roots of the cubic $w^{3}-3 w^{2}+3 w-1=(w-1)^{3}$. Hence $x=y=z=1$.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl", "problem_match": "# Problem 4", "solution_match": "# Solution"}}
{"year": "1973", "tier": "T1", "problem_label": "5", "problem_type": null, "exam": "USAMO", "problem": "Show that the cube roots of three distinct primes cannot be terms in an arithmetic progression (whether consecutive or not).", "solution": "Suppose the primes are $p, q, r$ so that $q^{1 / 3}=p^{1 / 3}+m d, r^{1 / 3}=p^{1 / 3}+n d$, where $m$ and $n$ (but not necessarily $d$ ) are integers. Then $n q^{1 / 3}-m r^{1 / 3}=(n-m) p^{1 / 3}$. Cubing: $n^{3} q-3 n^{2} m q^{2 / 3} r^{1 / 3}+3 n m^{2} q^{1 / 3} r^{2 / 3}-m^{3} r=(n-m)^{3} p$, or $q^{1 / 3} r^{1 / 3}\\left(m r^{1 / 3}-n q^{1 / 3}\\right)=\\left((n-m)^{3} p+m^{3} r-n^{3} q\\right) /(3 m n)$. But $_{m r^{1 / 3}}-n q^{1 / 3}=(m-n) p^{1 / 3}$, so we have $(p q r)^{1 / 3}=((n-$ $\\left.\\mathrm{m})^{3} \\mathrm{p}+\\mathrm{m}^{3} \\mathrm{r}-\\mathrm{n}^{3} \\mathrm{q}\\right) /(3 \\mathrm{mn}(\\mathrm{m}-\\mathrm{n}))(*)$.\n\nIt is now clear that we do not need $\\mathrm{p}, \\mathrm{q}, \\mathrm{r}$ prime, just that $\\mathrm{pqr}$ is not a cube, for then by the usual argument it must be irrational so that $(*)$ is impossible.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl", "problem_match": "# Problem 5", "solution_match": "# Solution"}}
{"year": "1974", "tier": "T1", "problem_label": "1", "problem_type": null, "exam": "USAMO", "problem": "$p(x)$ is a polynomial with integral coefficients. Show that there are no solutions to the equations $p(a)=b$, $\\mathrm{p}(\\mathrm{b})=\\mathrm{c}, \\mathrm{p}(\\mathrm{c})=\\mathrm{a}$, with $\\mathrm{a}, \\mathrm{b}, \\mathrm{c}$ distinct integers.", "solution": "Suppose there $\\mathrm{a}, \\mathrm{b}, \\mathrm{c}$ satisfy the equations. Then $\\mathrm{p}(\\mathrm{x})=(\\mathrm{x}-\\mathrm{a}) \\mathrm{q}(\\mathrm{x})+\\mathrm{b}=(\\mathrm{x}-\\mathrm{b}) \\mathrm{r}(\\mathrm{x})+\\mathrm{c}=(\\mathrm{x}-\\mathrm{c}) \\mathrm{s}(\\mathrm{x})+\\mathrm{a}$ for some polynomials $q(x), r(x), s(x)$ with integer coefficients. Hence $(b-a) q(b)+b=p(b)=c$, so $(b-a)$ divides $(\\mathrm{c}-\\mathrm{b})$. Similarly, $(\\mathrm{c}-\\mathrm{b})$ divides $(\\mathrm{a}-\\mathrm{c})$, and $(\\mathrm{a}-\\mathrm{c})$ divides $(\\mathrm{b}-\\mathrm{a})$. But $(\\mathrm{b}-\\mathrm{a})$ divides $(\\mathrm{c}-\\mathrm{b})$ divides $(\\mathrm{a}-\\mathrm{c})$ implies that $(\\mathrm{b}-\\mathrm{a})$ divides $(\\mathrm{a}-\\mathrm{c})$. So we have $(\\mathrm{b}-\\mathrm{a})$ and $(\\mathrm{a}-\\mathrm{c})$ dividing each other. Hence $(\\mathrm{b}-\\mathrm{a})= \\pm(\\mathrm{a}-\\mathrm{c})$.\n\nIf $\\mathrm{b}-\\mathrm{a}=\\mathrm{a}-\\mathrm{c}$, then $\\mathrm{b}-\\mathrm{c}=(\\mathrm{b}-\\mathrm{a})+(\\mathrm{a}-\\mathrm{c})=2(\\mathrm{a}-\\mathrm{c})$. But $(\\mathrm{b}-\\mathrm{a})$ divides $2(\\mathrm{a}-\\mathrm{c})$ (and both are non-zero since $\\mathrm{a}, \\mathrm{b}, \\mathrm{c}$ are distinct), so that is impossible. If $\\mathrm{b}-\\mathrm{a}=-(\\mathrm{a}-\\mathrm{c})$, the $\\mathrm{b}=\\mathrm{c}$, contradicting the fact that they are distinct. So there are no solutions.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl", "problem_match": "# Problem 1", "solution_match": "# Solution"}}
{"year": "1974", "tier": "T1", "problem_label": "2", "problem_type": null, "exam": "USAMO", "problem": "Show that for any positive reals $\\mathrm{x}, \\mathrm{y}, \\mathrm{z}$ we have $\\mathrm{x}^{\\mathrm{x}} \\mathrm{y}^{\\mathrm{y}} \\mathrm{z}^{\\mathrm{z}} \\geq(\\mathrm{xyz})^{\\mathrm{a}}$, where $\\mathrm{a}$ is the arithmetic mean of $\\mathrm{x}, \\mathrm{y}, \\mathrm{z}$.", "solution": "Without loss of generality $x \\geq y \\geq z$. We have $x^{x} y^{y} \\geq x^{y} y^{x}$, because that is equivalent to $(x / y)^{x} \\geq(x / y)^{y}$ which is obviously true. Similarly $y^{y} z^{z} \\geq y^{z} z^{y}$ and $z^{z} x^{x} \\geq z^{x} x^{z}$. Multiplying these three together we get $\\left(x^{x} y^{y} z^{z}\\right)^{x} \\geq$ $x^{y+z} y^{z+x} z^{x+y}$. Multiplying both sides by $x^{x} y^{y} z^{z}$ gives $\\left(x^{x} y^{y} z^{z}\\right)^{3} \\geq(x y z)^{3 a}$. Taking cube roots gives the required result.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl", "problem_match": "# Problem 2", "solution_match": "# Solution"}}
{"year": "1974", "tier": "T1", "problem_label": "3", "problem_type": null, "exam": "USAMO", "problem": "Two points in a thin spherical shell are joined by a curve shorter than the diameter of the shell. Show that the curve lies entirely in one hemisphere.", "solution": "Suppose the shell has diameter 2. Let $\\mathrm{M}$ be the midpoint of the curve. Let $\\mathrm{O}$ be the center of the shell and $\\mathrm{X}$ the midpoint of MO. Let $S$ be the circle center $X$ radius $\\sqrt{3}) / 2$ in the plane normal to OM. Then $S$ lies in the shell and every point of $\\mathrm{S}$ is a distance (in space) of 1 from M. Hence the curve cannot cross $\\mathrm{S}$ (because if it crossed at $\\mathrm{Y}$, we would have $\\mathrm{SY} \\geq 1$ along the curve, but the curve has length $<2$, so $\\mathrm{SY}<1$. So we have a stronger result than required.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl", "problem_match": "# Problem 3", "solution_match": "# Solution"}}
{"year": "1974", "tier": "T1", "problem_label": "4", "problem_type": null, "exam": "USAMO", "problem": "A, B, C play a series of games. Each game is between two players, The next game is between the winner and the person who was not playing. The series continues until one player has won two games. He wins the series. A is the weakest player, C the strongest. Each player has a fixed probability of winning against a given opponent. A chooses who plays the first game. Show that he should choose to play himself against B.", "solution": "It must be wrong to choose $\\mathrm{B}$ against $\\mathrm{C}$, for then after the first game (whatever its outcome) A would be playing one of the other players $(\\mathrm{X})$, and that player would already have won a game. That is a worse position than playing that person as the first game, because if he loses the game then $\\mathrm{X}$ has won the series, whereas if he lost to $\\mathrm{X}$ on the first game, there is still a chance $\\mathrm{A}$ could win the series. [If he wins the game as the second game, then he is certainly no better off than he would be after winning the match as the first game.]\n\nUse $\\mathrm{XbY}$ to denote that $\\mathrm{X}$ beats $\\mathrm{Y}$. If $\\mathrm{A}$ chooses to play $\\mathrm{B}$ in the first game, then he wins the series if either (1) $\\mathrm{AbB}, \\mathrm{AbC}$, (2) $\\mathrm{AbB}, \\mathrm{CbA}, \\mathrm{BbC}, \\mathrm{AbB}$, or (3) $\\mathrm{BbA}, \\mathrm{CbB}, \\mathrm{AbC}$, $\\mathrm{AbB}$. If $\\mathrm{A}$ plays $\\mathrm{C}$ in the first game, then he wins the series if (1') $\\mathrm{AbC}, \\mathrm{AbB},\\left(2^{\\prime}\\right) \\mathrm{AbC}, \\mathrm{BbA}, \\mathrm{CbB}, \\mathrm{AbC},\\left(3^{\\prime}\\right) \\mathrm{CbA}, \\mathrm{BbC}, \\mathrm{AbB}, \\mathrm{AbC}$. Evidently the probability of (1) is the same as the probability of (1'). If we compare (2) and (3) they are the same except that in (2) A must beat B and in (3') A must beat C. Similarly, if we compare (3) and (2') they are the same except that in (3) A must beat B and in (2') A must beat C. We assume that, since C is a stronger player than $\\mathrm{B}, \\mathrm{A}$ is more likely to beat $\\mathrm{B}$ than $\\mathrm{C}$. Hence $\\operatorname{prob}(2)>\\operatorname{prob}\\left(3^{\\prime}\\right)$ and $\\operatorname{prob}(3)>\\operatorname{prob}\\left(2^{\\prime}\\right)$. Thus A should choose to play B in the first game.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl", "problem_match": "# Problem 4", "solution_match": "# Solution"}}
{"year": "1974", "tier": "T1", "problem_label": "5", "problem_type": null, "exam": "USAMO", "problem": "A point inside an equilateral triangle with side 1 is a distance $\\mathrm{a}, \\mathrm{b}, \\mathrm{c}$ from the vertices. The triangle $\\mathrm{ABC}$ has $\\mathrm{BC}=\\mathrm{a}, \\mathrm{CA}=\\mathrm{b}, \\mathrm{AB}=\\mathrm{c}$. The sides subtend equal angles at a point inside it. Show that sum of the distances of the point from the vertices is 1 .", "solution": "Let $\\mathrm{D}$ be the point inside $\\mathrm{ABC}$, so that $\\angle \\mathrm{ADB}=\\angle \\mathrm{BDC}=120^{\\circ}$. The key is to start from $\\mathrm{ABC}$ and to rotate the triangle $\\mathrm{BDC}$ through $60^{\\circ}$ away from the triangle $\\mathrm{ADB}$. After that everything is routine.\n\nSuppose $\\mathrm{D}$ goes to $\\mathrm{D}^{\\prime}$ and $\\mathrm{C}$ to $\\mathrm{C}^{\\prime}$. Then $\\mathrm{BD}=\\mathrm{BD}^{\\prime}$ and $\\angle \\mathrm{DBD}^{\\prime}=60^{\\circ}$, so $\\mathrm{BDD}^{\\prime}$ is equilateral. Hence $\\angle \\mathrm{D}^{\\prime} \\mathrm{DB}$ $=60^{\\circ} . \\angle \\mathrm{BDA}=120^{\\circ}$, so $\\mathrm{ADD}^{\\prime}$ is a straight line. Also $\\angle \\mathrm{DD} ' \\mathrm{~B}=60^{\\circ}$ and $\\angle \\mathrm{C}^{\\prime} \\mathrm{D}^{\\prime} \\mathrm{B}=120^{\\circ}$, so $\\mathrm{DD}^{\\prime} \\mathrm{C}^{\\prime}$ is a straight line. Thus $A C^{\\prime}$ has length DA + DB + DC.\n\nNote that $\\mathrm{BC}=\\mathrm{BC}^{\\prime}$ and $\\angle \\mathrm{CBC}^{\\prime}=60^{\\circ}$, so $\\mathrm{CBC}^{\\prime}$ is equilateral. Hence $\\angle \\mathrm{CC} \\mathrm{C}^{\\prime} \\mathrm{B}=60^{\\circ}$. Now take $\\mathrm{Y}$ such that $\\mathrm{AC}^{\\prime} \\mathrm{Y}$ is equilateral, $\\mathrm{Y}$ is on the opposite side of $\\mathrm{AC}^{\\prime}$ to $\\mathrm{C}$. Then $\\angle \\mathrm{BC}^{\\prime} \\mathrm{Y}=60^{\\circ}-\\angle \\mathrm{AC} \\mathrm{C}^{\\prime} \\mathrm{B}=\\angle \\mathrm{CC}^{\\prime} \\mathrm{A}$. Also $\\mathrm{BC}^{\\prime}=\\mathrm{CC}^{\\prime}$ and $\\mathrm{YC}^{\\prime}=\\mathrm{AC}^{\\prime}$, so triangles $\\mathrm{BC}^{\\prime} \\mathrm{Y}$ and $\\mathrm{CC}^{\\prime} \\mathrm{A}$ are congruent. Hence $\\mathrm{BY}=\\mathrm{CA}=\\mathrm{b}$. Also $\\mathrm{BC}^{\\prime}=\\mathrm{BC}=$ $\\mathrm{a}$ and $\\mathrm{BA}=\\mathrm{c}$. Thus $\\mathrm{B}$ is a point inside an equilateral triangle and distances $\\mathrm{a}, \\mathrm{b}, \\mathrm{c}$ from the vertices. Hence the triangle must have side 1 . So $\\mathrm{DA}+\\mathrm{DB}+\\mathrm{DC}=\\mathrm{AC}^{\\prime}=1$.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl", "problem_match": "# Problem 5", "solution_match": "# Solution"}}
{"year": "1975", "tier": "T1", "problem_label": "1", "problem_type": null, "exam": "USAMO", "problem": "Show that for any non-negative reals $\\mathrm{x}, \\mathrm{y},[5 \\mathrm{x}]+[5 \\mathrm{y}] \\geq[3 \\mathrm{x}+\\mathrm{y}]+[\\mathrm{x}+3 \\mathrm{y}]$. Hence or otherwise show that (5a)! $(5 b)!/(a!b!(3 a+b)!(a+3 b)!)$ is integral for any positive integers $a, b$.", "solution": "If is obviously sufficient to prove that $[5 x]+[5 y] \\geq[3 x+y]+[x+3 y]$ for $0<x<y<1$. If $2 x \\geq y$, then $[5 x] \\geq$ $[3 x+y]$ and $[5 y] \\geq[x+3 y]$, so the result holds. So assume $2 x<y$. It is now a question of examining a lot of cases.\n\nIf $y<2 / 5$, then $3 x+y<5 y / 2<1$, so $[3 x+y]=0$, and $[3 y+x]<=[5 y]$, so the result holds. If $2 / 5 \\leq y<3 / 5$ and $x<1 / 5$, then $[5 y]=2,[3 x+y]=0$ or 1 and $[x+3 y]=1$. If $2 / 5 \\leq y<3 / 5$ and $x \\geq 1 / 5$, the $[5 x]+[5 y]=3,[3 x+y]$\n$=1,[x+3 y]=1$ or 2 . If $3 / 5 \\leq y<4 / 5$, then $[5 y]=3,[3 x+y]=0$ or $1,[x+3 y]=2$ or 3 . If $y \\geq 4 / 5$ and $x<1 / 5$, then $[5 x]+[5 y]=4,[3 x+y]=0$ or $1,[x+3 y]=2$ or 3 . If $y \\geq 4 / 5$ and $1 / 5 \\leq x$, then $[5 x]+[5 y]=5,[3 x+y]=2$, $[\\mathrm{x}+3 \\mathrm{y}]=2$ or 3 .\n\nThe highest power of a prime $p$ dividing $n!$ is $[n / p]+\\left[n / p^{2}\\right]+\\left[n / p^{3}\\right]+\\ldots$. Thus it is sufficient to show that $[5 \\mathrm{~m} / \\mathrm{p}]+[5 \\mathrm{n} / \\mathrm{p}] \\geq[(3 \\mathrm{~m}+\\mathrm{n}) / \\mathrm{p}]+[(\\mathrm{m}+3 \\mathrm{n}) / \\mathrm{p}]$, which follows immediately from the previous result, putting $\\mathrm{x}$ $=\\mathrm{m} / \\mathrm{p}, \\mathrm{y}=\\mathrm{n} / \\mathrm{p}$.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl", "problem_match": "# Problem 1", "solution_match": "# Solution"}}
{"year": "1975", "tier": "T1", "problem_label": "2", "problem_type": null, "exam": "USAMO", "problem": "Show that for any tetrahedron the sum of the squares of the lengths of two opposite edges is at most the sum of the squares of the other four.", "solution": "Let the vertices be A, B, C, D. We will show that $\\mathrm{AB}^{2}+\\mathrm{CD}^{2} \\leq \\mathrm{AC}^{2}+\\mathrm{AD}^{2}+\\mathrm{BC}^{2}+\\mathrm{BD}^{2}$. Use vectors with origin A. Let the vectors from A to B, C, D be B, C, D respectively. We have to show that $\\mathbf{B}^{2}+(\\mathbf{C}-$ D) $)^{2} \\leq \\mathbf{C}^{2}+\\mathbf{D}^{2}+(\\mathbf{B}-\\mathbf{C})^{2}+(\\mathbf{B}-\\mathbf{D})^{2}$. Rearranging, this is equivalent to $(\\mathbf{B}-\\mathbf{C}-\\mathbf{D})^{2} \\geq 0$.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl", "problem_match": "# Problem 2", "solution_match": "# Solution"}}
{"year": "1975", "tier": "T1", "problem_label": "3", "problem_type": null, "exam": "USAMO", "problem": "A polynomial $p(x)$ of degree $n$ satisfies $p(0)=0, p(1)=1 / 2, p(2)=2 / 3, \\ldots, p(n)=n /(n+1)$. Find $p(n+1)$.", "solution": "The polynomial $(\\mathrm{x}+1) \\mathrm{p}(\\mathrm{x})-\\mathrm{x}$ has degree $\\mathrm{n}+1$ and zeros at $0,1,2, \\ldots, \\mathrm{n}$, so it must be $\\mathrm{k} \\mathrm{x}(\\mathrm{x}-1)(\\mathrm{x}-2) \\ldots(\\mathrm{x}$ $-n)$. Also it has value 1 at $x=-1$, so $k(n+1)!(-1)^{n+1}$. Hence $(n+2) p(n+1)-(n+1)=(-1)^{n+1}$. So for $n$ odd, $p(n+1)=1$, and for $n$ even, $p(n+1)=n /(n+2)$.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl", "problem_match": "# Problem 3", "solution_match": "# Solution"}}
{"year": "1975", "tier": "T1", "problem_label": "4", "problem_type": null, "exam": "USAMO", "problem": "Two circles intersect at two points, one of them $\\mathrm{X}$. Find $\\mathrm{Y}$ on one circle and $\\mathrm{Z}$ on the other, so that $\\mathrm{X}, \\mathrm{Y}$ and $\\mathrm{Z}$ are collinear and $\\mathrm{XY} . \\mathrm{XZ}$ is as large as possible.", "solution": "Let the circle through $\\mathrm{XY}$ have center $\\mathrm{O}$ and the other circle have center $\\mathrm{O}$ '. $\\mathrm{XY}=2 \\mathrm{OX} \\sin \\mathrm{XOY}, \\mathrm{XZ}=2$ $\\mathrm{O}^{\\prime} \\mathrm{X} \\sin \\mathrm{XOZ}$, so we wish to maximise $2 \\sin \\mathrm{XOY} \\sin \\mathrm{XO}$ 'Z. But $1 / 2 \\angle \\mathrm{XOY}=90^{\\circ}-\\angle \\mathrm{OXY}, 1 / 2 \\angle \\mathrm{XO} Z=$ $90^{\\circ}-\\angle \\mathrm{OXZ}$, so $1 / 2 \\angle \\mathrm{XOY}+1 / 2 \\angle \\mathrm{XO} \\mathrm{Z}=\\angle \\mathrm{OXO}$ ', which is fixed. Thus $\\angle \\mathrm{XOY}+\\angle \\mathrm{XO} ' \\mathrm{Z}$ is fixed. But $2 \\sin \\mathrm{XOY} \\sin \\mathrm{XO} \\mathrm{Z}=\\cos \\left(\\mathrm{XOY}-\\mathrm{XO}^{\\prime} \\mathrm{Z}\\right)-\\cos (\\mathrm{XOY}+\\mathrm{XO} \\mathrm{O})$ ), so we maximise by taking $\\mathrm{XOY}=\\mathrm{XO} \\mathrm{Z}$.\n\nNote that in this case $\\angle \\mathrm{OYZ}=\\angle \\mathrm{O}$ 'ZY, so if we extend $\\mathrm{YO}$ and $\\mathrm{ZO}$ ' to meet at $\\mathrm{C}$, then $\\mathrm{CY}=\\mathrm{CZ}$ and hence $\\mathrm{C}$ is the center of a circle containing the two circles and touching them at $\\mathrm{Y}$ and $\\mathrm{Z}$. Also $\\angle \\mathrm{CZY}=\\angle \\mathrm{OXY}$, so O'Z is parallel to OX. Similarly OY is parallel to O'X, which shows how to construct $\\mathrm{Y}$ and $\\mathrm{Z}$.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl", "problem_match": "# Problem 4", "solution_match": "# Solution"}}
{"year": "1975", "tier": "T1", "problem_label": "5", "problem_type": null, "exam": "USAMO", "problem": "A pack of $\\mathrm{n}$ cards, including three aces, is well shuffled. Cards are turned over in turn. Show that the expected number of cards that must be turned over to reach the second ace is $(\\mathrm{n}+1) / 2$.", "solution": "For each arrangement A of the cards, let $\\mathrm{A}^{\\prime}$ be the reflection about the middle of the pack, so that if a card is in position $\\mathrm{m}$ in $\\mathrm{A}$, then it is in position ( $\\mathrm{n}+1-\\mathrm{m}$ ) in $\\mathrm{A}^{\\prime}$. Then all possible arrangements can be grouped into pairs ( $A, A^{\\prime}$ ) (note that $A$ cannot equal $A^{\\prime}$ ). If the position of the second ace in $A$ is $m$, then it is $n+1-m$ in $A^{\\prime}$, so the average over $\\mathrm{A}$ and $\\mathrm{A}^{\\prime}$ is $(\\mathrm{n}+1) / 2$. Hence that is also the average over all the arrangements.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl", "problem_match": "# Problem 5", "solution_match": "# Solution"}}
{"year": "1976", "tier": "T1", "problem_label": "1", "problem_type": null, "exam": "USAMO", "problem": "The squares of a $4 \\times 7$ chess board are colored red or blue. Show that however the coloring is done, we can find a rectangle with four distinct corner squares all the same color. Find a counter-example to show that this is not true for a $4 \\times 6$ board.", "solution": "A counter-example for $4 \\mathrm{x} 6$ is:\n\n| $\\mathrm{R}$ | $\\mathrm{B}$ | $\\mathrm{R}$ | $\\mathrm{B}$ | $\\mathrm{R}$ | $\\mathrm{B}$ |\n| :--- | :--- | :--- | :--- | :--- | :--- |\n| $\\mathrm{R}$ | $\\mathrm{B}$ | $\\mathrm{B}$ | $\\mathrm{R}$ | $\\mathrm{B}$ | $\\mathrm{R}$ |\n| $\\mathrm{B}$ | $\\mathrm{R}$ | $\\mathrm{R}$ | $\\mathrm{B}$ | $\\mathrm{B}$ | $\\mathrm{R}$ |\n| $\\mathrm{B}$ | $\\mathrm{R}$ | $\\mathrm{B}$ | $\\mathrm{R}$ | $\\mathrm{R}$ | $\\mathrm{B}$ |\n\nEvery column has two blue and two red squares and no two columns have the red squares in the same two rows or the blue squares in the same two rows, so there can be no rectangles.\n\nSuppose there is a counter-example for a $4 \\times 7$ rectangle. Suppose it has three red squares in the first column. Then in those rows each remaining column can have at most one red square, so four remaining columns each have at least two blue squares in those three columns. Hence two of those columns have blue squares in the same two rows and hence a blue cornered rectangle. Contradiction. Similarly if there are three blue squares in the first column. So each column must have two red and two blue squares. But there are only 6 ways of choosing 2 items from 4, so two columns must have red squares in the same rows. Contradiction.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl", "problem_match": "# Problem 1", "solution_match": "# Solution"}}
{"year": "1976", "tier": "T1", "problem_label": "2", "problem_type": null, "exam": "USAMO", "problem": "$\\mathrm{AB}$ is a fixed chord of a circle, not a diameter. $\\mathrm{CD}$ is a variable diameter. Find the locus of the intersection of $\\mathrm{AC}$ and $\\mathrm{BD}$.", "solution": "Let the lines meet at $\\mathrm{X}$ and suppose $\\mathrm{X}$ lies outside the circle. $\\angle \\mathrm{AXB}=\\angle \\mathrm{AXD}$ (same angle) $=180 \\mathrm{deg}-\\angle$ $\\mathrm{XAD}-\\angle \\mathrm{XDA}=90^{\\circ}-\\angle \\mathrm{XDA}$ (CD is a diameter, so angle $\\left.\\mathrm{CAD}=90^{\\circ}\\right)=90^{\\circ}-\\angle \\mathrm{BDA}$ (same angle). But $\\angle$ BDA is constant, so $\\angle \\mathrm{AXB}$ is constant and hence $\\mathrm{X}$ lies on a circle through $\\mathrm{A}$ and $\\mathrm{B}$.\n\nLet $\\mathrm{O}$ be the center of the circle $\\mathrm{ABC}$ and $\\mathrm{O}^{\\prime}$ the center of the circle $\\mathrm{ABX}$. We have $\\angle \\mathrm{AOB}=2 \\angle \\mathrm{ADB}, \\angle$ $\\mathrm{AO}^{\\prime} \\mathrm{B}=2 \\angle \\mathrm{AXB}$, so $\\angle \\mathrm{AOB}+\\angle \\mathrm{AO}^{\\prime} \\mathrm{B}=180^{\\circ}$. Hence $\\angle \\mathrm{OAO}^{\\prime}+\\angle \\mathrm{OBO}^{\\prime}=180^{\\circ}$. But $\\angle \\mathrm{OAO}^{\\prime}=\\angle \\mathrm{OBO}^{\\prime}$, so $\\angle \\mathrm{OAO}^{\\prime}=90^{\\circ}$. In other words, the circles are orthogonal.\n\nIf $\\mathrm{X}$ lies inside the circle center $\\mathrm{O}$, then $\\angle \\mathrm{AXB}=\\angle \\mathrm{XAD}+\\angle \\mathrm{XDA}=\\angle \\mathrm{CAD}+\\angle \\mathrm{ADB}$ (same angles) $=$ $90^{\\circ}+\\angle \\mathrm{ADB}$ (CD diameter). So $\\mathrm{X}$ lies on the same circle.\n\nConversely, suppose $\\mathrm{X}$ lies on the circle O'. Extend XA, XB to meet the circle center O at C and D respectively. If $\\mathrm{X}$ lies outside the circle center $\\mathrm{O}$, assume $\\mathrm{C}$ does not lie inside the circle center $\\mathrm{O}^{\\prime}$ (if not use D). Then $\\angle \\mathrm{CAD}=\\angle \\mathrm{AXD}+\\angle \\mathrm{ADX}=\\angle \\mathrm{AXB}+\\angle \\mathrm{ADB}$ (same angles) $=(\\angle \\mathrm{AO} \\mathrm{B}+\\angle \\mathrm{AOB}) / 2=90^{\\circ}$. Hence $\\mathrm{CD}$ is a diameter.\n\nIf $\\mathrm{X}$ lies inside the circle $\\mathrm{O}^{\\prime}$, then $\\angle \\mathrm{ABX}=90^{\\circ}+\\angle \\mathrm{ADB}$. But $\\angle \\mathrm{AXB}=\\angle \\mathrm{ADB}+\\angle \\mathrm{XBD}$. So $\\angle \\mathrm{CBD}=\\angle$ $\\mathrm{XBD}=90^{\\circ}$. Hence CD is a diameter.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl", "problem_match": "# Problem 2", "solution_match": "# Solution"}}
{"year": "1976", "tier": "T1", "problem_label": "3", "problem_type": null, "exam": "USAMO", "problem": "Find all integral solutions to $a^{2}+b^{2}+c^{2}=a^{2} b^{2}$.", "solution": "Answer: $0,0,0$.\n\nSquares must be 0 or $1 \\bmod 4$. Since the rhs is a square, each of the squares on the 1 hs must be $0 \\bmod 4$. So a, $\\mathrm{b}, \\mathrm{c}$ are even. Put $\\mathrm{a}=2 \\mathrm{a}_{1}, \\mathrm{~b}=2 \\mathrm{~b}_{1}, \\mathrm{c}=2 \\mathrm{c}_{1}$. Then $\\mathrm{a}_{1}{ }^{2}+\\mathrm{b}_{1}{ }^{2}+\\mathrm{c}_{1}{ }^{2}=$ square. Repeating, we find that $\\mathrm{a}, \\mathrm{b}, \\mathrm{c}$ must each be divisible by an arbitrarily large power of 2. So they must all be zero.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl", "problem_match": "# Problem 3", "solution_match": "# Solution"}}
{"year": "1976", "tier": "T1", "problem_label": "4", "problem_type": null, "exam": "USAMO", "problem": "A tetrahedron ABCD has edges of total length 1. The angles at A (BAC etc) are all $90^{\\circ}$. Find the maximum volume of the tetrahedron.", "solution": "Answer: $(2 \\sqrt{2}-7) / 162$.\n\nLet the edges at $A$ have lengths $x, y, z$. Then the volume is $x y z / 6$ and the perimeter is $x+y+z+\\sqrt{ }\\left(x^{2}+y^{2}\\right)+$ $\\sqrt{ }\\left(y^{2}+z^{2}\\right)+\\sqrt{ }\\left(z^{2}+x^{2}\\right)=1$. By AM/GM we have $x+y+z \\geq 3 k$, where $k=(x y z)^{1 / 3}$. Also $x^{2}+y^{2} \\geq 2 x y$, so $\\sqrt{ }\\left(x^{2}+y^{2}\\right)+\\sqrt{ }\\left(y^{2}+z^{2}\\right)+\\sqrt{ }\\left(z^{2}+x^{2}\\right) \\geq \\sqrt{ }(2 x y)+\\sqrt{ }(2 y x)+\\sqrt{ }(2 z x)$. By AM/GM that is $>=3 \\sqrt{2}$ k. So we have $1 \\geq$ $3(1+\\sqrt{2}) \\mathrm{k}$. Hence $\\mathrm{k}^{3} \\leq(5 \\sqrt{2}-7) / 27$. Hence the volume is at most $(5 \\sqrt{2}-7) / 162$. This is achieved if $\\mathrm{x}=\\mathrm{y}=\\mathrm{z}$ $=(\\sqrt{ } 2-1) / 3$. Then the other three sides are $(2-\\sqrt{ } 2) / 3$ and the perimeter is 1 .", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl", "problem_match": "# Problem 4", "solution_match": "# Solution"}}
{"year": "1976", "tier": "T1", "problem_label": "5", "problem_type": null, "exam": "USAMO", "problem": "The polynomials $a(x), b(x), c(x), d(x)$ satisfy $a\\left(x^{5}\\right)+x b\\left(x^{5}\\right)+x^{2} c\\left(x^{5}\\right)=\\left(1+x+x^{2}+x^{3}+x^{4}\\right) d(x)$. Show that $\\mathrm{a}(\\mathrm{x})$ has the factor $(\\mathrm{x}-1)$.", "solution": "Take $\\mathrm{k}$ to be a complex 5 th root of 1 , so that $1+\\mathrm{k}+\\mathrm{k}^{2}+\\mathrm{k}^{3}+\\mathrm{k}^{4}=0$. Putting $\\mathrm{x}=\\mathrm{k}, \\mathrm{k}^{2}, \\mathrm{k}^{3}, \\mathrm{k}^{4}$ in the given equation we get:\n\n$\\mathrm{a}(1)+\\mathrm{k} b(1)+\\mathrm{k}^{2} \\mathrm{c}(1)=0$\n\n$a(1)+k^{2} b(1)+k^{4} c(1)=0$\n\n$\\mathrm{a}(1)+\\mathrm{k}^{3} \\mathrm{~b}(1)+\\mathrm{k} c(1)=0$\n\n$a(1)+k^{4} b(1)+k^{3} c(1)=0$\n\nMultiplying by $-k,-k^{2},-k^{3},-k^{4}$ respectively, we get\n\n$-k a(1)-k^{2} b(1)-k^{3} c(1)=0$\n\n$-k^{2} a(1)-k^{4} b(1)-k c(1)=0$\n\n$-k^{3} a(1)-k b(1)-k^{4} c(1)=0$\n\n$-k^{4} a(1)-k^{3} b(1)-k^{2} c(1)=0$\n\nAdding all eight equations gives $5 \\mathrm{a}(1)=0$. Hence $\\mathrm{a}(\\mathrm{x})$ has the root $\\mathrm{x}=1$ and hence the factor $(\\mathrm{x}-1)$.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl", "problem_match": "# Problem 5", "solution_match": "# Solution"}}
{"year": "1977", "tier": "T1", "problem_label": "1", "problem_type": null, "exam": "USAMO", "problem": "For which positive integers $\\mathrm{a}, \\mathrm{b}$ does $\\left(\\mathrm{x}^{\\mathrm{a}}+\\ldots+\\mathrm{x}+1\\right)$ divide $\\left(\\mathrm{x}^{\\mathrm{ab}}+\\mathrm{x}^{\\mathrm{ab}-\\mathrm{b}}+\\ldots+\\mathrm{x}^{2 b}+\\mathrm{x}^{\\mathrm{b}}+1\\right)$ ?", "solution": "Answer: $\\mathrm{a}+1$ and $\\mathrm{b}$ relatively prime.\n\nThe question is when $\\left(\\mathrm{x}^{\\mathrm{a}+1}-1\\right) /(\\mathrm{x}-1)$ divides $\\left(\\mathrm{x}^{\\mathrm{b}(a+1)}-1\\right) /\\left(\\mathrm{x}^{\\mathrm{b}}-1\\right)$ or when $\\left(\\mathrm{x}^{\\mathrm{a}+1}-1\\right)\\left(\\mathrm{x}^{\\mathrm{b}}-1\\right)$ divides $\\left(\\mathrm{x}^{\\mathrm{b}(\\mathrm{a}+1)}-\\right.$ 1)( $x-1)$. Now both $\\left(x^{a+1}-1\\right)$ and $\\left(x^{b}-1\\right)$ divide ( $\\left.x^{b(a+1)}-1\\right)$. They both have a factor $(x-1)$, so if that is their only common factor, then their product divides $\\left(x^{b(a+1)}-1\\right)(x-1)$. That is true if $a+1$ and $b$ are relatively prime, for the roots of $x^{k}-1$ are the kth roots of 1 . Thus if $a+1$ and $b$ are relatively prime, then the only (complex) number which is an (a+1)th root of 1 and a bth root of 1 is 1 .\n\nBut suppose $d$ is a common factor of $a+1$ and $b$, then $\\exp (2 \\pi i / d)$ is a root of both $x^{a+1}-1$ and $x^{b}-1$. It is a root of $\\mathrm{x}^{\\mathrm{b}(a+1)}-1$, but only with multiplicity 1 , so $\\left(\\mathrm{x}^{\\mathrm{a+1}}-1\\right)\\left(\\mathrm{x}^{\\mathrm{b}}-1\\right)$ does not divide $\\left(\\mathrm{x}^{\\mathrm{b}(a+1)}-1\\right)(\\mathrm{x}-1)$.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl", "problem_match": "# Problem 1", "solution_match": "# Solution"}}
{"year": "1977", "tier": "T1", "problem_label": "2", "problem_type": null, "exam": "USAMO", "problem": "The triangles ABC and DEF have AD, BE and CF parallel. Show that $[\\mathrm{AEF}]+[\\mathrm{DBF}]+[\\mathrm{DEC}]+[\\mathrm{DBC}]+$ $[\\mathrm{AEC}]+[\\mathrm{ABF}]=3[\\mathrm{ABC}]+3[\\mathrm{DEF}]$, where $[\\mathrm{XYZ}]$ denotes the signed area of the triangle XYZ. Thus $[\\mathrm{XYZ}]$ is + area XYZ if the order $\\mathrm{X}, \\mathrm{Y}, \\mathrm{Z}$ is anti-clockwise and - area $\\mathrm{XYZ}$ if the order $\\mathrm{X}, \\mathrm{Y}, \\mathrm{Z}$ is clockwise. So, in particular, $[\\mathrm{XYZ}]=[\\mathrm{YZX}]=-[\\mathrm{YXZ}]$.", "solution": "The starting point is that $[\\mathrm{ABC}]=[\\mathrm{XBC}]+[\\mathrm{AXC}]+[\\mathrm{ABX}](*)$ for any point $\\mathrm{X}$. If $\\mathrm{X}$ is inside the triangle, then all the rotations have the same sense. If $\\mathrm{X}$ is outside, then they do not. But it is easy to check that $(*)^{*}$ ) always holds.\n\nSo $[\\mathrm{ABC}]=[\\mathrm{DBC}]+[\\mathrm{ADC}]+[\\mathrm{ABD}],[\\mathrm{DEF}]=[\\mathrm{AEF}]+[\\mathrm{DAF}]+[\\mathrm{DEA}]$. Now, ignoring sign, the triangles $\\mathrm{ABD}$ and DEA have equal area, because they have a common base AD and the same height (since AD is parallel to $\\mathrm{BE})$. But the sign is opposite, so $[\\mathrm{ABD}]+[\\mathrm{DEA}]=0$. Similarly, $[\\mathrm{ADC}]+[\\mathrm{DAF}]=0$, so $[\\mathrm{ABC}]+$ $[\\mathrm{DEF}]=[\\mathrm{DBC}]+[\\mathrm{AEF}]$. Adding the two similar equations (obtained from $\\mathrm{E}$ with $[\\mathrm{ABC}]$ and $\\mathrm{B}$ with $[\\mathrm{DEF}]$, and from $\\mathrm{F}$ with $[\\mathrm{ABC}]$ and $\\mathrm{C}$ with $[\\mathrm{DEF}]$ ) gives the required result.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl", "problem_match": "# Problem 2", "solution_match": "# Solution"}}
{"year": "1977", "tier": "T1", "problem_label": "3", "problem_type": null, "exam": "USAMO", "problem": "Prove that the product of the two real roots of $x^{4}+x^{3}-1=0$ is a root of $x^{6}+x^{4}+x^{3}-x^{2}-1=0$.", "solution": "Let the roots of the quartic be $\\mathrm{a}, \\mathrm{b}, \\mathrm{c}, \\mathrm{d}$. We show that ab, ac, ad, bc, bd, cd are the roots of the sextic. We have $a+b+c+d=-1, a b+a c+a d+b c+b d+c d=0,1 / a+1 / b+1 / c+1 / d=0, a b c d=-1$.\n\nLet $x^{6}+a_{5} x^{5}+a_{4} x^{4}+a_{3} x^{3}+a_{2} x^{2}+a_{1} x+a_{0}$ be the sextic with roots $a b, a c, a d, b c$, bd, cd. Since their sum is zero, we have $\\mathrm{a}_{5}=0$. Their product is (abcd) ${ }^{3}$, so $\\mathrm{a}_{0}=-1$.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl", "problem_match": "# Problem 3", "solution_match": "# Solution"}}
{"year": "1977", "tier": "T1", "problem_label": "4", "problem_type": null, "exam": "USAMO", "problem": "$\\mathrm{ABCD}$ is a tetrahedron. The midpoint of $\\mathrm{AB}$ is $\\mathrm{M}$ and the midpoint of $\\mathrm{CD}$ is $\\mathrm{N}$. Show that $\\mathrm{MN}$ is perpendicular to $\\mathrm{AB}$ and $\\mathrm{CD}$ iff $\\mathrm{AC}=\\mathrm{BD}$ and $\\mathrm{AD}=\\mathrm{BC}$.", "solution": "Use vectors. Take any origin $\\mathrm{O}$ and write the vector $\\mathrm{OX}$ as $\\mathbf{X}$. Then $\\mathrm{MN}$ perpendicular to $\\mathrm{AB}$ and $\\mathrm{CD}$ is equivalent to:\n\n$(\\mathbf{A}+\\mathbf{B}-\\mathbf{C}-\\mathbf{D}) .(\\mathbf{A}-\\mathbf{B})=0$ and\n\n$(\\mathbf{A}+\\mathbf{B}-\\mathbf{C}-\\mathbf{D}) .(\\mathbf{C}-\\mathbf{D})=0$.\n\nExpanding and adding the two equations gives $(\\mathbf{A}-\\mathbf{D})^{2}=(\\mathbf{B}-\\mathbf{C})^{2}$ or $\\mathrm{AD}=\\mathrm{BC}$. Subtracting gives $\\mathrm{AC}=\\mathrm{BD}$.\n\nConversely, $\\mathrm{AD}=\\mathrm{BC}$ and $\\mathrm{AC}=\\mathrm{BD}$ gives:\n\n$(\\mathbf{A}-\\mathbf{D})^{2}=(\\mathbf{B}-\\mathbf{C})^{2}$ and\n\n$(\\mathbf{A}-\\mathbf{C})^{2}=(\\mathbf{B}-\\mathbf{D})^{2}$.\n\nAdding gives $(\\mathbf{A}+\\mathbf{B}-\\mathbf{C}-\\mathbf{D}) .(\\mathbf{A}-\\mathbf{B})=0$, so MN is perpendicular to AB. Subtracting gives MN\n\nperpendicular to $\\mathrm{CD}$.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl", "problem_match": "# Problem 4", "solution_match": "# Solution"}}
{"year": "1977", "tier": "T1", "problem_label": "5", "problem_type": null, "exam": "USAMO", "problem": "The positive reals $\\mathrm{v}, \\mathrm{w}, \\mathrm{x}, \\mathrm{y}, \\mathrm{z}$ satisfy $0<\\mathrm{h} \\leq \\mathrm{v}, \\mathrm{w}, \\mathrm{x}, \\mathrm{y}, \\mathrm{z} \\leq \\mathrm{k}$. Show that $(\\mathrm{v}+\\mathrm{w}+\\mathrm{x}+\\mathrm{y}+\\mathrm{z})(1 / \\mathrm{v}+1 / \\mathrm{w}+1 / \\mathrm{x}$ $+1 / \\mathrm{y}+1 / \\mathrm{z}) \\leq 25+6(\\sqrt{ }(\\mathrm{h} / \\mathrm{k})-\\sqrt{ }(\\mathrm{k} / \\mathrm{h}))^{2}$. When do we have equality?", "solution": "Fix four of the variables and allow the other to vary. Suppose, for example, we fix all but $x$. Then the expression on the lhs has the form $(r+x)(s+1 / x)=(r s+1)+s x+r / x$, where $r$ and $s$ are fixed. But this is convex. That is to say, as $\\mathrm{x}$ increases if first decreases, then increases. So its maximum must occur at $\\mathrm{x}=\\mathrm{h}$ or $\\mathrm{x}=\\mathrm{k}$. This is true for each variable.\n\nSuppose all five are $\\mathrm{h}$ or all five are $\\mathrm{k}$, then the $\\mathrm{lh}$ is 25 , so the inequality is true and strict unless $\\mathrm{h}=\\mathrm{k}$. If four are $h$ and one is $k$, then the $\\mathrm{hs}$ is $17+4(\\mathrm{~h} / \\mathrm{k}+\\mathrm{k} / \\mathrm{h})$. Similarly if four are $\\mathrm{k}$ and one is $\\mathrm{h}$. If three are $\\mathrm{h}$ and two are $\\mathrm{k}$, then the lhs is $13+6(\\mathrm{~h} / \\mathrm{k}+\\mathrm{k} / \\mathrm{h})$. Similarly if three are $\\mathrm{k}$ and two are $\\mathrm{h}$.\n\n$\\mathrm{h} / \\mathrm{k}+\\mathrm{k} / \\mathrm{h} \\geq 2$ with equality iff $\\mathrm{h}=\\mathrm{k}$, so if $\\mathrm{h}<\\mathrm{k}$, then three of one and two of the other gives a larger lhs than four of one and one of the other. Finally, we note that the rhs is in fact $13+6(\\mathrm{~h} / \\mathrm{k}+\\mathrm{k} / \\mathrm{h})$, so the inequality is true with equality iff either (1) $\\mathrm{h}=\\mathrm{k}$ or (2) three of $\\mathrm{v}, \\mathrm{w}, \\mathrm{x}, \\mathrm{y} \\mathrm{z}$ are $\\mathrm{h}$ and two are $\\mathrm{k}$ or vice versa.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl", "problem_match": "# Problem 5", "solution_match": "# Solution"}}
{"year": "1978", "tier": "T1", "problem_label": "1", "problem_type": null, "exam": "USAMO", "problem": "The sum of 5 real numbers is 8 and the sum of their squares is 16 . What is the largest possible value for one of the numbers?", "solution": "Answer: $16 / 5$.\n\nLet the numbers be $\\mathrm{v}, \\mathrm{w}, \\mathrm{x}, \\mathrm{y}, \\mathrm{z}$. We have $(\\mathrm{v}-6 / 5)^{2}+(\\mathrm{w}-6 / 5)^{2}+\\ldots+(\\mathrm{z}-6 / 5)^{2}=\\left(\\mathrm{v}^{2}+\\ldots+\\mathrm{z}^{2}\\right)-12 / 5(\\mathrm{v}+\\ldots$ $+z)+36 / 5=16-96 / 5+36 / 5=4$. Hence $|v-6 / 5| \\leq 2$, so $v \\leq 16 / 5$. This value can be realized by putting $v=$ $16 / 5$ and setting the other numbers to $6 / 5$.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl", "problem_match": "# Problem 1", "solution_match": "# Solution"}}
{"year": "1978", "tier": "T1", "problem_label": "2", "problem_type": null, "exam": "USAMO", "problem": "Two square maps cover exactly the same area of terrain on different scales. The smaller map is placed on top of the larger map and inside its borders. Show that there is a unique point on the top map which lies exactly above the corresponding point on the lower map. How can this point be constructed?", "solution": "The point is obviously unique, because the two maps have different scales (but if $\\mathrm{P}$ and $\\mathrm{Q}$ where two fixed points the distance between them would be the same on both maps).\n\nLet the small map square be $A^{\\prime} B^{\\prime} C^{\\prime} D^{\\prime}$ and the large be $A B C D$, where $X$ and $X^{\\prime}$ are corresponding points. We deal first with the special case where $A^{\\prime} B^{\\prime}$ is parallel to $A B$. In this case let $A A^{\\prime}$ and $B B^{\\prime}$ meet at $O$. Then triangles $\\mathrm{OAB}$ and $\\mathrm{OA}^{\\prime} \\mathrm{B}^{\\prime}$ are similar, so $\\mathrm{O}$ must represent the same point. So assume $\\mathrm{A}^{\\prime} \\mathrm{B}^{\\prime}$ is not parallel to AB.\n\nLet the lines $\\mathrm{A}^{\\prime} \\mathrm{B}^{\\prime}$ and $\\mathrm{AB}$ meet at $\\mathrm{W}$, the lines $\\mathrm{B}^{\\prime} \\mathrm{C}^{\\prime}$ and $\\mathrm{BC}$ meet at $\\mathrm{X}$, the lines $\\mathrm{C}^{\\prime} \\mathrm{D}^{\\prime}$ and $\\mathrm{CD}$ meet at $\\mathrm{Y}$, and the lines D'A' and DA meet at Z. We claim that the segments WY and XZ meet at a point O inside the smaller square. W cannot lie between $\\mathrm{A}^{\\prime}$ and $\\mathrm{B}^{\\prime}$ (or one of the vertices $\\mathrm{A}^{\\prime}, \\mathrm{B}^{\\prime}$ of the smaller square would lie outside the larger square). If it lies on the opposite side of $\\mathrm{A}^{\\prime}$ to $\\mathrm{B}^{\\prime}$, then $\\mathrm{Y}$ must lie on the opposite side of $\\mathrm{C}^{\\prime}$ to $\\mathrm{D}^{\\prime}$. Thus the segment WY must cut the side $\\mathrm{A}^{\\prime} \\mathrm{D}^{\\prime}$ at some point $\\mathrm{Z}^{\\prime}$ and the side $\\mathrm{B}^{\\prime} \\mathrm{C}^{\\prime}$ at some point $\\mathrm{X}^{\\prime}$. The same conclusion holds if $\\mathrm{W}$ lies on the opposite side of $\\mathrm{B}^{\\prime}$ to $\\mathrm{A}^{\\prime}$, because then $\\mathrm{Y}$ must lie on the opposite side of $\\mathrm{D}^{\\prime}$ to $\\mathrm{C}^{\\prime}$. Similarly, the segment XZ must cut the side $\\mathrm{A}^{\\prime} \\mathrm{B}^{\\prime}$ at some point $\\mathrm{W}^{\\prime}$ and the side $\\mathrm{C}^{\\prime} \\mathrm{D}^{\\prime}$ at some point $Y^{\\prime}$. But now the segments $X^{\\prime} Z^{\\prime}$ and $W^{\\prime} Y^{\\prime}$ join pairs of points on opposite sides of the small square and so they must meet at some point $\\mathrm{O}$ inside the small square.\n\nNow the triangles WOW' and YOY' are similar (WW' and YY' are parallel). Hence OW/OY $=\\mathrm{OW}^{\\prime} / \\mathrm{OY}^{\\prime}$. So if we set up coordinate systems with $\\mathrm{AB}$ as the $\\mathrm{x}$-axis and $\\mathrm{AD}$ as the $\\mathrm{y}$-axis (for the large square) and $\\mathrm{A}^{\\prime} \\mathrm{B}^{\\prime}$ as the $x^{\\prime}$-axis and $A^{\\prime} D^{\\prime}$ as the $y^{\\prime}$-axis (for the small square) so that corresponding points have the same coordinates, then the $y$ coordinate of $\\mathrm{O}$ equals the $\\mathrm{y}^{\\prime}$ coordinate of O. Similarly, XOX' and ZOZ' are similar, so $\\mathrm{OX} / \\mathrm{OZ}=\\mathrm{OX}^{\\prime} / \\mathrm{OZ}$ ', so the $\\mathrm{x}$-coordinate of $\\mathrm{O}$ equals its $\\mathrm{x}^{\\prime}$-coordinate. In other words, $\\mathrm{O}$ represents the same point on both maps.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl", "problem_match": "# Problem 2", "solution_match": "# Solution"}}
{"year": "1978", "tier": "T1", "problem_label": "3", "problem_type": null, "exam": "USAMO", "problem": "You are told that all integers from 33 to 73 inclusive can be expressed as a sum of positive integers whose reciprocals sum to 1 . Show that the same is true for all integers greater than 73.", "solution": "The trick is consider the integers $2 a_{1}, 2 a_{2}, \\ldots, 2 a_{m}$ given that $a_{1}, a_{2}, \\ldots, a_{m}$ is a solution for $n$. The sum of their reciprocals is $1 / 2$. So if we throw in two $4 \\mathrm{~s}$, we get a solution for $2 n+8$. Similarly, adjoining 3 and 6 gives a solution for $2 \\mathrm{n}+9$. It is now a simple induction. For the starter set gives 74 thru 155 , then those give 156 thru 319 , and so on. In general, $n$ thru $2 n+7$ gives $2 n+8$ thru $4 n+23=2(2 n+8)+7$.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl", "problem_match": "# Problem 3", "solution_match": "# Solution"}}
{"year": "1978", "tier": "T1", "problem_label": "4", "problem_type": null, "exam": "USAMO", "problem": "Show that if the angle between each pair of faces of a tetrahedron is equal, then the tetrahedron is regular.\n\nDoes a tetrahedron have to be regular if five of the angles are equal?", "solution": "Answer: no.\n\nLet the tetrahedron be $\\mathrm{ABCD}$. Let the insphere have center $\\mathrm{O}$ and touch the sides at $\\mathrm{W}, \\mathrm{X}, \\mathrm{Y}, \\mathrm{Z}$. The $\\mathrm{OW}$, $\\mathrm{OX}, \\mathrm{OY}, \\mathrm{OZ}$ are the normals to the faces. But the angle between each pair of normals is equal. So $\\mathrm{OW}, \\mathrm{OX}$,\n$\\mathrm{OY}$ and $\\mathrm{OZ}$ are vectors of equal length at equal angles. Hence each side of $\\mathrm{WXYZ}$ is equal (eg $\\mathrm{WX}=2 \\mathrm{OW}$ $\\sin (\\mathrm{WOX} / 2)$ ). So $\\mathrm{WXYZ}$ is a regular tetrahedron. But the faces of $\\mathrm{ABCD}$ are just the tangent planes at $\\mathrm{W}$, $\\mathrm{X}, \\mathrm{Y}, \\mathrm{Z}$. So if we rotate through an angle $120^{\\circ}$ about the line $\\mathrm{OW}$, then $\\mathrm{X}$ goes to $\\mathrm{Y}, \\mathrm{Y}$ to $\\mathrm{Z}$ and $\\mathrm{Z}$ to $\\mathrm{X}$. Hence $\\mathrm{AB}=\\mathrm{AC}, \\mathrm{AC}=\\mathrm{AD}, \\mathrm{AD}=\\mathrm{AB}$ and $\\mathrm{BC}=\\mathrm{CD}=\\mathrm{DB}$ (assuming appropriate labeling). Similarly, for rotation about the other axes. So $A B C D$ has equal edges and hence is regular.\n\nConsider the four normals OW, OX, OY, OZ. We can move $\\mathrm{X}, \\mathrm{Y}, \\mathrm{Z}$ slightly closer together so that $\\mathrm{XYZ}$ remains an equilateral triangle. Then move $\\mathrm{W}$ so that $\\mathrm{WX}=\\mathrm{WY}=\\mathrm{XY}$. So five of the distances are equal, but the sixth is unequal. The reason for slightly is that all the angles between pairs of normals must remain less than $180^{\\circ}$. The corresponding tetrahedron will now have the angles between only five pairs of faces equal.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl", "problem_match": "# Problem 4", "solution_match": "# Solution"}}
{"year": "1978", "tier": "T1", "problem_label": "5", "problem_type": null, "exam": "USAMO", "problem": "There are 9 delegates at a conference, each speaking at most three languages. Given any three delegates, at least 2 speak a common language. Show that there are three delegates with a common language.", "solution": "Suppose not. Then A can shares a language with at most 3 delegates, because if he shared a language with 4 , he would have to share the same language with 2 of them (since he can only speak 3 languages). So there are 5 delegates who do not share a language with A. Let one of them be B. By the same argument, there must be at least one of the other 4 (call her C) who does not share a language with B. But now no two of A, B, C share a language. Contradiction.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl", "problem_match": "# Problem 5", "solution_match": "# Solution"}}
{"year": "1979", "tier": "T1", "problem_label": "1", "problem_type": null, "exam": "USAMO", "problem": "Find all sets of 14 or less fourth powers which sum to 1599 .", "solution": "Answer: none.\n\nThe only 4th powers less than 1599 are $1,16,81,256,625,1296\\left(7^{4}=2401\\right)$. Note that $1,81,625=1 \\mathrm{mod}$ 16 and $16,256,1296=0 \\bmod 16$. But $1599=15 \\bmod 16$, so it cannot be expressed as a sum of 14 or less fourth powers.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl", "problem_match": "# Problem 1", "solution_match": "# Solution"}}
{"year": "1979", "tier": "T1", "problem_label": "2", "problem_type": null, "exam": "USAMO", "problem": "$\\mathrm{N}$ is the north pole. A and $\\mathrm{B}$ are points on a great circle through $\\mathrm{N}$ equidistant from $\\mathrm{N} . \\mathrm{C}$ is a point on the equator. Show that the great circle through $\\mathrm{C}$ and $\\mathrm{N}$ bisects the angle $\\mathrm{ACB}$ in the spherical triangle $\\mathrm{ABC}$ (a spherical triangle has great circle arcs as sides).", "solution": "Let $\\mathrm{S}_{\\mathrm{A}}, \\mathrm{S}_{\\mathrm{B}}, \\mathrm{S}_{\\mathrm{N}}$ be the great circles through $\\mathrm{A}$ and $\\mathrm{C}, \\mathrm{B}$ and $\\mathrm{C}$, and $\\mathrm{N}$ and $\\mathrm{C}$ respectively. Let $\\mathrm{C}^{\\prime}$ be the point directly opposite $\\mathrm{C}$ on the sphere. Then any great circle through $\\mathrm{C}$ also goes through $\\mathrm{C}^{\\prime}$. So, in particular, $\\mathrm{S}_{\\mathrm{A}}$, $\\mathrm{S}_{\\mathrm{B}}$ and $\\mathrm{S}_{\\mathrm{N}}$ go through $\\mathrm{C}^{\\prime}$.\n\nTwo great circles through $\\mathrm{C}$ meet at the same angle at $\\mathrm{C}$ and at $\\mathrm{C}^{\\prime}$, so the spherical angles $\\mathrm{ACN}$ and $\\mathrm{AC}^{\\prime} \\mathrm{N}$ are equal. Now rotate the sphere through an angle $180^{\\circ}$ about the diameter through $\\mathrm{N}$. Then great circles through $\\mathrm{N}$ map into themselves, so $\\mathrm{C}$ and $\\mathrm{C}^{\\prime}$ change places ( $\\mathrm{C}$ is on the equator). Also $\\mathrm{A}$ and $\\mathrm{B}$ change places (they are equidistant from $\\mathrm{N}$ ). $\\mathrm{S}^{\\mathrm{A}}$ must go into another great circle through $\\mathrm{C}$ and $\\mathrm{C}^{\\prime}$. But since $\\mathrm{A}$ maps to $\\mathrm{B}$, it must be $\\mathrm{S}_{\\mathrm{B}}$. Hence the spherical angle $\\mathrm{AC}^{\\prime} \\mathrm{N}=$ angle $\\mathrm{BCN}$ (since one rotates into the other). Hence $\\mathrm{ACN}$ and $\\mathrm{BCN}$ are equal.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl", "problem_match": "# Problem 2", "solution_match": "# Solution"}}
{"year": "1979", "tier": "T1", "problem_label": "3", "problem_type": null, "exam": "USAMO", "problem": "$a_{1}, a_{2}, \\ldots, a_{n}$ is an arbitrary sequence of positive integers. A member of the sequence is picked at random. Its value is a. Another member is picked at random, independently of the first. Its value is $b$. Then a third, value c. Show that the probability that $\\mathrm{a}+\\mathrm{b}+\\mathrm{c}$ is divisible by 3 is at least $1 / 4$.", "solution": "Let the prob of a value $=0,1,2 \\bmod 3$ be $\\mathrm{p}, \\mathrm{q}, \\mathrm{r}$ respectively. So $\\mathrm{p}+\\mathrm{q}+\\mathrm{r}=1$ and $\\mathrm{p}, \\mathrm{q}, \\mathrm{r}$ are non-negative. If $\\mathrm{a}=\\mathrm{b} \\bmod 3$, then to get $\\mathrm{a}+\\mathrm{b}+\\mathrm{c}=0 \\bmod 3$, we require $\\mathrm{c}=\\mathrm{a} \\bmod 3$. So the prob of such events is $\\mathrm{p}^{3}+\\mathrm{q}^{3}+$ $r^{3}$. If the first two values are different mod 3, then the third must be different again, so the prob. is 6 pqr. Thus we have to show that $\\mathrm{p}^{3}+\\mathrm{q}^{3}+\\mathrm{r}^{3}+6 \\mathrm{pqr}>=1 / 4$.\n\n$1=(p+q+r)^{3}=p^{3}+q^{3}+r^{3}+6 p q r+3\\left(p^{2} q+p q^{2}+p^{2} r+p r^{2}+q^{2} r+q^{2}\\right)$. So we have to show that $\\left(p^{2} q+\\right.$ $\\left.\\mathrm{pq}^{2}+\\mathrm{p}^{2} \\mathrm{r}+\\mathrm{pr}^{2}+\\mathrm{q}^{2} \\mathrm{r}+\\mathrm{qr}^{2}\\right)<=1 / 4$. Or $\\mathrm{p}^{2}(\\mathrm{q}+\\mathrm{r})+\\mathrm{p}\\left(\\mathrm{q}^{2}+\\mathrm{r}^{2}\\right)+\\mathrm{qr}(\\mathrm{q}+\\mathrm{r})<=1 / 4$, or $\\mathrm{p}^{2}(1-\\mathrm{p})+\\mathrm{p}(\\mathrm{q}+\\mathrm{r})^{2}+\\mathrm{qr}(\\mathrm{q}+$ $r-2 p) \\leq 1 / 4$, or $p^{2}(1-p)+p(1-p)^{2}+q r(1-3 p)=p(1-p)+q r(1-3 p) \\leq 1 / 4$.\n\nIf $p \\geq 1 / 3$, then we maximise $p(1-p)+q r(1-3 p)$ by taking $q r=0$ and $p=1 / 2$ to maximise $p(1-p)$. Thus the maximum is $1 / 4$, achieved at $p=1 / 2, q=1 / 2, r=0$ and $p=1 / 2, q=0, r=1 / 2$. But because of the symmetry, there is no loss of generality in assuming that $p \\geq q \\geq r$, so $p$ must be $\\geq 1 / 3$. So the maximum value is $1 / 4$.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl", "problem_match": "# Problem 3", "solution_match": "# Solution"}}
{"year": "1979", "tier": "T1", "problem_label": "4", "problem_type": null, "exam": "USAMO", "problem": "$\\mathrm{P}$ lies between the rays $\\mathrm{OA}$ and $\\mathrm{OB}$. Find $\\mathrm{Q}$ on $\\mathrm{OA}$ and $\\mathrm{R}$ on $\\mathrm{OB}$ collinear with $\\mathrm{P}$ so that $1 / \\mathrm{PQ}+1 / \\mathrm{PR}$ is as large as possible.", "solution": "Let the line through $\\mathrm{P}$ parallel to $\\mathrm{OA}$ meet $\\mathrm{OB}$ at $\\mathrm{C}$. Note that $\\mathrm{C}$ is fixed. Let the line through $\\mathrm{C}$ parallel to $\\mathrm{QR}$ meet $\\mathrm{OP}$ at $\\mathrm{D}$. $\\mathrm{D}$ varies as $\\mathrm{Q}$ varies. Triangles $\\mathrm{ODC}, \\mathrm{OPR}$ are similar, so $\\mathrm{CD} / \\mathrm{PR}=\\mathrm{OD} / \\mathrm{OP}$. Also triangles OPQ and $\\mathrm{PDC}$ are similar, so $\\mathrm{CD} / \\mathrm{PQ}=\\mathrm{DP} / \\mathrm{OP}$. Adding, $\\mathrm{CD} / \\mathrm{PR}+\\mathrm{CD} / \\mathrm{PQ}=1$. Hence we maximise $1 / \\mathrm{PQ}+1 / \\mathrm{PR}$ by making $\\mathrm{CD}$ as small as possible. That is achieved by making angle $\\mathrm{CDP}=90^{\\circ}$ and hence QR perpendicular to OP.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl", "problem_match": "# Problem 4", "solution_match": "# Solution"}}
{"year": "1979", "tier": "T1", "problem_label": "5", "problem_type": null, "exam": "USAMO", "problem": "$\\mathrm{X}$ has $\\mathrm{n}$ members. Given $\\mathrm{n}+1$ subsets of $\\mathrm{X}$, each with 3 members, show that we can always find two which have just one member in common.", "solution": "We use induction on $n$. The result is true for $n<5$, because there are at most $n$ distinct subsets of 3 elements. Suppose it is true for all sets $\\mathrm{X}$ with $<\\mathrm{n}$ members. Let $\\mathrm{X}$ have $\\mathrm{n}$ members and consider a collection of $\\mathrm{n}+1$ subsets with 3 members. If every member of $X$ was in at most 3 of the subsets, there would be at most $n$ subsets, so some member A is in at least 4 of the subsets. Suppose one of them is $\\{A, B, C\\}$. There are at least three others, so B (say) must be in at least two of the others. Suppose they are $\\{$ A, B, D $\\}$ and $\\{A, B, E\\}$. Now any other subset containing A must contain B, because otherwise it would have to contain C, D and E, which is impossible. Similarly, any other subset containing B must contain A. So suppose there are $m$ subsets containing $A$ and $B$. If $\\{A, B, K\\}$ is such a subset, then $K$ cannot belong to any other subsets (because if it also belonged to $\\mathrm{S}$, then $\\mathrm{S}$ and $\\{\\mathrm{A}, \\mathrm{B}, \\mathrm{K}\\}$ would have just one member in common). So consider the $n-(m+2)$ people other than those who belong to sets of the type $\\{A, B, K\\}$. There can be at most $\\mathrm{n}-\\mathrm{m}-2$ subsets involving them. Hence at most $\\mathrm{n}-2$ in total. So the result is true for $\\mathrm{n}$.\n\nNote that we do worse than $n$ subsets if we allow any member to be in 4 subsets. But at least for $n=4 m$ we can achieve $n$ subsets. Just take $m$ groups of 4 and take all subsets with 3 members of each group.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl", "problem_match": "# Problem 5", "solution_match": "# Solution"}}
{"year": "1980", "tier": "T1", "problem_label": "1", "problem_type": null, "exam": "USAMO", "problem": "A balance has unequal arms and pans of unequal weight. It is used to weigh two objects of unequal weight. The first object balances against a weight A, when placed in the left pan and against a weight a, when placed in the right pan. The corresponding weights for the second object are $\\mathrm{B}$ and $\\mathrm{b}$. A third object balances against a weight C, when placed in the left pan. What is its true weight?", "solution": "The effect of the unequal arms and pans is that if an object of weight $x$ in the left pan balances an object of weight $\\mathrm{y}$ in the right pan, then $\\mathrm{x}=\\mathrm{hy}+\\mathrm{k}$ for some constants $\\mathrm{h}$ and $\\mathrm{k}$. Thus if the first object has true weight\n$\\mathrm{x}$, then $\\mathrm{x}=\\mathrm{hA}+\\mathrm{k}, \\mathrm{a}=\\mathrm{h} x+\\mathrm{k}$. So $\\mathrm{a}=\\mathrm{h}^{2} \\mathrm{~A}+(\\mathrm{h}+1) \\mathrm{k}$. Similarly, $\\mathrm{b}=\\mathrm{h}^{2} \\mathrm{~B}+(\\mathrm{h}+1) \\mathrm{k}$. Subtracting gives $\\mathrm{h}^{2}=(a-$ b) /(A - B). and hence $(h+1) k=a-h^{2} A=(b A-a B) /(A-B)$.\n\nThe true weight of the third object is thus $h C+k=\\sqrt{ }((a-b) /(A-B)) C+(b A-a B) /(A-B) 1 /(\\sqrt{ }(a-b) /(A-B))$ $+1)$.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl", "problem_match": "# Problem 1", "solution_match": "# Solution"}}
{"year": "1980", "tier": "T1", "problem_label": "2", "problem_type": null, "exam": "USAMO", "problem": "Find the maximum possible number of three term arithmetic progressions in a monotone sequence of $n$ distinct reals.", "solution": "Answer: $\\mathrm{m}^{2}$ for $\\mathrm{n}=2 \\mathrm{~m}+1, \\mathrm{~m}(\\mathrm{~m}-1)$ for $\\mathrm{n}=2 \\mathrm{~m}$.\n\nLet the reals be $\\mathrm{a}_{1}, \\ldots, \\mathrm{a}_{\\mathrm{n}}$. Suppose $\\mathrm{n}=2 \\mathrm{~m}+1$ and the middle term is $\\mathrm{a}_{\\mathrm{k}}$. If $\\mathrm{k}<\\mathrm{m}+1$, then we are constrained by the shortage of first terms. If $k>m+1$ we are constrained by the shortage of third terms. Thus if $k=1$, $\\mathrm{a}_{\\mathrm{k}}$ cannot be the middle term. If $\\mathrm{k}=2$, there is only one candidate for the first term. If $\\mathrm{k}=3$, there are two candidates for the middle terms and so on. Thus the total number of possible progressions certainly cannot exceed: $1+2+\\ldots+\\mathrm{m}+\\mathrm{m}-1+\\mathrm{m}-2+\\ldots+1=\\mathrm{m}(\\mathrm{m}+1) / 2+\\mathrm{m}(\\mathrm{m}-1) / 2=\\mathrm{m}^{2}$. But this bound is achieved by the sequence $1,2,3, \\ldots, \\mathrm{n}$.\n\nSimilarly, if $\\mathrm{n}=2 \\mathrm{~m}$, then the upper bound is $1+2+\\ldots+\\mathrm{m}-1+\\mathrm{m}-1+\\ldots+1=\\mathrm{m}(\\mathrm{m}-1)$. Again, this is achieved by the sequence $1,2, \\ldots, \\mathrm{n}$.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl", "problem_match": "# Problem 2", "solution_match": "# Solution"}}
{"year": "1980", "tier": "T1", "problem_label": "3", "problem_type": null, "exam": "USAMO", "problem": "A $+B+C$ is an integral multiple of $\\pi . x, y, z$ are real numbers. If $x \\sin A+y \\sin B+z \\sin C=x^{2} \\sin 2 A+$ $y^{2} \\sin 2 B+z^{2} \\sin 2 C=0$, show that $x^{n} \\sin n A+y^{n} \\sin n B+z^{n} \\sin n C=0$ for any positive integer $n$.", "solution": "The juxtaposition of $\\mathrm{x}^{2}$ and $\\sin 2 \\mathrm{~A}$ strongly suggests considering $\\cos \\mathrm{A}+\\mathrm{i} \\sin \\mathrm{A}$. So put $\\mathrm{u}=\\mathrm{x}(\\cos \\mathrm{A}+\\mathrm{i} \\sin$ $A), v=y(\\cos B+i \\sin B), w=z(\\cos C+i \\sin C)$. Put $a_{n}=u^{n}+v^{n}+w^{n}$. So $a_{1}$ and $a_{2}$ are real. Hence also uv + $\\mathrm{vw}+\\mathrm{wu}=\\left(\\mathrm{a}_{1}{ }^{2}-\\mathrm{a}_{2}\\right) / 2$ is real. Also $u v w=\\mathrm{xyz} \\exp (\\mathrm{i}(\\mathrm{A}+\\mathrm{B}+\\mathrm{C}))= \\pm \\mathrm{xyz}$, since $\\mathrm{A}+\\mathrm{B}+\\mathrm{C}$ is an integral multiple of $\\pi$. Thus $\\mathrm{u}, \\mathrm{v}, \\mathrm{w}$ are roots of some cubic $\\mathrm{p}^{3}+\\mathrm{ap}^{2}+\\mathrm{bp}+\\mathrm{c}=0$ with real coefficients. Putting $\\mathrm{p}=\\mathrm{u}, \\mathrm{v}, \\mathrm{w}$ and adding, we get $a_{3}+a a_{2}+b a_{1}+3 c=0$, so $a_{3}$ is real. Also multiplying through by $p^{n}$, then putting $p=u, v, w$ and adding, we get: $a_{n+3}+a a_{n+2}+b a_{n+1}+c a_{n}=0$. So by a trivial induction $a_{n}$ is real for all positive $n$. Hence result.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl", "problem_match": "# Problem 3", "solution_match": "# Solution"}}
{"year": "1980", "tier": "T1", "problem_label": "4", "problem_type": null, "exam": "USAMO", "problem": "The insphere of a tetrahedron touches each face at its centroid. Show that the tetrahedron is regular.", "solution": "Let the tetrahedron be $\\mathrm{ABCD}$. Let $\\mathrm{G}$ be the centroid of $\\mathrm{ABC}$ and $\\mathrm{H}$ the centroid of $\\mathrm{ACD}$. Let $\\mathrm{AM}$ be a median in $\\mathrm{ABC}$ and $\\mathrm{AN}$ a median in $\\mathrm{ACD}$. Then $\\mathrm{AG}$ and $\\mathrm{AH}$ are tangents to the insphere, so they are equal. $\\mathrm{CG}$ and $\\mathrm{CH}$ are also tangents and hence equal. So the triangles $\\mathrm{ACG}$ and $\\mathrm{ACH}$ are congruent. Hence $\\angle \\mathrm{AGC}$ $=\\angle \\mathrm{AHC}$ and so $\\angle \\mathrm{CGM}=\\angle \\mathrm{CHN}$. But $\\mathrm{GM}=\\mathrm{AG} / 2=\\mathrm{AH} / 2=\\mathrm{GN}$, so the triangles $\\mathrm{CGM}$ and $\\mathrm{CHN}$ are also congruent. Hence $\\mathrm{CM}=\\mathrm{CN}$. Hence $\\mathrm{CB}=\\mathrm{CD}$. So every pair of adjacent edges is equal. Hence all the edges are equal and the tetrahedron is regular.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl", "problem_match": "# Problem 4", "solution_match": "# Solution"}}
{"year": "1980", "tier": "T1", "problem_label": "5", "problem_type": null, "exam": "USAMO", "problem": "If $x, y, z$ are reals such that $0 \\leq x, y, z \\leq 1$, show that $x /(y+z+1)+y /(z+x+1)+z /(x+y+1) \\leq 1-(1-$ $\\mathrm{x})(1-\\mathrm{y})(1-\\mathrm{z})$.", "solution": "Consider $\\mathrm{x} /(\\mathrm{y}+\\mathrm{z}+1)+\\mathrm{y} /(\\mathrm{z}+\\mathrm{x}+1)+\\mathrm{z} /(\\mathrm{x}+\\mathrm{y}+1)+(1-\\mathrm{x})(1-\\mathrm{y})(1-\\mathrm{z})$ as a function of $\\mathrm{x}$, with $\\mathrm{y}$ and $\\mathrm{z}$ fixed. Each term is convex, so the whole function is convex. Hence its maximum value occurs at its endpoints. The same is true for $\\mathrm{x}$ and $\\mathrm{y}$, so we need only check the eight possible values $\\mathrm{x}, \\mathrm{y}, \\mathrm{z}=0$ or 1 . In fact, we easily find the expression has value 1 at all eight points. The result follows.\n\nA function is convex if for any three points $\\mathrm{a}<\\mathrm{b}<\\mathrm{c}$, the point $(\\mathrm{b}, \\mathrm{f}(\\mathrm{b})$ ) lies on or below the chord joining the points (a, $\\mathrm{f}(\\mathrm{a})$ ) and $(\\mathrm{c}, \\mathrm{f}(\\mathrm{c}))$. Analytically, this means that if $\\mathrm{b}=\\mathrm{ha}+(1-\\mathrm{h}) \\mathrm{c}$, where $0 \\leq \\mathrm{h} \\leq 1$, then $\\mathrm{f}(\\mathrm{b}) \\leq$ $h \\mathrm{f}(\\mathrm{a})+(1-\\mathrm{h}) \\mathrm{f}(\\mathrm{c})$. The linear nature of this relation implies immediately that a sum of convex functions is convex and that a positive multiple of a convex function is convex. Linear functions are obviously convex. It is obvious from the graph that the function $\\mathrm{a} /(\\mathrm{b}+\\mathrm{x})$ is convex. To prove it analytically we must show that $a /(b+h x+(1-h) y)) \\leq h a /(b+x)+(1-h) a /(b+y)$ or $a\\left(b^{2}+b x+b y+x y\\right) \\leq h a(b+y)(b+h x+(1-h) y)+(1-$\n\nh)a(b+x)(b+hx+(1-h)y). After cancelling some terms, we have to show that $x y<=h(1-h) x^{2}+(1-h)^{2} x y+h^{2} x y$ $+h(1-h) y^{2}$. This is obviously true for $h=1$ or 0 . Otherwise we may divide by $1-h$, then $h$ to get $2 x y \\leq x^{2}+y^{2}$, which is true.\n\nTo see that the maximum value of a convex function must occur at its endpoints just draw a chord between the endpoints. All other points of the curve must lie below the chord.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl", "problem_match": "# Problem 5", "solution_match": "# Solution"}}
{"year": "1981", "tier": "T1", "problem_label": "1", "problem_type": null, "exam": "USAMO", "problem": "Prove that if $n$ is not a multiple of 3 , then the angle $\\pi / n$ can be trisected with ruler and compasses.", "solution": "The key is to use $\\pi / 3$. Since $n$ and 3 are relatively prime we can find integers a and $b$ such that $3 a+n b=1$. Hence $a \\pi / n+b \\pi / 3=\\pi /(3 n)$. So take a circle with arcs subtending $\\pi / n$ and $\\pi / 3$ at the center. Start at a point $X$ on the circumference, mark off $|a|$ arcs subtending $\\pi / n$ in one direction, then mark off $|b|$ arcs subtending $\\pi / 3$ in the other direction. We end up at a point $\\mathrm{Y}$ with $\\mathrm{XY}$ subtending $\\pi / 3 \\mathrm{n}$ at the center.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl", "problem_match": "# Problem 1", "solution_match": "# Solution"}}
{"year": "1981", "tier": "T1", "problem_label": "2", "problem_type": null, "exam": "USAMO", "problem": "What is the largest number of towns that can meet the following criteria. Each pair is directly linked by just one of air, bus or train. At least one pair is linked by air, at least one pair by bus and at least one pair by train. No town has an air link, a bus link and a trian link. No three towns, A, B, C are such that the links between $\\mathrm{AB}, \\mathrm{AC}$ and $\\mathrm{BC}$ are all air, all bus or all train.", "solution": "Answer: 4. [eg A and B linked by bus. C and D both linked to A and B by air and to each other by train.]\n\nSuppose A is linked to three other towns by air. Let them be B, C, D. B has at most one other type of link. Suppose it is bus. Then $\\mathrm{B}$ must be linked to $\\mathrm{C}$ and $\\mathrm{D}$ by bus. But now there is a problem with the $\\mathrm{CD}$ link. If it is air, then A, C, D are all linked by air. If it is bus, then B, C, D are all linked by bus. If it is train, then C has links of all three types. So A cannot be linked to three others by air. Similarly it cannot be linked to three others by train, or to three others by bus. Since it has at most two types of link, it can be linked to at most 4 towns ( 2 by one type of link and 2 by another). So certainly there cannot be more than 5 towns.\n\nSuppose 5 is possible. A must be linked to two towns by one type of link and two by another. Without loss of generality we may suppose $\\mathrm{A}$ is linked to $\\mathrm{B}$ and $\\mathrm{C}$ by bus and to $\\mathrm{D}$ and $\\mathrm{E}$ by train. Now suppose $\\mathrm{BE}$ is an air link. Then B cannot have a train link (or it will have all three). It cannot be linked to $C$ by bus (or $A B C$ is all bus), so $\\mathrm{BC}$ must be air. Similarly, ED must be air (it cannot be train or ADE is all train, or bus because then $\\mathrm{E}$ has air, bus and train). But now there is a problem with the $\\mathrm{CE}$ link. It cannot be air (or $\\mathrm{BCE}$ is all air). It cannot be train, because $\\mathrm{C}$ already has air and bus links. It cannot be bus, because $\\mathrm{E}$ already has train and air links. So BE cannot be an air link. So it must be a bus or train link. Without loss of generality, we may assume it is a bus link.\n\nE has a bus and a train link, so it cannot have air links. It cannot be linked to $\\mathrm{D}$ by train (or AED is all train). So ED must be bus. Now DB cannot be air (or D has air, train and bus). It cannot be bus (or DBE is all bus). So it must be train. So $\\mathrm{BC}$ cannot be air (or $\\mathrm{B}$ has train, bus and air). It cannot be bus (or $\\mathrm{ABC}$ is all bus). So $\\mathrm{BC}$ must be train. $\\mathrm{CD}$ cannot be air (or $\\mathrm{C}$ has air, bus and train). It cannot be train (or $\\mathrm{BCD}$ is all train). So CD must be bus. Finally, $\\mathrm{CE}$ cannot be air (or $\\mathrm{C}$ has air, bus and train). It cannot be bus (or CDE is all bus). So CE must be train. So none of the links are air. Contradiction. Hence 5 towns is not possible. But 4 is, as given in the Answer above.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl", "problem_match": "# Problem 2", "solution_match": "# Solution"}}
{"year": "1981", "tier": "T1", "problem_label": "3", "problem_type": null, "exam": "USAMO", "problem": "Show that for any triangle, $3(\\sqrt{3}) / 2 \\geq \\sin 3 A+\\sin 3 B+\\sin 3 C \\geq-2$. When does equality hold?", "solution": "Answer: 1st inequality, $\\mathrm{A}=\\mathrm{B}=20^{\\circ}, \\mathrm{C}=140^{\\circ}$. 2nd inequality, $\\mathrm{A}=0^{\\circ}, \\mathrm{B}=\\mathrm{C}=90^{\\circ}$ (which is degenerate).\n\nFor $\\mathrm{x}$ between 0 and $180^{\\circ}, \\sin 3 \\mathrm{x}$ is negative iff $60^{\\circ}<\\mathrm{x}<120^{\\circ}$. So at most two angles $\\mathrm{x}$ in a triangle can have $\\sin 3 x$ negative. Obviously $\\sin 3 x \\geq-1$, so $\\sin 3 A+\\sin 3 B+\\sin 3 C \\geq-2$. We can only get equality in the degenerate case $\\mathrm{A}=0^{\\circ}, \\mathrm{B}=\\mathrm{C}=90^{\\circ}$ (or with the angles relabeled).\n\nWe can certainly achieve $3(\\sqrt{3}) / 2$ as shown above. But $3(\\sqrt{3}) / 2>2$, so we must have all three sines positive. [If only two are positive, then their sum is at most 2.]. $\\sin 3 \\mathrm{x}$ is positive for $\\mathrm{x}<60^{\\circ}$ and $\\mathrm{x}>120^{\\circ}$. So one angle must be $>120^{\\circ}$. Assume, for definiteness, that $\\mathrm{A} \\leq \\mathrm{B}<60^{\\circ}, \\mathrm{C}>120^{\\circ}$. Put $\\mathrm{C}^{\\prime}=\\mathrm{C}-120^{\\circ}$. Then $3 \\mathrm{~A}+3 \\mathrm{~B}$ $+3 C^{\\prime}=180^{\\circ}$. Now $\\sin \\mathrm{x}$ is a concave function for $0 \\leq \\mathrm{x} \\leq 180^{\\circ}$, or - $\\sin \\mathrm{x}$ is a convex function. So by Jensen's inequality $-\\sin x-\\sin y-\\sin z \\geq 3-\\sin (x+y+z) / 3$. Hence $\\sin 3 A+\\sin 3 B+\\sin 3 C \\leq 3 \\sin 60^{\\circ}=3(\\sqrt{3}) / 2$.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl", "problem_match": "# Problem 3", "solution_match": "# Solution"}}
{"year": "1981", "tier": "T1", "problem_label": "4", "problem_type": null, "exam": "USAMO", "problem": "A convex polygon has $\\mathrm{n}$ sides. Each vertex is joined to a point $\\mathrm{P}$ not in the same plane. If $\\mathrm{A}, \\mathrm{B}, \\mathrm{C}$ are adjacent vertices of the polygon take the angle between the planes PBA and PBC. The sum of the $n$ such angles equals the sum of the $\\mathrm{n}$ angles subtended at $\\mathrm{P}$ by the sides of the polygon (such as the angle APB). Show that $\\mathrm{n}=3$.", "solution": "$\\mathrm{n}=3$ is certainly possible. For example, take $\\angle \\mathrm{APB}=\\angle \\mathrm{APC}=\\angle \\mathrm{BPC}=90^{\\circ}$ (so that the lines $\\mathrm{PA}, \\mathrm{PB}, \\mathrm{PC}$ are mutually perpendicular). Then the three planes through $\\mathrm{P}$ are also mutually perpendicular, so the two sums are both $270^{\\circ}$.\n\nWe show that $\\mathrm{n}>3$ is not possible.\n\nThe sum of the $\\mathrm{n}$ angles APB etc at $\\mathrm{P}$ is less than $360^{\\circ}$. This is almost obvious. Take another plane which meets the lines PA, PB, PC etc at $\\mathrm{A}^{\\prime}, \\mathrm{B}^{\\prime}, \\mathrm{C}^{\\prime}, \\ldots$ and so that the foot of the perpendicular from $\\mathrm{P}$ to the plane lies inside the $\\mathrm{n}-\\mathrm{g}$ - $\\mathrm{A}^{\\prime} \\mathrm{B}^{\\prime} \\mathrm{C}^{\\prime}$... then as we move $\\mathrm{P}$ down the perpendicular the angles $\\mathrm{A}^{\\prime} \\mathrm{PB}$ ' etc all increase. But when it reaches the plane their sum is $360^{\\circ}$. However, I do not immediately see how to make that rigorous. Instead, take any point $\\mathrm{O}$ inside the $\\mathrm{n}$-gon $\\mathrm{ABC} \\ldots$... We have $\\angle \\mathrm{PBA}+\\angle \\mathrm{PBC}>\\angle \\mathrm{ABC}$. Adding the $\\mathrm{n}$ such equations we get $\\sum\\left(180^{\\circ}-\\mathrm{APB}\\right)>\\sum \\mathrm{ABC}=(\\mathrm{n}-2) 180^{\\circ}$. So $\\sum \\mathrm{APB}<360^{\\circ}$.\n\nThe sum of the $\\mathrm{n}$ angles between the planes is at least $(\\mathrm{n}-2) 180^{\\circ}$. If we take a sphere center $\\mathrm{P}$. Then the lines $\\mathrm{PA}, \\mathrm{PB}$ intersect it at $\\mathrm{A}^{\\prime \\prime}, \\mathrm{B}^{\\prime \\prime}, \\ldots$ which form a spherical polygon. The angles of this polygon are the angles between the planes. We can divide the polygon into $\\mathrm{n}-2$ triangles. The angles in a spherical triangle sum to at least $180^{\\circ}$. So the angles in the spherical polygon are at least $(n-2) 180^{\\circ}$. So we have $(n-2) 180^{\\circ}<$ $360^{\\circ}$ and hence $\\mathrm{n}<4$.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl", "problem_match": "# Problem 4", "solution_match": "# Solution"}}
{"year": "1981", "tier": "T1", "problem_label": "5", "problem_type": null, "exam": "USAMO", "problem": "Show that for any positive real $\\mathrm{x},[\\mathrm{nx}] \\geq \\sum_{1}{ }^{\\mathrm{n}}[\\mathrm{kx}] / \\mathrm{k}$.", "solution": "If $x$ is an integer, then we have equality. So it is sufficient to prove the result for $0<x<1$. The rhs only increases at $\\mathrm{x}=\\mathrm{a} / \\mathrm{b}$, where $\\mathrm{a}, \\mathrm{b}$ are coprime positive integers with $1 \\leq \\mathrm{b} \\leq \\mathrm{n}$, and $0 \\leq \\mathrm{a} \\leq \\mathrm{b}$. So it is sufficient to consider $\\mathrm{x}$ of this form. In fact we can assume $0<\\mathrm{a}<\\mathrm{b}$ since the equality is obvious for $\\mathrm{x}=0$ or 1 . We may write:\n\n$\\mathrm{a}=\\mathrm{q}_{1} \\mathrm{~b}+\\mathrm{r}_{1}$\n\n$2 a=q_{2} b+r_{2}$\n\n$3 \\mathrm{a}=\\mathrm{q}_{3} \\mathrm{~b}+\\mathrm{r}_{3}$\n\n...\n\n$\\mathrm{na}=\\mathrm{q}_{\\mathrm{n}} \\mathrm{b}+\\mathrm{r}_{\\mathrm{n}}$\n\nwith each $0 \\leq r_{i}<b$.\n\nThus $\\mathrm{kx}=\\mathrm{q}_{\\mathrm{k}}$ and we have to prove that $\\mathrm{q}_{1}+\\mathrm{q}_{2} / 2+\\ldots+\\mathrm{q}_{\\mathrm{n}} / \\mathrm{n} \\leq \\mathrm{q}_{\\mathrm{n}}$.\n\nWe claim that $r_{1}, r_{2}, \\ldots, r_{b-1}$ is just a permutation of $1,2, \\ldots, b-1$. For if $r_{i}=r_{j}$ with $i<j$, then $(j-i) a=\\left(q_{j}-\\right.$ $\\left.\\mathrm{q}_{\\mathrm{i}}\\right) \\mathrm{b}$, but a and $\\mathrm{b}$ are coprime and $\\mathrm{j}-\\mathrm{i}<\\mathrm{b}$, so that is impossible. So we may use the rearrangement inequality to give $\\mathrm{r}_{1} / 1+\\mathrm{r}_{2} / 2+\\ldots+\\mathrm{r}_{\\mathrm{b}-1} / \\mathrm{b}-1 \\geq 1 / 1+2 / 2+\\ldots+(\\mathrm{b}-1) /(\\mathrm{b}-1)=\\mathrm{b}-1$. The inequality remains true if we add some positive terms to the lhs, so we have $r_{1} / 1+r_{2} / 2+\\ldots+r_{n} / n \\geq b-1$. Hence $r_{1} / b+r_{2} /(2 b)+\\ldots+r_{n} /(n b) \\geq(b-$ $1) / b$.\n\nSo $\\left(\\mathrm{q}_{1}+\\mathrm{q}_{2} / 2+\\ldots+\\mathrm{q}_{\\mathrm{n}} / \\mathrm{n}\\right)+(\\mathrm{b}-1) / \\mathrm{b} \\leq\\left(\\mathrm{q}_{1}+\\mathrm{r}_{1} / \\mathrm{b}\\right)+\\left(\\mathrm{q}_{2}+\\mathrm{r}_{2} /(2 \\mathrm{~b})\\right)+\\ldots\\left(\\mathrm{q}_{\\mathrm{n}}+\\mathrm{r}_{\\mathrm{n}} /(\\mathrm{nb})\\right)=\\mathrm{a} / \\mathrm{b}+\\mathrm{a} / \\mathrm{b}+\\ldots+\\mathrm{a} / \\mathrm{b}=$ $\\mathrm{na} / \\mathrm{b}=\\left(\\mathrm{q}_{\\mathrm{n}} \\mathrm{b}+\\mathrm{r}_{\\mathrm{n}}\\right) / \\mathrm{b} \\leq \\mathrm{q}_{\\mathrm{n}}+(\\mathrm{b}-1) / \\mathrm{b}$. Subtracting $(\\mathrm{b}-1) / \\mathrm{b}$ from both sides gives the required result.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl", "problem_match": "# Problem 5", "solution_match": "# Solution"}}
{"year": "1982", "tier": "T1", "problem_label": "1", "problem_type": null, "exam": "USAMO", "problem": "A graph has 1982 points. Given any four points, there is at least one joined to the other three. What is the smallest number of points which are joined to 1981 points?", "solution": "Answer: 1979.\n\nSuppose there were 4 points not joined to 1981 points. Let one of them be A. Take a point B not joined to A. Now if $\\mathrm{X}$ and $\\mathrm{Y}$ are any two other points, $\\mathrm{X}$ and $\\mathrm{Y}$ must be joined, because otherwise none of the points $\\mathrm{A}$, B, X, Y could be joined to the other 3. There must be two other points C and D not joined to 1981 points. We have just shown that $\\mathrm{C}$ must be joined to every point except possibly $\\mathrm{A}$ and $\\mathrm{B}$. So it must be not joined to one of those. Similarly D. But now none of A, B, C, D is joined to the other 3. Contradiction. So there cannot be 4 points not joined to 1981 points. But there can be 3. Just take the graph to have all edges except AB and AC.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl", "problem_match": "# Problem 1", "solution_match": "# Solution"}}
{"year": "1982", "tier": "T1", "problem_label": "2", "problem_type": null, "exam": "USAMO", "problem": "Show that if $m, n$ are positive integers such that $\\left(x^{m+n}+y^{m+n}+z^{m+n}\\right) /(m+n)=\\left(x^{m}+y^{m}+z^{m}\\right) / m\\left(x^{n}+y^{n}+\\right.$ $\\left.\\mathrm{z}^{\\mathrm{n}}\\right) / \\mathrm{n}$ for all real $\\mathrm{x}, \\mathrm{y}, \\mathrm{z}$ with sum 0 , then $\\{\\mathrm{m}, \\mathrm{n}\\}=\\{2,3\\}$ or $\\{2,5\\}$.", "solution": "Put $\\mathrm{z}=-\\mathrm{x}-\\mathrm{y}$. If $\\mathrm{m}$ and $\\mathrm{n}$ are both odd, the lhs has a term in $\\mathrm{x}^{\\mathrm{m}+\\mathrm{n}}$ but the rhs does not. So at least one of $\\mathrm{m}$ and $n$ is even. Suppose both are even. Then comparing terms in $x^{m+n}$ we get $2 /(m+n)=4 / m n$. Put $m=2 M, n$ $=2 N$, then $M N=M+N$. So $M$ must divide $N$ and $N$ must divide $M$. Hence $M=N=2$. So $m=n=4$. But put $\\mathrm{x}=\\mathrm{y}=1, \\mathrm{z}=-2$, the lhs is $(1+1+256) / 8=129 / 4$ and the rhs is $((1+1+16) / 4)^{2}=81 / 4$. So one of $\\mathrm{m}, \\mathrm{n}$ must be odd and the other even. Without loss of generality we may take m odd.\n\nComparing the terms in $x^{\\mathrm{m}+\\mathrm{n}-1} \\mathrm{y}$, the coefficient on the lhs is 1 . On the rhs it is $1 \\mathrm{x} 2 / \\mathrm{n}$. So we must have $\\mathrm{n}=2$. Put $x=y=1, z=-2$. We get lhs $=\\left(1+1-2^{\\mathrm{m}+2}\\right) /(\\mathrm{m}+2)$ and rhs $=(1+1+4) / 2 x\\left(1+1-2^{\\mathrm{m}}\\right) / \\mathrm{m}$. So $(6-\\mathrm{m}) 2^{\\mathrm{m}-}$ $1=2 m+6$. We cannot have $m>6$ or the lhs is negative. Trying $m=1,3,5$ we find that 3 and 5 are the only solutions.\n\nArguably, we still have to verify that $\\mathrm{m}=3, \\mathrm{n}=2$ and $\\mathrm{m}=5, \\mathrm{n}=2$ are solutions. That is just tedious algebra. [We do have equality for those values.]", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl", "problem_match": "# Problem 2", "solution_match": "# Solution"}}
{"year": "1982", "tier": "T1", "problem_label": "3", "problem_type": null, "exam": "USAMO", "problem": "$\\mathrm{D}$ is a point inside the equilateral triangle $\\mathrm{ABC}$. $\\mathrm{E}$ is a point inside $\\mathrm{DBC}$. Show that area $\\mathrm{DBC} /($ perimeter $\\mathrm{DBC})^{2}>$ area $\\mathrm{EBC} /(\\text { perimeter } \\mathrm{EBC})^{2}$.", "solution": "Let us find an expression for $\\mathrm{t}=($ area $\\mathrm{DBC}) /(\\text { perimeter } \\mathrm{DBC})^{2}$. Let angle $\\mathrm{B}=2 \\mathrm{x}$, angle $\\mathrm{C}=2 \\mathrm{y}$ and angle $\\mathrm{D}$ $=2 z$ and let the inradius be $r$. Then area $D B C=r / 2 x$ perimeter $D B C$. Also $B C=r \\cot x+r \\cot y$, and similarly for the other sides, so perimeter $=2 r(\\cot x+\\cot y+\\cot z)$. Hence $t=1 /(4(\\cot x+\\cot y+\\cot z))$. Putting $\\mathrm{z}=90^{\\circ}-\\mathrm{x}-\\mathrm{y}$, we get $1 / 4 \\mathrm{t}=\\cot \\mathrm{x}+\\cot \\mathrm{y}+(\\cot \\mathrm{x}+\\cot \\mathrm{y}) /(\\cot \\mathrm{x} \\cot \\mathrm{y}-1)$. Now EBC has both $\\mathrm{x}$ and $\\mathrm{y}$ smaller than $\\mathrm{DBC}$, and cot $\\mathrm{x}$ is a decreasing function of $\\mathrm{x}$ (certainly for $\\mathrm{x}$ in the range 0 to $30^{\\circ}$ ). So writing\n$u=\\cot x, v=\\cot y$, it is sufficient to show that $u+v+(u+v) /(u v-1)$ is an increasing function of $u$. [It is symmetric, so it follows that it is also an increasing function of $v$.] We have $x, y<30^{\\circ}$, and hence $u, v>\\sqrt{3}$, so we need the result at least for $\\mathrm{u}, \\mathrm{v}>\\sqrt{ }$.\n\nWe have $u+v+(u+v) /(u v-1)=u+v+1 / v(u v-1) /(u v-1)+(v+1 / v) /(u v-1)$. In considering the dependence on $u$, we can ignore the terms that do not depend on $u$, so we have $u+\\left(1+1 / v^{2}\\right) /(u-1 / v)$. Put $k$ $=\\mathrm{u}-1 / \\mathrm{v}$, then we have to consider $\\mathrm{k}+\\mathrm{h}^{2} / \\mathrm{k}$, where $\\mathrm{h}^{2}=1+1 / \\mathrm{v}^{2}$. But this is an increasing function for $\\mathrm{k}>\\mathrm{h}$ (see below). Thus $u+v+(u+v) /(u v-1)$ is an increasing function of $u$ for $u-1 / v>\\left(1+1 / v^{2}\\right)^{1 / 2}$, or for $u>$ $1 / v+\\left(1+1 / v^{2}\\right)^{1 / 2}$. This bound is highest for the smallest $\\mathrm{v}$ in other words for $\\mathrm{v}=\\sqrt{ }$, when it is $\\sqrt{3}$. So for $\\mathrm{v}>$ $\\sqrt{3}, u+v+(u+v) /(u v-1)$ is an increasing function of $u$ for $u>\\sqrt{3}$.\n\n[To see that $\\mathrm{k}+\\mathrm{h}^{2} / \\mathrm{k}$ is increasing for $\\mathrm{k}>\\mathrm{h}$, take $\\mathrm{k}^{\\prime}>\\mathrm{k}>\\mathrm{h}$. Then $\\mathrm{k}^{\\prime}+\\mathrm{h}^{2} / \\mathrm{k}^{\\prime}-\\mathrm{k}-\\mathrm{h}^{2} / \\mathrm{k}=\\left(\\mathrm{k}^{\\prime}-\\mathrm{k}\\right)-\\mathrm{h}^{2}\\left(1 / k-1 / \\mathrm{k}^{\\prime}\\right)$ $\\left.=\\left(\\mathrm{k}^{\\prime}-\\mathrm{k}\\right)\\left(1-\\mathrm{h}^{2} / \\mathrm{kk}^{\\prime}\\right)>0.\\right]$", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl", "problem_match": "# Problem 3", "solution_match": "# Solution"}}
{"year": "1982", "tier": "T1", "problem_label": "4", "problem_type": null, "exam": "USAMO", "problem": "Show that there is a positive integer $\\mathrm{k}$ such that, for every positive integer $\\mathrm{n}, \\mathrm{k} 2^{\\mathrm{n}}+1$ is composite.", "solution": "Answer: $\\mathrm{k}=542258$ (for example).\n\nSuppose $\\mathrm{p}$ is a prime dividing $2^{\\mathrm{b}}-1,0 \\leq \\mathrm{a}<\\mathrm{b}$ and $\\mathrm{k}=-2^{\\mathrm{b}-\\mathrm{a}}$ mod $\\mathrm{p}$. Then if $\\mathrm{n}=\\mathrm{a}$ mod $\\mathrm{b}$, we have $\\mathrm{k} 2^{\\mathrm{n}}=-2^{\\mathrm{b}-}$ ${ }^{a} 2^{a+h b}=-2^{(h+1) b}=-1 \\bmod p$, so $k 2^{n}+1$ is divisible by p. So we would like to find a collection of pairs (a, b), such that every positive integer $n$ satisfies $n=a \\bmod b$ for at least one member of the collection. We need the corresponding $\\mathrm{p}$ distinct so that we can be sure of finding $\\mathrm{k}$ by the Chinese Remainder theorem which satisfies $k=-2^{b-a} \\bmod p$ for all members of the collection. For the congruences $n=a \\bmod b$ to cover all the integers, we need the $1 \\mathrm{~cm}$ of the $\\mathrm{b}$ to be small relative to their size, so we look for an $1 \\mathrm{~cm}$ with many factors.\n\n$6=2.3$ does not work because $2^{2}-1=3,2^{3}-1=7,2^{6}-1=3^{2} 7$, we cannot find distinct primes p. $10=2.5$ does not work because there are not enough factors to cover all integers. The mod 2 residue covers $1 / 2$, the $\\bmod 5$ residue covers $1 / 5$ and the $\\bmod 10$ residue covers $1 / 10$, but that adds up to less than 1 . Similarly 12 does not work. We must drop one of $2,3,6$. But then the rest cover at most 11 residue classes mod 12 . So we try 24 . Again we drop 6 , but we have:\n\n```\n\\(2: 2^{2}-1=3\\)\n\\(3: 2^{3}-1=7\\)\n\\(4: 2^{5}-1\\) has factor 5\n\\(8: 2^{8}-1\\) has factor 17\n12: \\(2^{12}-1\\) has factor 13\n\\(24: 2^{24}-1\\) has factor 241\n```\n\nWe now find, for example, the following covering set:\n\n```\n0 mod 2 covers the even residues\n1 mod 3 covers 1, 7, 13, 19\n3 mod 4 covers 3, 11, 15, 23\n5 mod 8 covers 5, 21\n5 mod 12 covers 17\n9 mod 24 covers 9\n```\n\nSo we now need $\\mathrm{k}$ which is\n\n```\n-4 mod 3\n-4 mod 7\n-2 mod 5\n-8 mod 17\n-128 mod 13\n-215 mod 241\n```", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl", "problem_match": "# Problem 4", "solution_match": "# Solution"}}
{"year": "1982", "tier": "T1", "problem_label": "5", "problem_type": null, "exam": "USAMO", "problem": "$\\mathrm{O}$ is the center of a sphere $\\mathrm{S}$. Points $\\mathrm{A}, \\mathrm{B}, \\mathrm{C}$ are inside $\\mathrm{S}, \\mathrm{OA}$ is perpendicular to $\\mathrm{AB}$ and $\\mathrm{AC}$, and there are two spheres through $A, B$, and $\\mathrm{C}$ which touch $\\mathrm{S}$. Show that the sum of their radii equals the radius of $\\mathrm{S}$.", "solution": "Let $\\mathrm{D}$ be the circumcenter of $\\mathrm{ABC}$. The triangle $\\mathrm{ABC}$ is in the plane normal to OA. The two spheres both contain through the circumcircle of $\\mathrm{ABC}$, so their centers must lie on a line $\\mathrm{L}$ normal to the plane $\\mathrm{ABC}$ and hence parallel to OA. Take the plane through OA and the line L. Suppose the centers are P and Q. The sphere center $\\mathrm{P}$ must touch $\\mathrm{S}$ at a point $\\mathrm{X}$ on the line OP. Similarly, the sphere center $\\mathrm{Q}$ must touch $\\mathrm{S}$ at a point $\\mathrm{Y}$ on the line $\\mathrm{OQ}$. Since the spheres pass through $A$, we have $\\mathrm{PA}=\\mathrm{PX}$ and hence $\\mathrm{OP}+\\mathrm{PA}=\\mathrm{R}$, the radius of the sphere $\\mathrm{S}$. Similarly $\\mathrm{OQ}+\\mathrm{QA}=\\mathrm{R}$. Indeed if $\\mathrm{K}$ is any point on the line $\\mathrm{L}$ such that $\\mathrm{OK}+\\mathrm{KA}=$ $R$, then the sphere center $K$ will touch $S$ and pass through A. Since it will also have KD perpendicular to DA, it will contain all points on the circumcircle of $A B C$. But the locus of points such that $O K+K A=R$ is an ellipse with foci $\\mathrm{O}$ and $\\mathrm{A}$. So it meets the line $\\mathrm{L}$ in just two points, which must therefore be $\\mathrm{P}$ and $\\mathrm{Q}$. Moreover, since the line $\\mathrm{L}$ is parallel to $\\mathrm{OA}$ the points must be equidistant from the midpoint of OA (which is the center of the ellipse). Hence $\\mathrm{OP}=\\mathrm{AQ}$ and so $\\mathrm{AQ}+\\mathrm{AP}=\\mathrm{R}$, as required.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl", "problem_match": "# Problem 5", "solution_match": "# Solution"}}
{"year": "1983", "tier": "T1", "problem_label": "1", "problem_type": null, "exam": "USAMO", "problem": "If six points are chosen sequentially at random on the circumference of a circle, what is the probability that the triangle formed by the first three is disjoint from that formed by the second three.", "solution": "Answer: $3 / 10$.\n\nOnly the order is important. We are interested in permutations of 123456 where the 123 are together (allowing wrapping). wlog the 1 is in first position. So the triangles are disjoint in the cases $123 \\mathrm{xxx}, 132 \\mathrm{xxx}$, $1 \\mathrm{xxx} 23,1 \\mathrm{xxx} 32,12 \\mathrm{xxx} 3,13 \\mathrm{xxx} 2$. So the probability is $6 \\cdot 6 / 5!=3 / 10$.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl", "problem_match": "# Problem 1", "solution_match": "# Solution"}}
{"year": "1983", "tier": "T1", "problem_label": "2", "problem_type": null, "exam": "USAMO", "problem": "Show that the five roots of the quintic $a_{5} x^{5}+a_{4} x^{4}+a_{3} x^{3}+a_{2} x^{2}+a_{1} x+a_{0}=0$ are not all real if $2 a_{4}{ }^{2}<5 a_{5} a_{3}$.", "solution": "Let the roots be $\\mathrm{r}_{\\mathrm{i}}$. If the condition holds, then $2 \\sum \\mathrm{r}_{\\mathrm{i}}<5 \\sum \\mathrm{r}_{\\mathrm{i}} \\mathrm{r}_{\\mathrm{j}}$. Expanding, $2 \\sum \\mathrm{r}_{\\mathrm{i}}^{2}+4 \\sum \\mathrm{r}_{\\mathrm{i}} \\mathrm{r}_{\\mathrm{j}}<5 \\sum \\mathrm{r}_{\\mathrm{i}} \\mathrm{r}_{\\mathrm{j}}$, or $2 \\sum$ $r_{i}^{2}<\\sum r_{i} r_{j}$. But if $r_{i}$ and $r_{j}$ are real we have we have $2 r_{i} r_{j} \\leq r_{i}^{2}+r_{j}^{2}$. So if all the roots are real, adding the 10 similar equations gives $2 \\sum r_{i} r_{j} \\leq 4 \\sum r_{i}^{2}$. Contradiction. Hence not all the roots are real.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl", "problem_match": "# Problem 2", "solution_match": "# Solution"}}
{"year": "1983", "tier": "T1", "problem_label": "3", "problem_type": null, "exam": "USAMO", "problem": "$\\mathrm{S}_{1}, \\mathrm{~S}_{2}, \\ldots, \\mathrm{S}_{\\mathrm{n}}$ are subsets of the real line. Each $\\mathrm{S}_{\\mathrm{i}}$ is the union of two closed intervals. Any three $\\mathrm{S}_{\\mathrm{i}}$ have a point in common. Show that there is a point which belongs to at least half the $\\mathrm{S}_{\\mathrm{i}}$.", "solution": "We can write $\\mathrm{S}_{\\mathrm{i}}=\\left[\\mathrm{a}_{\\mathrm{i}}, \\mathrm{b}_{\\mathrm{i}}\\right] \\cup\\left[\\mathrm{c}_{\\mathrm{i}}, \\mathrm{d}_{\\mathrm{i}}\\right]$, where $\\mathrm{a}_{\\mathrm{i}} \\leq \\mathrm{b}_{\\mathrm{i}} \\leq \\mathrm{c}_{\\mathrm{i}}<=\\mathrm{d}_{\\mathrm{i}}$. Put $\\mathrm{a}=\\max \\mathrm{a}_{\\mathrm{i}}, \\mathrm{d}=\\min \\mathrm{d}_{\\mathrm{i}}$. Then a belongs to some $\\mathrm{S}_{\\mathrm{h}}$, and d belongs to some $\\mathrm{S}_{\\mathrm{k}}$. Suppose there is some $\\mathrm{S}_{\\mathrm{i}}$ which does not contain a or $\\mathrm{d}$. Then $\\mathrm{b}_{\\mathrm{i}}<\\mathrm{a}$, so any point in $S_{i}$ and $S_{h}$ does not belong to $\\left[a_{i}, b_{i}\\right]$. Similarly $c_{i}>b$, so that any point in $S_{i}$ and $S_{k}$ does not belong to $\\left[\\mathrm{c}_{\\mathrm{i}}, \\mathrm{d}_{\\mathrm{i}}\\right]$. But that means that $\\mathrm{S}_{\\mathrm{i}}, \\mathrm{S}_{\\mathrm{h}}$ and $\\mathrm{S}_{\\mathrm{k}}$ cannot have a point in common. Contradiction. So every $\\mathrm{S}_{\\mathrm{i}}$ must contain a or $\\mathrm{d}$. Hence either a or $\\mathrm{d}$ belongs to at least half of them.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl", "problem_match": "# Problem 3", "solution_match": "# Solution"}}
{"year": "1983", "tier": "T1", "problem_label": "4", "problem_type": null, "exam": "USAMO", "problem": "Show that one can construct (with ruler and compasses) a length equal to the altitude from $\\mathrm{A}$ of the tetrahedron $A B C D$, given the lengths of all the sides. [So for each pair of vertices, one is given a pair of points in the plane the appropriate distance apart.]", "solution": "Let the altitude from $\\mathrm{A}$ be $\\mathrm{AH}$ with $\\mathrm{H}$ in the plane $\\mathrm{BCD}$. The plane normal to $\\mathrm{BC}$ through $\\mathrm{A}$ also contains $\\mathrm{H}$.\n\nSuppose it meets BC at X. Then HX and AX are both perpendicular to BC.\n\nSince we have the side lengths we can construct a cardboard cutout of the tetrahedron: the base BCD and the face BCA next to it, also the face $\\mathrm{CDA}^{\\prime}$ (and the face BDA\", although we do not need it. If we folded along the lines $\\mathrm{BC}, \\mathrm{CD}$ and $\\mathrm{BD}$, then $\\mathrm{A}, \\mathrm{A}^{\\prime}$ and $\\mathrm{A}^{\\prime \\prime}$ would become coincident and we would get the tetrahedron.) We have just shown that in the plane $\\mathrm{AH}$ is a straight line perpendicular to $\\mathrm{BC}$ (and meeting it at $\\mathrm{X}$ ). So we draw this line and also the line through $\\mathrm{A}^{\\prime}$ perpendicular to $\\mathrm{CD}$, giving $\\mathrm{H}$ as their point of intersection. Thus we have $\\mathrm{AH}$ and $\\mathrm{HX}$ and we know that in the tetrahedron AH is perpendicular to HX. So draw a circle diameter $\\mathrm{AX}$ and take a circle center $\\mathrm{H}$ radius HX meeting the circle at $\\mathrm{K}$. Then $\\mathrm{AK}$ is the required length.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl", "problem_match": "# Problem 4", "solution_match": "# Solution"}}
{"year": "1983", "tier": "T1", "problem_label": "5", "problem_type": null, "exam": "USAMO", "problem": "Prove that an open interval of length $1 / \\mathrm{n}$ in the real line contains at most $(\\mathrm{n}+1) / 2$ rational points $\\mathrm{p} / \\mathrm{q}$ with $1 \\leq$ $\\mathrm{q} \\leq \\mathrm{n}$.", "solution": "This is a variant on the familiar result that $\\mathrm{m}+1$ integers from $\\{1,2, \\ldots, 2 \\mathrm{~m}\\}$ must include one which divides another. To prove that, take the largest odd divisor of each of the $\\mathrm{m}+1$ integers. That gives us $\\mathrm{m}+1$ odd numbers from $\\{1,3, \\ldots, 2 \\mathrm{~m}-1\\}$, so by the pigeonhole principle we must have some odd integer $\\mathrm{b}$ twice. If the corresponding integers are $2^{\\mathrm{h}} \\mathrm{b}$ and $2^{\\mathrm{k}} \\mathrm{b}$, then one must divide the other.\n\nNow if $1 \\leq \\mathrm{q} \\leq \\mathrm{n}$ and $1 \\leq \\mathrm{kq} \\leq \\mathrm{n}$, then $\\left|\\mathrm{p} / \\mathrm{q}-\\mathrm{p}^{\\prime} / \\mathrm{kq}\\right| \\geq 1 / \\mathrm{kq} \\geq 1 / \\mathrm{n}$. But we cannot have two such points in an open interval of length $1 / \\mathrm{n}$. Obviously we cannot have two points with the same denominator, so if $\\mathrm{n}=$ $2 \\mathrm{~m}$, there are at most $\\mathrm{m}$ points and if $\\mathrm{n}=2 \\mathrm{~m}+1$ there are at most $\\mathrm{m}+1$ points.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl", "problem_match": "# Problem 5", "solution_match": "# Solution"}}
{"year": "1984", "tier": "T1", "problem_label": "1", "problem_type": null, "exam": "USAMO", "problem": "Two roots of the real quartic $x^{4}-18 x^{3}+a x^{2}+200 x-1984=0$ have product -32 . Find a.", "solution": "Let the two roots satisfy the quadratic $\\mathrm{x}^{2}+\\mathrm{hx}-32=0$ (we have not yet shown that $\\mathrm{h}$ is real). The other two roots have product $-1984 /-32=62$. Let them satisfy the quadratic $x^{2}+k x+62$. So $x^{4}-18 x^{3}+a x^{2}+200 x-$ $1984=\\left(\\mathrm{x}^{2}+\\mathrm{hx}-32\\right)\\left(\\mathrm{x}^{2}+\\mathrm{kx}+62\\right)=\\mathrm{x}^{4}+(\\mathrm{h}+\\mathrm{k}) \\mathrm{x}^{3}+(\\mathrm{hk}+30) \\mathrm{x}^{2}+(62 \\mathrm{~h}-32 \\mathrm{k}) \\mathrm{x}-1984=0$. Equating coefficients: $h+k=-18,62 h-32 k=200$. Solving, $h=-4, k=-14$. Hence $a=86$.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl", "problem_match": "# Problem 1", "solution_match": "# Solution"}}
{"year": "1984", "tier": "T1", "problem_label": "2", "problem_type": null, "exam": "USAMO", "problem": "Can one find a set of $n$ distinct positive integers such that the geometric mean of any (non-empty, finite) subset is an integer? Can one find an infinite set with this property?", "solution": "Answer: yes, no.\n\nTake each member to be an $n!$ power (for example, $\\left.1^{n!}, 2^{n!}, \\ldots, n^{n!}\\right)$.\n\nSuppose we could find an infinite set. Take any two members $\\mathrm{m}$ and $\\mathrm{n}$. Then for sufficiently large $\\mathrm{k}$, $(m / n)^{1 / k}$ must be irrational. But now if we take any other $a_{1}, a_{2}, \\ldots, a_{k-1}$ in the set, $\\left(m a_{1} \\ldots a_{k}\\right)^{1 / k}$ and $\\left(n a_{1} \\ldots\\right.$ $\\left.a_{k}\\right)^{1 / k}$ cannot both be integers. Contradiction.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl", "problem_match": "# Problem 2", "solution_match": "# Solution"}}
{"year": "1984", "tier": "T1", "problem_label": "3", "problem_type": null, "exam": "USAMO", "problem": "$A, B, C, D, X$ are five points in space, such that $A B, B C, C D, D A$ all subtend the acute angle $\\theta$ at $X$. Find the maximum and minimum possible values of $\\angle \\mathrm{AXC}+\\angle \\mathrm{BXD}$ (for all such configurations) in terms of $\\theta$.", "solution": "Answer: minimum 0 (see below); maximum $2 \\cos ^{-1}(2 \\cos \\theta-1)$, achieved by a square pyramid.\n\nIf we take $\\mathrm{A}, \\mathrm{B}, \\mathrm{C}, \\mathrm{D}$ to lie in a plane so that $\\mathrm{AC}$ and $\\mathrm{BD}$ meet at $\\mathrm{X}$ with the angle $\\mathrm{AXC}=\\theta$, then $\\mathrm{AB}, \\mathrm{BC}$, $\\mathrm{CD}$, DA all subtend the angle $\\theta$ at $\\mathrm{X}$, but $\\mathrm{AC}$ and $\\mathrm{BD}$ subtend the angle zero. So the minimum is 0 .\n\nA little playing around suggests that we should take $\\mathrm{ABCD}$ to be a square, with $\\mathrm{X}$ on the normal through the center, so that $\\mathrm{XABCD}$ is a square pyramid. We calculate the result for this case. Suppose $\\mathrm{XA}=\\mathrm{XB}=\\mathrm{XC}=$ $\\mathrm{XD}=1, \\angle \\mathrm{AXB}=\\angle \\mathrm{BXC}=\\angle \\mathrm{CXD}=\\angle \\mathrm{DXA}=\\theta$. Then $\\mathrm{AB}=2 \\sin \\theta / 2$, so $\\mathrm{AC}=2 \\sqrt{2} \\sin \\theta / 2$. So $\\sin$ $\\mathrm{AXC} / 2=\\sqrt{2} \\sin \\theta / 2$. Hence $\\cos \\mathrm{AXC}=1-2 \\sin ^{2} A X C / 2=1-4 \\sin ^{2} \\theta / 2=2 \\cos \\theta-1$. Hence $\\angle \\mathrm{AXC}+\\angle$ BXD $=2 \\cos ^{-1}(2 \\cos \\theta-1)$. We note that this increases monotonically from 0 (at $\\theta=0$ ) to $2 \\pi$ (at $\\theta=\\pi / 2$ ). For $\\theta>0$, we have $\\cos \\theta<1$, hence $2 \\cos \\theta-1<\\cos \\theta$ and hence $2 \\cos ^{-1}(2 \\cos \\theta-1)>2 \\theta$. In other words, except for $\\theta=0$, we can certainly do better than $2 \\theta$.\n\nTake vectors origin $\\mathrm{X}$. Write $\\mathrm{XA}=\\mathbf{A}$ etc. Since we are only interested in the angles, it is convenient to take a to be the unit vector in the direction $\\mathbf{A}$. Then we have $\\mathbf{a . b}=\\mathbf{b} . \\mathbf{c}=\\mathbf{c} . \\mathbf{d}=\\mathbf{d} . \\mathbf{a}=\\cos \\theta$. So $(\\mathbf{a}-\\mathbf{c}) .(\\mathbf{b}-\\mathbf{d})=$ 0 . So either $\\mathbf{a}=\\mathbf{c}$, or $\\mathbf{b}=\\mathbf{d}$ or $\\mathbf{a}-\\mathbf{c}$ is perpendicular to $\\mathbf{b}-\\mathbf{d}$. Suppose $\\mathbf{a}=\\mathbf{c}$. Then $X$ lies on the line $A C$. In this case we certainly have $A D$ and $C D$ subtending the same angle at $X$, and $A B$ and $B C$ subtending the same angle at $\\mathrm{X}$. But $\\angle \\mathrm{AXC}=0 . \\mathrm{XABD}$ is a tetrahedron with $\\angle \\mathrm{AXD}=\\angle \\mathrm{AXB}=\\theta . \\angle \\mathrm{BXD} \\leq \\angle \\mathrm{AXB}+\\angle$ AXD (with equality only if A, B, D, $X$ lie in a plane). So $\\angle B X D+\\angle A X C \\leq 2 \\theta$ and we have just seen that this is not maximal. Similarly if $\\mathbf{b}=\\mathbf{d}$. So we may assume that $\\mathbf{a}-\\mathbf{c}$ is perpendicular to $\\mathbf{b}-\\mathbf{d}$ and both are non-zero.\n\nWe also have $(\\mathbf{a}+\\mathbf{c}) \\cdot(\\mathbf{a}-\\mathbf{c})=\\mathrm{a}^{2}-\\mathbf{c}^{2}=0$ and $(\\mathbf{a}+\\mathbf{c}) .(\\mathbf{b}-\\mathbf{d})=0$. So $\\mathbf{a}+\\mathbf{c}$ is normal to the plane containing $\\mathbf{a}-$ $\\mathbf{c}$ and $\\mathbf{b}-\\mathbf{d}$. Similarly, $\\mathbf{b}+\\mathbf{d}$. Hence $\\mathbf{a}+\\mathbf{c}$ and $\\mathbf{b}+\\mathbf{d}$ are multiples of each other and $(\\mathbf{a}+\\mathbf{c}) .(\\mathbf{b}+\\mathbf{d})=|\\mathbf{a}+\\mathbf{c}|$ $|\\mathbf{b}+\\mathbf{d}| \\cdot$ But lhs $=4 \\cos \\theta$ and rhs $=(2 \\cos \\mathrm{AXC} / 2)(2 \\cos \\mathrm{BXD} / 2)$. So $\\cos \\theta=\\cos \\mathrm{AXC} / 2 \\cos \\mathrm{BXD} / 2=1 / 2$ $\\cos (\\mathrm{AXC} / 2+\\mathrm{BXD} / 2)+1 / 2 \\cos (\\mathrm{AXC} / 2-\\mathrm{BXD} / 2)$. We want to maximise $\\mathrm{AXC} / 2+\\mathrm{BXD} / 2$ and hence to minimise $\\cos (\\mathrm{AXC} / 2+\\mathrm{BXD} / 2)$, so we must maximise $\\cos (\\mathrm{AXC} / 2$ - $\\mathrm{BXD} / 2)$ and hence take $\\angle \\mathrm{AXC}=\\angle$ BXD. That then gives $\\cos (\\mathrm{AXC} / 2+\\mathrm{BXD} / 2)=2 \\cos \\theta-1$, so $\\mathrm{AXC}+\\mathrm{BXD}=2 \\cos ^{-1}(2 \\cos \\theta-1)$. So the square pyramid is indeed maximal.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl", "problem_match": "# Problem 3", "solution_match": "# Solution"}}
{"year": "1984", "tier": "T1", "problem_label": "4", "problem_type": null, "exam": "USAMO", "problem": "A maths exam has two papers, each with at least one question and 28 questions in total. Each pupil attempted 7 questions. Each pair of questions was attempted by just two pupils. Show that one pupil attempted either nil or at least 4 questions in the first paper.", "solution": "Each pupil attempts 7 questions and hence 21 pairs of questions. There are $28 \\cdot 27 / 2=378$ pairs of questions in total and each is attempted by 2 pupils. So there must be $378 \\cdot 2 / 21=36$ pupils. Suppose n pupils solved question 1. Each solved 6 pairs involving question 1, so there must be $3 \\mathrm{n}$ pairs involving question 1. But there are 27 pairs involving question 1 , so $\\mathrm{n}=9$. The same applies to any other question. So each question was solved by 9 pupils.\n\nSuppose the result is false. Suppose there are $\\mathrm{m}$ questions in the first paper, that the number of pupils solving $1,2,3$ questions in the first paper is $\\mathrm{a}, \\mathrm{b}, \\mathrm{c}$ respectively. So $\\mathrm{a}+\\mathrm{b}+\\mathrm{c}=36, \\mathrm{a}+2 \\mathrm{~b}+3 \\mathrm{c}=9 \\mathrm{~m}$. Now consider pairs of problems in the first paper. There are $\\mathrm{m}(\\mathrm{m}-1) / 2$ such pairs. Pupils solving just 1 solve no pairs, those solving 2 solve 1 pair and those solving 3 solve 3 pairs, so we have $b+3 c=m(m-1)$. Solving for $b$ we get $b$ $=-2 m^{2}+29 m-108=-2(m-29 / 4)^{2}-23 / 8<0$. Contradiction. So the result must be true.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl", "problem_match": "# Problem 4", "solution_match": "# Solution"}}
{"year": "1984", "tier": "T1", "problem_label": "5", "problem_type": null, "exam": "USAMO", "problem": "A polynomial of degree $3 n$ has the value 2 at $0,3,6, \\ldots, 3 n$, the value 1 at $1,4,7, \\ldots, 3 n-2$ and the value 0 at $2,5,8, \\ldots, 3 n-1$. Its value at $3 n+1$ is 730 . What is $n$ ?", "solution": "Answer: $\\mathrm{n}=4$.\n\nThe $(3 n+1)$ th differences of the polynomial are zero. Call it $p(x)$, so we have $p(3 n+1)-(3 n+1) C 1 p(3 n)+$ $(3 n+1) C 2 p(3 n-1)-\\ldots+(-1)^{3 n+1} p(0)=0$, where $r C s$ is the binomial coefficient. Hence $p(3 n+1)=2($\n\n$(3 n+1) C 1-(3 n+1) C 4+\\ldots)+((3 n+1) C 3-(3 n+1) C 6+\\ldots)$. Putting $n=1$, we get: $p(4)=2(4 C 1-4 C 4)+$ $4 C 3=6+4=10$. So $n$ is not 1 . Putting $n=2$, we get: $p(7)=2(7 C 1-7 C 4+7 C 7)+(7 C 3-7 C 6)=2(7-35$ $+1)+(35-7)=-26$. So $n$ is not 2. Putting $n=3$, we get: $p(10)=2(10 C 1-10 C 4+10 C 7-10 C 10)+(10 C 3$ $-10 C 6+10 C 9)=2(10-210+120-1)+(120-210+10)=-162-100=-262$. So $n$ is not 3 . Putting $n=4$, we get: $p(13)=2(13 C 1-13 C 4+13 C 7-13 C 10+13 C 13)+(13 C 3-13 C 6+13 C 9-13 C 12)=2(13-715+$ $1716-286+1)+(286-1716+715-13)=1458-728=730$. So $n=3$ works.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl", "problem_match": "# Problem 5", "solution_match": "# Solution"}}
{"year": "1985", "tier": "T1", "problem_label": "1", "problem_type": null, "exam": "USAMO", "problem": "Do there exist 1985 distinct positive integers such that the sum of their squares is a cube and the sum of their cubes is a square?", "solution": "Answer: yes.\n\nTake any $\\mathrm{n}$ integers $\\mathrm{a}_{\\mathrm{i}}$. Suppose that $\\sum \\mathrm{a}_{\\mathrm{i}}^{2}=\\mathrm{b}$ and $\\sum \\mathrm{a}_{\\mathrm{i}}^{3}=\\mathrm{c}$. Now multiply each $\\mathrm{a}_{\\mathrm{i}}$ by $\\mathrm{b}^{4} \\mathrm{c}^{3}$. The sum of their squares becomes $b^{9} c^{6}$ which is a cube and the sum of their cubes becomes $b^{12} c^{10}$ which is a square.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl", "problem_match": "# Problem 1", "solution_match": "# Solution"}}
{"year": "1985", "tier": "T1", "problem_label": "2", "problem_type": null, "exam": "USAMO", "problem": "Find all real roots of the quartic $\\mathrm{x}^{4}-(2 \\mathrm{~N}+1) \\mathrm{x}^{2}-x+\\mathrm{N}^{2}+N-1=0$ correct to 4 decimal places, where $\\mathrm{N}=$ $10^{10}$.", "solution": "Answer: 99999.9984 and 100000.0016 .\n\nWe can write the equation as $\\left(x^{2}-N-1 / 2\\right)^{2}=x+5 / 4$. For $x<-5 / 4$ the lhs is positive and the rhs is negative, so there are no roots with $x<-5 / 4$. If $x$ lies between $-5 / 4$ and 0 , then the lhs is obviously much larger than the rhs, so again there is no root. Thus there are no negative roots. Descartes' rule of signs (see below) tells us that there are at most 2 positive roots.\n\nIf $x=10^{5}$, then the lhs is $1 / 4$ and the rhs is much larger (approx $10^{5}$ ). If $x=10^{5} \\pm 1$, then the lhs is $\\left( \\pm 210^{5}+\\right.$ $1 / 2)^{2}$ which is approximately $410^{10}$ and much larger than the rhs. So there is a root either side of $10^{5}$. Put $\\mathrm{x}=$ $10^{5} \\pm \\mathrm{k}$. Then we want $\\left( \\pm 2 \\mathrm{k} \\cdot 10^{5}+\\mathrm{k}^{2}-1 / 2\\right)^{2}=10^{5}+5 / 4$, or $\\left(4 \\mathrm{k}^{2} 10^{5}-1\\right) 10^{5} \\pm 2 \\mathrm{k}\\left(1-2 \\mathrm{k}^{2}\\right) 10^{5}-5 / 4=0$. So evidently we need approximately $4 \\mathrm{k}^{2}=10^{-5}$, or $\\mathrm{k}= \\pm 0.0016$. So it looks as though the roots are $10^{5} \\pm 0.0016$ $=99999.9984$ and 100000.0016.\n\nPut $\\mathrm{x}=10^{5} \\pm 0.00155$. Then $\\left(\\mathrm{x}^{2}-10^{10}-1 / 2\\right)^{2}-\\mathrm{x}-5 / 4=\\left( \\pm 310-1 / 2+0.00155^{2}\\right)^{2}-10^{5} \\pm 0.00155-5 / 4<$ $311^{2}-10^{5}<0$. Put $x=10^{5} \\pm 0.00165$, then $\\left(x^{2}-10^{10}-1 / 2\\right)^{2}-x-5 / 4=\\left( \\pm 330-1 / 2+0.00165^{2}\\right)^{2}-10^{5} \\pm$ $0.00165-5 / 4>329^{2}-10^{5}-2>0$. So indeed one root lies between $10^{5}-0.00165$ and $10^{5}-0.00155$ and the other root lies between $10^{5}+0.00155$ and $10^{5}+0.00165$.\n\n## Descartes' rule of signs.\n\nThis states that if the number of sign changes in the coefficients of the polynomial is $d$, then the number of positive roots is $\\mathrm{d}$ or $\\mathrm{d}$ less an even number. So, for example, if the polynomial is $\\mathrm{x}^{5}+14.3 \\mathrm{x}^{4}-34 \\mathrm{x}^{2}-\\mathrm{x}+$ 3.2, then there are two sign changes ( +14.3 to -34 and -1 to 3.2 ), so there are either 0 or 2 positive roots. Note that we ignore zero coefficients. If $r$ is a positive root of $p(-x)=0$, then $-r$ is a negative root of $p(x)=0$. So if we substitute - $\\mathrm{x}$ for $\\mathrm{x}$ in the polynomial and the number of sign changes is then $\\mathrm{d}^{\\prime}$, then we can conclude that the number of negative roots of the polynomial is either $\\mathrm{d}^{\\prime}$ or $\\mathrm{d}^{\\prime}$ less an even number. With the example above we get $-x^{5}+14.3 x^{4}-34 x^{2}+x+3.2$, which has 3 sign changes. So the polynomial has 1 or 3 negative roots.\n\nThe proof is not difficult. The key idea is to show that if $k$ is a positive root, so that $p(x)=(x-k) q(x)$, then (1) $p(x)$ has at least one more sign change than $q(x)$, and (2) the difference between the number of sign changes is odd (note that the signs of the constant coefficients of $p(x)$ and $q(x)$ are different).", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl", "problem_match": "# Problem 2", "solution_match": "# Solution"}}
{"year": "1985", "tier": "T1", "problem_label": "3", "problem_type": null, "exam": "USAMO", "problem": "A tetrahedron has at most one edge longer than 1 . What is the maximum total length of its edges?", "solution": "Answer: $5+\\sqrt{3}$.\n\nSuppose $\\mathrm{AB}$ is the edge which may be longer than 1. Then if we rotate the triangle $\\mathrm{ACD}$ about $\\mathrm{CD}$ until $\\mathrm{A}$ lies in the same plane as $B C D$ and is on the opposite side of $\\mathrm{CD}$ to $\\mathrm{B}$, then we maximise $\\mathrm{AB}$ without changing any of the other side lengths. So the maximal configuration must be planar.\n\nNow suppose we regard $\\mathrm{C}$ and $\\mathrm{D}$ as fixed and the other points as variable. Suppose $\\mathrm{CD}=2 \\mathrm{x}(<=1)$. Then $\\mathrm{A}$ and $\\mathrm{B}$ must both lie inside the circle center $\\mathrm{C}$ radius 1 and inside the circle center $\\mathrm{D}$ radius 1 and hence inside their intersection which is bounded by the two arcs XY (assuming they meet at $\\mathrm{X}$ and $\\mathrm{Y}$ ). Obviously we maximise $\\mathrm{AC}+\\mathrm{AD}+\\mathrm{BC}+\\mathrm{BD}$ by taking $\\mathrm{A}$ at $\\mathrm{X}$ and $\\mathrm{B}$ at $\\mathrm{Y}$ (or vice versa). We claim that that choice also maximises $\\mathrm{AB}$. Suppose that is true. Then it also maximises $\\mathrm{AC}+\\mathrm{AD}+\\mathrm{BC}+\\mathrm{BD}+\\mathrm{AB}$ at $4+2\\left(1-\\mathrm{x}^{2}\\right)^{1 / 2}$. So we now have to vary $\\mathrm{CD}$ to maximise $2 \\mathrm{x}+4+2\\left(1-\\mathrm{x}^{2}\\right)^{1 / 2}$. We show that $\\mathrm{x}+\\left(1-\\mathrm{x}^{2}\\right)^{1 / 2}$ is increasing for $\\mathrm{x}$ $<=1 / 2$ and hence that the maximum is at $x=1 / 2$. Put $x=\\sin$. Then we have $x+\\left(1-x^{2}\\right)^{1 / 2}=\\sqrt{2} \\sin (t+\\pi / 4)$ which is indeed increasing for $\\mathrm{x} \\leq \\pi / 6$.\n\nIt remains to prove the claim. Take the circle diameter XY. Then the two ares both lie inside this circle. [Two circles intersect in at most two points, so each arc must lie entirely inside or entirely outside the circle center $\\mathrm{O}$ and it obviously does not lie outside. ] But AB lies inside a chord of this circle The length of the chord cannot exceed the diameter of the circle (which is $\\mathrm{XY}$ ) and hence $\\mathrm{AB} \\leq \\mathrm{XY}$.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl", "problem_match": "# Problem 3", "solution_match": "# Solution"}}
{"year": "1985", "tier": "T1", "problem_label": "4", "problem_type": null, "exam": "USAMO", "problem": "A graph has $n>2$ points. Show that we can find two points $A$ and $B$ such that at least $[n / 2]-1$ of the remaining points are joined to either both or neither of $\\mathrm{A}$ and $\\mathrm{B}$.", "solution": "Consider the number of pairs $(X,\\{Y, Z\\})$, where $\\mathrm{X}, \\mathrm{Y}, \\mathrm{Z}$ are distinct points such that $\\mathrm{X}$ is joined to just one of $Y, Z$. If $X$ is joined to just $k$ points, then there are just $k(n-1-k) \\leq(n-1)^{2} / 4$ such pairs $(X,\\{Y, Z\\})$. Hence in total there are at most $n(n-1)^{2} / 4$ such pairs. But there are $n(n-1) / 2$ possible $\\{\\mathrm{Y}, \\mathrm{Z}\\}$. So we must be able to find one of them $\\{A, B\\}$ which belongs to at most $[(n-1) / 2]$ such pairs. Hence there are at least $n-2$ $-[(n-1) / 2]=[n / 2]-1$ points $X$ which are joined to both of A and B or to neither of A and B. [If confused by the [] , consider $n=2 m$ and $n=2 m+1$ separately! $]$", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl", "problem_match": "# Problem 4", "solution_match": "# Solution"}}
{"year": "1985", "tier": "T1", "problem_label": "5", "problem_type": null, "exam": "USAMO", "problem": "$0<a_{1} \\leq a_{2} \\leq a_{3} \\leq \\ldots$ is an unbounded sequence of integers. Let $b_{n}=m$ if $a_{m}$ is the first member of the sequence to equal or exceed $n$. Given that $a_{19}=85$, what is the maximum possible value of $a_{1}+a_{2}+\\ldots+$ $\\mathrm{a}_{19}+\\mathrm{b}_{1}+\\mathrm{b}_{2}+\\ldots+\\mathrm{b}_{8} ?$", "solution": "We show that the only possible value of the sum is $85.20=1700$.\n\nThat is certainly the value if all $a_{i}=85$, for then all $b_{j}=1$ and so the sum is $19 \\cdot 85+85 \\cdot 1=85 \\cdot 20$. Now consider the general case. Suppose that we increase some $\\mathrm{a}_{\\mathrm{i}}$ by 1 from $\\mathrm{k}$ to $\\mathrm{k}+1$ (whilst preserving the property that $a_{i}$ is increasing, so we must have $a_{i}<a_{i+1}$ before the increase). The effect of the increase is to change $b_{k+1}$ from $i+1$ to $i$, but not to change any other $b_{j}$. This is obvious if $a_{i-1}<a_{i}$ and $a_{i+1}>a_{i}+1$. If $a_{i-1}=a_{i}$, then before the change $b_{k}=i-1$ (not i) and that is still true after the change. Equally, if $a_{i}=a_{i+1}$ after the change, then it is still true that $\\mathrm{b}_{\\mathrm{k}+1}$ changes from $\\mathrm{i}+1$ to $\\mathrm{i}$. Thus the overall effect of the increase is not to change the sum of the $\\mathrm{a}_{\\mathrm{i}}$ plus the sum of the $\\mathrm{b}_{\\mathrm{j}}$. But by a series of such changes we convert any initial sequence to all $\\mathrm{a}_{\\mathrm{i}}=85$.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl", "problem_match": "# Problem 5", "solution_match": "# Solution"}}
{"year": "1986", "tier": "T1", "problem_label": "1", "problem_type": null, "exam": "USAMO", "problem": "Do there exist 14 consecutive positive integers each divisible by a prime less than 13 ? What about 21 consecutive positive integers each divisible by a prime less than 17 ?", "solution": "Answer: no, yes.\n\nThere are 7 odd numbers. At most one can be a multiple of 7 and one a multiple of 11. It is only possible to have two of them multiples of 5 if either the largest or smallest is a multiple of 5 . But it is only possible to have three of them multiples of 3 if both the largest and the smallest are multiples of 3 . So at most four numbers are multiples of 3 or 5 . That leaves one odd number unaccounted for.\n\nA little juggling shows that we want the odd numbers to be divisible by: $3,7,5,3,11,13,3,5,7,3$ (or the 11 and 13 can be interchanged). We need the 6th odd to be 13 mod 143 to get the 11 and 13 correct, and hence the 1 st in the sequence to be 2 mod 143. But the 1st must be even, so it must be 2 mod 286 . The 2 nd must be divisible by 3 , so the 1 st must be $2 \\bmod 858$. But we need the 4 th to be divisible by 7 , so the 1 st must be $3434 \\bmod 6006$. Then the 6 th must be divisible by 5 , so the 1 st must be $9440 \\bmod 30030$. Thus an example is $9440,9441, \\ldots, 9460$.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl", "problem_match": "# Problem 1", "solution_match": "# Solution"}}
{"year": "1986", "tier": "T1", "problem_label": "2", "problem_type": null, "exam": "USAMO", "problem": "Five professors attended a lecture. Each fell asleep just twice. For each pair there was a moment when both were asleep. Show that there was a moment when three of them were asleep.", "solution": "This is a slightly tricky application of the pigeonhole principle.\n\nFor each pair take the first moment when they are both asleep. There are ten pairs, so ten such moments. If two coincide, then we are done because at that moment at least three professors were asleep. So suppose they are all distinct and form a set S. Each such moment must also be one of the 10 occasions when a professor falls asleep. But consider the earliest member of S. Two professors were asleep at that moment so two fell asleep at or before that moment. Thus each of the remaining 9 members of $\\mathrm{S}$ must be one of the 8 later occasons when a professor fell asleep. So they cannot all be distinct. Contradiction.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl", "problem_match": "# Problem 2", "solution_match": "# Solution"}}
{"year": "1986", "tier": "T1", "problem_label": "3", "problem_type": null, "exam": "USAMO", "problem": "What is the smallest $\\mathrm{n}>1$ for which the average of the first $\\mathrm{n}$ (non-zero) squares is a square?", "solution": "Answer: 337.\n\nWe need $1 / 6(n+1)(2 n+1)$ a square. We need $n=6 m+1$ for it to be an integer. So $(3 m+1)(4 m+1)$ must be a square. But $3 m+1$ and $4 m+1$ are relatively prime, so each must be a square. Suppose $3 m+1=a^{2}$ and $4 m+1=$ $(a+k)^{2}$, then (subtracting) $m=2 a k+k^{2}$, so $a^{2}-6 a k-\\left(3 k^{2}+1\\right)=0$, so $a=3 k+\\sqrt{ }\\left(12 k^{2}+1\\right)$. By inspection the smallest solution of this is $\\mathrm{k}=2$, giving $\\mathrm{a}=13$ and hence $\\mathrm{m}=56$ and $\\mathrm{n}=337$.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl", "problem_match": "# Problem 3", "solution_match": "# Solution"}}
{"year": "1986", "tier": "T1", "problem_label": "4", "problem_type": null, "exam": "USAMO", "problem": "A T-square allows you to construct a straight line through two points and a line perpendicular to a given line through a given point. Circles $\\mathrm{C}$ and $\\mathrm{C}^{\\prime}$ intersect at $\\mathrm{X}$ and $\\mathrm{Y}$. $\\mathrm{XY}$ is a diameter of $\\mathrm{C}$. $\\mathrm{P}$ is a point on $\\mathrm{C}^{\\prime}$ inside\n\nC. Using only a $T$-square, find points $\\mathrm{Q}, \\mathrm{R}$ on $\\mathrm{C}$ such that $\\mathrm{QR}$ is perpendicular to $\\mathrm{XY}$ and $\\mathrm{PQ}$ is perpendicular to PR.", "solution": "$\\mathrm{C}^{\\text {' }}$ seems slightly odd. Why not just say that $\\mathrm{XY}$ is a diameter of $\\mathrm{C}, \\mathrm{P}$ is a point inside $\\mathrm{C}$ and we want points $\\mathrm{Q}, \\mathrm{R}$ on $\\mathrm{C}$ such that $\\mathrm{QR}$ is perpendicular etc.? So presumably it is there to help.\n\nNote that PQ perpendicular to PR means that P lies on the circle diameter QR. So take M to be the midpoint of $\\mathrm{QR}$. Then $M Q=M R=M P$. Guided by the hint about $C^{\\prime}$, we extend $P M$ to meet $C^{\\prime}$ at $S$. Then $Q M \\cdot M R=$ $\\mathrm{XM} \\cdot \\mathrm{MY}($ circle $\\mathrm{C})=\\mathrm{PM} \\cdot \\mathrm{MS}$ (circle $\\left.\\mathrm{C}^{\\prime}\\right)$. So M is also the midpoint of PS. So we want to find $\\mathrm{S}$ on $\\mathrm{C}^{\\prime}$ so that XY bisects SP.\n\nLet the line XP meet $\\mathrm{C}$ at U. Draw the line through U perpendicular to XY to meet $\\mathrm{C}$ again at V. Draw XV. Draw the line through $\\mathrm{P}$ perpendicular to $\\mathrm{XY}$ to meet $\\mathrm{XV}$ at $\\mathrm{T}$. Then $\\mathrm{XY}$ bisects $\\mathrm{PT}$, which is parallel to $\\mathrm{QR}$. Now draw the line through T perpendicular to PT (and hence parallel to XY) to meet $\\mathrm{C}^{\\prime}$ at $\\mathrm{S}$. XY must also bisect ST. So we have the required point M. Draw the line through M perpendicular to XY to meet the circle $\\mathrm{C}$ at $\\mathrm{Q}$ and $\\mathrm{R}$.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl", "problem_match": "# Problem 4", "solution_match": "# Solution"}}
{"year": "1986", "tier": "T1", "problem_label": "5", "problem_type": null, "exam": "USAMO", "problem": "A partition of $n$ is an increasing sequence of integers with sum $n$. For example, the partitions of 5 are: $1,1,1$, 1,$1 ; 1,1,1,2 ; 1,1,3 ; 1,4 ; 5 ; 1,2,2 ;$ and 2,3 . If $\\mathrm{p}$ is a partition, $\\mathrm{f}(\\mathrm{p})=$ the number of $1 \\mathrm{~s}$ in $\\mathrm{p}$, and $\\mathrm{g}(\\mathrm{p})=$ the number of distinct integers in the partition. Show that $\\sum f(p)=\\sum g(p)$, where the sum is taken over all partitions of $n$.", "solution": "Let $\\pi(n)$ be the number of partitions of $n$. If $p=a_{1}, a_{2}, \\ldots, a_{m}$ is a partition of $n$, then $h(p)=1, a_{1}, \\ldots, a_{m}$ is $a$ partition of $n+1$ which contains a 1 . Moreover, $h$ is obviously a bijection between partitions of $n$ and partitions of $\\mathrm{n}+1$ which contain a 1. But $h(p)$ also contains one more 1 than $p$, so $\\sum f(p)$ for $n+1$ is $\\sum f(p)$ for $\\mathrm{n}$ plus $\\pi(\\mathrm{n})$. Thus $\\sum \\mathrm{f}(\\mathrm{p})$ for $\\mathrm{n}$ is $1+\\pi(1)+\\pi(2)+\\ldots+\\pi(\\mathrm{n}-1)$. [Obviously $\\sum \\mathrm{f}(\\mathrm{p})$ for $\\mathrm{n}=1$ is 1 .]\n\nChecking, we find $\\pi(1)=1, \\pi(2)=2, \\pi(3)=3, \\pi(4)=5, \\pi(5)=7$. So this formula gives $\\sum f(p)$ for $n=5$ is $1+$ $1+2+3+5=12$, which is correct.\n\nNow fix $n$ and consider the number of pairs $(p, m)$, where $p$ is a partition of $n$ containing $m$. For each $p$ the number of such pairs is $g(p)$. So the total number of such pairs $\\sum g(p)$. But the number of pairs $(p, m)$ for fixed $m$ is just $\\pi(n-m)($ taking $\\pi(0)=0)$. So the total is also $\\sum \\pi(n-m)$. So $\\sum g(p)=1+\\pi(1)+\\ldots+\\pi(n-1)=$ $\\sum \\mathrm{f}(\\mathrm{p})$.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl", "problem_match": "# Problem 5", "solution_match": "# Solution"}}
{"year": "1987", "tier": "T1", "problem_label": "1", "problem_type": null, "exam": "USAMO", "problem": "Find all solutions to $\\left(m^{2}+n\\right)\\left(m+n^{2}\\right)=(m-n)^{3}$, where $m$ and $n$ are non-zero integers.", "solution": "Answer: $(m, n)=(-1,-1),(8,-10),(9,-6),(9,-21)$.\n\nIf $m, n>0$, then lhs is certainly positive, so we must have $m>n$ to make the rhs positive. But then $m^{2}+n>$ $\\mathrm{m}^{2}$ and $\\mathrm{n}^{2}+\\mathrm{m}>\\mathrm{m}$, so $\\mathrm{lh} \\mathrm{p}>\\mathrm{m}^{3}>$ rhs. Contradiction. So there are no solutions with $\\mathrm{m}$ and $\\mathrm{n}$ both positive.\n\nPut $M=|m|, N=|n|$. Consider next $m=M, n=-N$. So we have $\\left(M^{2}-N\\right)\\left(N^{2}+M\\right)=(M+N)^{3}$. Hence $M^{2} N^{2}+$ $M^{3}-N^{3}-M N=M^{3}+3 M^{2} N+3 M N^{2}+N^{3}$. $N$ is non-zero, so we can divide to get: $2 N^{2}+N\\left(3 M-M^{2}\\right)+$ $3 M^{2}+M=0$. Regarding this as a quadratic in $N$ we solve, getting $N=\\left(\\left(M^{2}-3 M\\right) \\pm \\sqrt{ }\\left(M^{4}-6 M^{3}+9 M^{2}-\\right.\\right.$ $\\left.\\left.24 M^{2}-8 M\\right)\\right) / 4$. Thus $M^{4}-6 M^{3}-15 M^{2}-8 M=M(M-1)^{2}(M-8)$ must be a square. Hence $M(M-8)$ must be a square. We have $(M-4)^{2}=M(M-8)+16$ and for $M \\geq 13$, we have $(M-5)^{2}=M^{2}-10 M+25<M^{2}-8 M$. So we need only consider $M \\leq 12$. But obviously we cannot have $M<8$, or $M(M-8)$ is negative. Checking the remaining values, we find $M=8$ and 9 are the only solutions. They give the solutions $(m, n)=(8,-10)$, $(9,-6),(9,-21)$.\n\nNext, consider the case $m=-M, n=N$. That clearly does not work. We get $\\left(M^{2}+N\\right)\\left(N^{2}-M\\right)=-(M+N)^{3}$. So $2 N^{2}+\\left(M^{2}+3 M\\right) N+3 M^{2}-M=0$, which has no solutions because $3 M^{2}>M$.\n\nFinally, consider the case $m=-M, n=-N$. Then we get $\\left(M^{2}-N\\right)\\left(N^{2}-M\\right)=(N-M)^{3}$, so $2 N^{2}-\\left(M^{2}+3 M\\right) N+$ $\\left(3 M^{2}-M\\right)=0$. Solving for $N$, we get $N=\\left(\\left(M^{2}+3 M\\right) \\pm \\sqrt{ }\\left(M^{4}+6 M^{3}+9 M^{2}-24 M^{2}+8 M\\right)\\right) / 4$. So we require\n$M\\left(M^{3}+6 M^{2}-15 M+8\\right)=M(M+8)(M-1)^{2}$ to be a square. Hence $M(M+8)$ is a square. We have $(M+$ $4)^{2}>M(M+8)$ and for $M \\geq 5$, we have $(M+3)^{2}=M^{2}+6 M+9<M^{2}+8 M$. So we need only check $M=1$, $2,3,4$. We find $M=1$ gives the only square, and that gives the solution ( $m, n)=(-1,-1)$.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl", "problem_match": "# Problem 1", "solution_match": "# Solution"}}
{"year": "1987", "tier": "T1", "problem_label": "2", "problem_type": null, "exam": "USAMO", "problem": "The feet of the angle bisectors of the triangle $\\mathrm{ABC}$ form a right-angled triangle. If the right-angle is at $\\mathrm{X}$, where $\\mathrm{AX}$ is the bisector of angle $\\mathrm{A}$, find all possible values for angle $\\mathrm{A}$.", "solution": "Answer: $120^{\\circ}$.\n\nUse vectors origin $A$. Write the vector $A B$ as $\\mathbf{B}$ etc. Using the familiar $B X / C X=A B / A C$ etc, we have $\\mathbf{Z}=$ $\\mathrm{b} \\mathbf{B} /(\\mathrm{a}+\\mathrm{b}), \\mathbf{Y}=\\mathrm{c} \\mathbf{C} /(\\mathrm{a}+\\mathrm{c}), \\mathbf{X}=(\\mathrm{b} \\mathbf{B}+\\mathrm{c} \\mathbf{C}) /(\\mathrm{b}+\\mathrm{c})$. Hence $\\mathbf{Z}-\\mathbf{X}=\\mathrm{b} \\mathbf{B}(\\mathrm{c}-\\mathrm{a}) /((\\mathrm{a}+\\mathrm{b})(\\mathrm{b}+\\mathrm{c}))-\\mathrm{c} \\mathbf{C} /(\\mathrm{b}+\\mathrm{c}), \\mathbf{Y}-\\mathbf{X}=-$ $b \\mathbf{B} /(b+c)+c \\mathbf{C}(b-a) /((a+c)(b+c))$\n\nWe have (Z-X). $(\\mathbf{Y}-\\mathbf{X})=0$, so after multiplying through by $(a+b)(a+c)(b+c)^{2}$, we get $b^{2} \\mathbf{B}^{2}\\left(a^{2}-c^{2}\\right)+c^{2} \\mathbf{C}^{2}\\left(a^{2}-b^{2}\\right)$ $+2 b c \\mathbf{B} \\cdot \\mathbf{C}\\left(a^{2}+b c\\right)=0$. But $\\mathbf{B}^{2}=c^{2}, \\mathbf{C}^{2}=b^{2}$, so $b c\\left(2 a^{2}-b^{2}-c^{2}\\right)+2$ B. $C\\left(a^{2}+b c\\right)=0$.\n\nBut $a^{2}=b^{2}+c^{2}-2$ B. $\\mathbf{C}$ (cosine rule), so $2 a^{2}-b^{2}-c^{2}=a^{2}-2$ B.C. Hence $b c\\left(a^{2}-2\\right.$ B.C. $)+2$ B. $\\mathbf{C}\\left(a^{2}+b c\\right)=0$, so $\\mathbf{B} \\cdot \\mathbf{C}=$ -bc/2. Hence $\\cos \\mathrm{A}=-1 / 2$, so $\\angle \\mathrm{A}=120^{\\circ}$.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl", "problem_match": "# Problem 2", "solution_match": "# Solution"}}
{"year": "1987", "tier": "T1", "problem_label": "3", "problem_type": null, "exam": "USAMO", "problem": "$X$ is the smallest set of polynomials $p(x)$ such that: (1) $p(x)=x$ belongs to $X$; and (2) if $r(x)$ belongs to $X$, then $x r(x)$ and $(x+(1-x) r(x))$ both belong to $X$. Show that if $r(x)$ and $s(x)$ are distinct elements of $X$, then $\\mathrm{r}(\\mathrm{x}) \\neq \\mathrm{s}(\\mathrm{x})$ for any $0<\\mathrm{x}<1$.", "solution": "If they are never equal, then we must be able to order them, so that $r(x)>s(x)$ for all $x$ in $(0,1)$ or $s(x)>r(x)$ for all $\\mathrm{x}$ in $(0,1)$. Let us use the notation $[+--+\\ldots]$. The operation $r(x)$ to $x r(x)$ is denoted as - , and the operation $r(x)$ to $x+(1-x) r(x)$ is denoted as + . Then the operations are listed in reverse order with the last carried out put first. So for example $[+-]$ means we first apply - to get $x^{2}$, then + to get $x+(1-x) x^{2}=x+$ $x^{2}-x^{3}$. We claim that we have the ordering $+>$ nothing $>-$, which we apply starting with the first term. So looking at the first few polynomials we have: $[++]>[+]$, because we compare the first terms which are equal, then we compare the second terms. We regard $[+]$ as having nothing for the second term, so $+>$ nothing. Then $[+]>[+-]$, because they have equal first terms, but unequal second terms (nothing $>-$ ). The starting polynomial is represented as [ ]. So we have $[+-]>[]$ because $+>$ nothing. Then []$>[-+]>$ $[-]>[--]$. Clearly this ordering, which is a type of lexicographic ordering is a total order, that is, for any two distinct polynomials in $\\mathrm{X}$ we will find that one is larger than the other.\n\nSo suppose $r(x)$ belongs to the set $X$. We have to establish that + > nothing $>-$. In other words, that $x+(1-$ x) $r(x)>x$ and that $x>x r(x)$. But that is obvious, because by a trivial induction we have $0<r(x)<1$ for all $r(x)$ in $X$ and $x$ in $(0,1)$. So, if follows that if $r(x)$ and $s(x)$ are in $X$ then $x+(1-x) r(x)>x ~ s(x)$. It is also obvious that if $[\\mathrm{a}]>[\\mathrm{b}]$, where $\\mathrm{a}$ and $\\mathrm{b}$ are some strings of + and,- then $[+\\mathrm{a}]>[+b]$ and $[-\\mathrm{a}]>[-b]$. But that is sufficient to establish the ordering.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl", "problem_match": "# Problem 3", "solution_match": "# Solution"}}
{"year": "1987", "tier": "T1", "problem_label": "4", "problem_type": null, "exam": "USAMO", "problem": "$\\mathrm{M}$ is the midpoint of $\\mathrm{XY}$. The points $\\mathrm{P}$ and $\\mathrm{Q}$ lie on a line through $\\mathrm{Y}$ on opposite sides of $\\mathrm{Y}$, such that $|\\mathrm{XQ}|=$ $2|\\mathrm{MP}|$ and $|\\mathrm{XY}| / 2<|\\mathrm{MP}|<3|\\mathrm{XY}| / 2$. For what value of $|\\mathrm{PY}| /|\\mathrm{QY}|$ is $|\\mathrm{PQ}|$ a minimum?", "solution": "Let the angle between the line through $\\mathrm{Y}$ and $\\mathrm{XY}$ be $\\theta$. Take $\\mathrm{Y}^{\\prime}$ on the line such that $\\mathrm{X} \\mathrm{Y}^{\\prime}=\\mathrm{XY}$. If $\\mathrm{P}$ is on the opposite side of $\\mathrm{Y}$ to $\\mathrm{Y}^{\\prime}$, then $\\mathrm{Q}$ is on the opposite side of $\\mathrm{Y}^{\\prime}$ to $\\mathrm{Y}$. As $\\mathrm{P}$ approaches $\\mathrm{Y}, \\mathrm{Q}$ approaches $\\mathrm{Y}^{\\prime}$ so the minimum value of $\\mathrm{PQ}$ is $\\mathrm{Y} \\mathrm{Y}^{\\prime}$, corresponding to $\\mathrm{PY} / \\mathrm{QY}=0$. But it is unrealised, since the problem requires $\\mathrm{MY}<\\mathrm{MP}$. If $\\mathrm{P}$ is on the same side of $\\mathrm{Y}$ as $\\mathrm{Y}^{\\prime}$, then as $\\mathrm{P}$ approaches the midpoint of $\\mathrm{Y} \\mathrm{Y}^{\\prime}, \\mathrm{Q}$\n\n![](https://cdn.mathpix.com/cropped/2024_05_09_ff3994484212c24fa6bfg-24.jpg?height=48&width=1665&top_left_y=2620&top_left_x=181)\nthe problem requires MY <MP. $\\mathrm{P}$ is allowed to be on either side of $\\mathrm{Y}$, so the unrealised minimum value of $\\mathrm{PQ}$ is $\\mathrm{Y}^{\\prime} / 2$ as $\\mathrm{PY/QY}$ approaches infinity.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl", "problem_match": "# Problem 4", "solution_match": "# Solution"}}
{"year": "1987", "tier": "T1", "problem_label": "5", "problem_type": null, "exam": "USAMO", "problem": "$\\mathrm{a}_{1}, \\mathrm{a}_{2}, \\ldots, \\mathrm{a}_{\\mathrm{n}}$ is a sequence of $0 \\mathrm{~s}$ and $1 \\mathrm{~s}$. $\\mathrm{T}$ is the number of triples $\\left(\\mathrm{a}_{i}, \\mathrm{a}_{j}, \\mathrm{a}_{\\mathrm{k}}\\right)$ with $\\mathrm{i}<\\mathrm{j}<\\mathrm{k}$ which are not equal to $(0,1,0)$ or $(1,0,1)$. For $1 \\leq i \\leq n, f(i)$ is the number of $j<i$ with $a_{j}=a_{i}$ plus the number of $j>i$ with $a_{j} \\neq a_{i}$. Show that $\\mathrm{T}=\\mathrm{f}(1)(\\mathrm{f}(1)-1) / 2+\\mathrm{f}(2)(\\mathrm{f}(2)-1) / 2+\\ldots+\\mathrm{f}(\\mathrm{n})(\\mathrm{f}(\\mathrm{n})-1) / 2$. If $\\mathrm{n}$ is odd, what is the smallest value of T?", "solution": "For $n$ odd, the smallest value of $T$ is $n(n-1)(n-3) / 8$ achieved by $01010 \\ldots 010$.\n\nSuppose a particular $a_{i}=0$. Let $S_{i}$ be the set of $a_{j}$ with $j<i$ and $a_{j}=a_{i}$ and $a_{j}$ with $j>i$ and $a_{j} \\neq a_{i}$. Suppose we take any two members of $S_{i}$ and consider the triple formed with $a_{i}$ itself. Surprisingly perhaps, they are all of the required form ( $\\mathrm{a}_{\\mathrm{i}}$ is bold $)$ :\n\n$$\n\\begin{array}{llllll}\n\\ldots & 0 & \\ldots & 0 & \\ldots & 0 \\\\\n\\ldots & 0 & \\ldots & 0 & \\ldots & 1 \\\\\n\\ldots & 0 & \\ldots & 1 & \\ldots & 1\n\\end{array}\n$$\n\nSimilarly, if $\\mathrm{a}_{\\mathrm{i}}=1$ :\n\n. . $\\quad 1$. . 1 . . 1\n\n. . 1 . . . 1 ... 0\n\n... 1 . . . 0 . . . 0\n\nConversely, any triple of the required form can be considered (uniquely) to be an $\\mathrm{a}_{\\mathrm{i}}$ with members of the triple before it equal to it and members of the triple after it unequal to it. Since $\\mathrm{f}(\\mathrm{m})(\\mathrm{f}(\\mathrm{m})-1)$ is the number of ways of choosing two items from $f(m)$, we have established that $T=\\sum f(m)(f(m)-1)$.\n\nWe show that $\\sum f(m)=n(n-1) / 2$ (and is hence independent of the details of the particular sequence $a_{i}-$ it depends only on its length). In the case of a sequence of all $0 \\mathrm{~s}$, we have $\\mathrm{f}(1)=0, \\mathrm{f}(2)=1, \\ldots, \\mathrm{f}(\\mathrm{n})=\\mathrm{n}-1$, so $\\sum$ $f(m)=n(n-1) / 2$. Now suppose we have a particular sequence and we change $a_{i}$ from 0 to 1 . Suppose there are a 0s before $a_{i}$, $b 1 s$ before $a_{i}, c 0 s$ after $a_{i}$ and $d 1 s$ after $a_{i}$. Then the value of $f(i)$ is changed from a $+d$ to $b+$ $c$, an increase of $(b-a)+(c-d)$. The value of $f(j)$ with $j<i$ is decreased by 1 if $f(j)$ is 0 and increased by 1 if $f(j)$ is 0 . So in total there is an increase of $(a-b)$ for such $j$. Similarly, for $j>i$, the value of $f(j)$ is decreased by 1 if $f(j)$ is 0 and increased by 1 if $f(j)$ is 1 , a net increase of $(d-c)$. Thus overall $\\sum f(m)$ is increased by (b a) $+(c-d)+(a-b)+(d-c)=0$. But by a sequence of such changes we can get to any sequence $a_{i}$, so all sequences (of length $n$ ) have the same value for $\\sum \\mathrm{f}(\\mathrm{m})$.\n\nThus $T$ is minimised when $\\sum \\mathrm{f}(\\mathrm{m})^{2}$ is minimised. But since $\\sum \\mathrm{f}(\\mathrm{m})$ is fixed, that is achieved when the $\\mathrm{f}(\\mathrm{m})$ are as equal as possible. If $n=2 k+1$, then $\\sum f(m)=n(n-1) / 2$, so the average value of $f(m)$ is exactly $k$, so we look for an arrangement with all $\\mathrm{f}(\\mathrm{m})=\\mathrm{k}$. It is not hard to see that $01010 \\ldots 1010$ works. So this gives $\\mathrm{T}=\\sum$ $\\mathrm{k}(\\mathrm{k}-1) / 2=\\mathrm{n}(\\mathrm{n}-1)(\\mathrm{n}-3) / 8$.\n\nThe question does not ask for it, but the even case is also straightforward. If $n=2 k$, then the best we can hope for is $k$ values of $k$ and $k$ values of $k-1$, so that $\\sum f(m)=k \\cdot k+k .(k-1)=k(2 k-1)=n(n-1) / 2$. That is achieved by $0101 \\ldots 0101$ (all the 0s have $\\mathrm{f}(\\mathrm{m})=\\mathrm{k}$ and all the $1 \\mathrm{~s}$ have $\\mathrm{f}(\\mathrm{m})=\\mathrm{k}-1)$. Hence $\\mathrm{T}=\\mathrm{k} \\cdot \\mathrm{k}(\\mathrm{k}-1) / 2+(\\mathrm{k}-$ 1). $(k-1)(k-2) / 2=(k-1)\\left(2 k^{2}-3 k+2\\right) / 2=(n-2)\\left(n^{2}-3 n+4\\right) / 8$.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl", "problem_match": "# Problem 5", "solution_match": "# Solution"}}
{"year": "1988", "tier": "T1", "problem_label": "1", "problem_type": null, "exam": "USAMO", "problem": "The repeating decimal $0 . \\mathrm{ab} \\ldots \\mathrm{kpq} \\ldots \\mathrm{u}=\\mathrm{m} / \\mathrm{n}$, where $\\mathrm{m}$ and $\\mathrm{n}$ are relatively prime integers, and there is at least one decimal before the repeating part. Show that $\\mathrm{n}$ is divisible by 2 or 5 (or both). [For example, $0.011 \\underline{36}=0.01136363636 \\ldots=1 / 88$ and 88 is divisible by 2.]", "solution": "Note that $\\mathrm{k}$ and $\\mathrm{u}$ are not equal (otherwise we should have regarded the repeating part as starting at $\\mathrm{k}$ ). We have $\\mathrm{m} / \\mathrm{n}=\\mathrm{ab} \\ldots \\mathrm{k} / 10^{\\mathrm{r}} \\mathrm{pq} \\ldots \\mathrm{u} /\\left(10^{\\mathrm{r}}\\left(10^{\\mathrm{s}}-1\\right)\\right)=\\left(\\mathrm{ab} \\ldots \\mathrm{k}\\left(10^{\\mathrm{s}}-1\\right)+\\mathrm{pq} \\ldots \\mathrm{u}\\right) /\\left(10^{\\mathrm{r}}\\left(10^{\\mathrm{s}}-1\\right)\\right)$. The numerator $=\\mathrm{u}-\\mathrm{k}$ $\\bmod 10$, which is non-zero, so the numerator is not divisible by 10 . But the denominator is divisible by 10 . Hence after reduction to lowest terms the denominator is divisible by 2 or 5 or both.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl", "problem_match": "# Problem 1", "solution_match": "# Solution"}}
{"year": "1988", "tier": "T1", "problem_label": "2", "problem_type": null, "exam": "USAMO", "problem": "The cubic $x^{3}+a x^{2}+b x+c$ has real coefficients and three real roots $r \\geq s \\geq t$. Show that $k=a^{2}-3 b \\geq 0$ and that $\\sqrt{ } \\mathrm{k} \\leq \\mathrm{r}-\\mathrm{t}$.", "solution": "$\\mathrm{a}^{2}-3 \\mathrm{~b}=(\\mathrm{r}+\\mathrm{s}+\\mathrm{t})^{2}-3(\\mathrm{rs}+\\mathrm{st}+\\mathrm{tr})=\\mathrm{r}^{2}+\\mathrm{s}^{2}+\\mathrm{t}^{2}-(\\mathrm{rs}+\\mathrm{st}+\\mathrm{tr})$. By Cauchy-Schwartz we have $(\\mathrm{rs}+\\mathrm{st}+\\mathrm{tr})^{2} \\leq$ $\\left(r^{2}+s^{2}+t^{2}\\right)^{2}$, so $r^{2}+s^{2}+t^{2} \\geq|r s+s t+\\operatorname{tr}| \\geq(r s+s t+$ tr $)$. Hence $a^{2}-3 b \\geq 0$.\n\n$a^{2}-3 b \\leq(r-t)^{2}$ is the same as $r^{2}+s^{2}+t^{2}-r s-s t-t r \\leq r^{2}-2 r t+t^{s}$ or $s^{2}+r t-r s-s t \\leq 0$ or $(r-s)(s-t) \\geq 0$, which is true since $r \\geq s$ and $s \\geq t$. So $a^{2}-3 b \\leq(r-t)^{2}$. Taking the non-negative square roots, we get the required result.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl", "problem_match": "# Problem 2", "solution_match": "# Solution"}}
{"year": "1988", "tier": "T1", "problem_label": "3", "problem_type": null, "exam": "USAMO", "problem": "Let $X$ be the set $\\{1,2, \\ldots, 20\\}$ and let $\\mathrm{P}$ be the set of all 9-element subsets of $\\mathrm{X}$. Show that for any map f: $\\mathrm{P}$ $\\rightarrow X$ we can find a 10 -element subset $Y$ of $X$, such that $f(Y-\\{k\\}) \\neq k$ for any $k$ in $Y$.", "solution": "Consider pairs $(S, k)$ with $S$ in $P$ and $k$ in $X$ such that $f(S)=k$. There are evidently $20 \\mathrm{C} 9$ such pairs, since we can choose any $S$ and $k$ is then fixed. Now consider the pairs $(\\mathrm{Y}, \\mathrm{k})$ such that $\\mathrm{Y}$ is a 10-element subset of $\\mathrm{X}$ containing $k$ and $f(Y-\\{k\\})=k$. The map $(Y, k)$ to $(Y-\\{k\\}, k)$ is an injection because if $(Y-\\{k\\}, k)=\\left(Y^{\\prime}-\\right.$ $\\left.\\left\\{\\mathrm{k}^{\\prime}\\right\\}, \\mathrm{k}^{\\prime}\\right)$, then $\\mathrm{k}=\\mathrm{k}^{\\prime}$ and hence $\\mathrm{Y}=\\mathrm{Y}^{\\prime}$. It is not necessarily a bijection because if there are any pairs ( $\\mathrm{S}, \\mathrm{k}$ ) with $\\mathrm{k}$ in $\\mathrm{S}$ then they do not correspond to any $(\\mathrm{Y}, \\mathrm{k})$. But certainly the number of pairs $(\\mathrm{Y}, \\mathrm{k})$ is at most the number of pairs $(\\mathrm{S}, \\mathrm{k})$. So there are at most $20 \\mathrm{C} 9$ pairs $(\\mathrm{Y}, \\mathrm{k})$. But there are $20 \\mathrm{C} 10$ subsets $\\mathrm{Y}$ with 10 elements, so at least 20C10 - 20C9 of them (more than 16000) do not belong to any pairs ( $\\mathrm{Y}, \\mathrm{k}$ ), in other words they are such that $f(Y-\\{k\\})$ is not $k$ for any $k$ in $Y$.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl", "problem_match": "# Problem 3", "solution_match": "# Solution"}}
{"year": "1988", "tier": "T1", "problem_label": "4", "problem_type": null, "exam": "USAMO", "problem": "$A B C$ is a triangle with incenter I. Show that the circumcenters of IAB, IBC, ICA lie on a circle whose center is the circumcenter of $\\mathrm{ABC}$.", "solution": "In fact they lie on the circumcircle of $\\mathrm{ABC}$.\n\nExtend AI to meet the circumcircle again at $\\mathrm{A}^{\\prime}$. We show that $\\mathrm{A}^{\\prime}$ is the circumcenter of $\\mathrm{BCI}$. Angle $\\mathrm{A}^{\\prime} \\mathrm{AC}=$ angle $\\mathrm{A}^{\\prime} \\mathrm{AB}$, so $\\mathrm{A}^{\\prime}$ is the midpoint of the arc $\\mathrm{BC}$, so $\\mathrm{A}^{\\prime} \\mathrm{B}=\\mathrm{A}^{\\prime} \\mathrm{C}$. Also $\\angle \\mathrm{A}^{\\prime} \\mathrm{CB}=\\angle \\mathrm{A}^{\\prime} \\mathrm{AB}=\\mathrm{A} / 2$, so $\\angle \\mathrm{A}^{\\prime} \\mathrm{CI}=$ $\\mathrm{A} / 2+\\mathrm{B} / 2$. But $\\angle \\mathrm{A}^{\\prime} \\mathrm{IC}=\\angle \\mathrm{IAC}+\\angle \\mathrm{ICA}=\\mathrm{A} / 2+\\mathrm{B} / 2$, so $\\mathrm{A}^{\\prime} \\mathrm{CI}$ is isosceles, so $\\mathrm{A}^{\\prime} \\mathrm{C}=\\mathrm{A}^{\\prime} \\mathrm{I}$.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl", "problem_match": "# Problem 4", "solution_match": "# Solution"}}
{"year": "1988", "tier": "T1", "problem_label": "5", "problem_type": null, "exam": "USAMO", "problem": "Let $p(x)$ be the polynomial $(1-x)^{a}\\left(1-x^{2}\\right)^{b}\\left(1-x^{3}\\right)^{c} \\ldots\\left(1-x^{32}\\right)^{k}$, where $a, b, \\ldots, k$ are integers. When expanded in powers of $x$, the coefficient of $x^{1}$ is -2 and the coefficients of $x^{2}, x^{3}, \\ldots, x^{32}$ are all zero. Find $\\mathrm{k}$.", "solution": "Answer: $2^{27}-2^{11}$\n\nWe have $p(x)=1-2 x+O\\left(x^{33}\\right)$. Hence $p(-x)=1+2 x+O\\left(x^{33}\\right)$. Multiplying $p(x) p(-x)=1-2^{2} x^{2}+O\\left(x^{33}\\right)$. Now $p(x) p(-x)$ cannot have any odd terms, so we can write it as a polynomial in $x^{2}, q\\left(x^{2}\\right)$. Hence $q\\left(x^{2}\\right)=1-$ $2^{2} x^{2}+O\\left(x^{34}\\right)$. Similarly, $r\\left(x^{4}\\right)=q\\left(x^{2}\\right) q\\left(-x^{2}\\right)=1-2^{4} x^{4}+O\\left(x^{36}\\right), s\\left(x^{8}\\right)=r\\left(x^{4}\\right) r\\left(-x^{4}\\right)=1-2^{8} x^{8}+O\\left(x^{40}\\right)$, and $t\\left(x^{16}\\right)=1-2^{16} x^{16}+O\\left(x^{48}\\right)$.\n\nNow go back to the definition of $p(x)$. When we take $p(x) p(-x)$, the term $(1-x)^{a}$ becomes $\\left(1-x^{2}\\right)^{a}$. All the even terms just double their exponent, so $\\left(1-x^{2}\\right)^{b}$ becomes $\\left(1-x^{2}\\right)^{2 b},\\left(1-x^{4}\\right)^{d}$ becomes $\\left(1-x^{4}\\right)^{2 d}$ and so on. The odd terms all keep the same exponent, so $\\left(1-\\mathrm{x}^{3}\\right)^{\\mathrm{c}}$ becomes $\\left(1-\\mathrm{x}^{6}\\right)^{\\mathrm{c}}$ and so on. Thus we get $\\mathrm{t}\\left(\\mathrm{x}^{16}\\right)=(1-$ $\\left.x^{16}\\right)^{n}\\left(1-x^{32}\\right)^{16 k} \\ldots$. The first exponent is a sum of several exponents from $p(x)$, but the details are unimportant. We know that $\\mathrm{t}\\left(\\mathrm{x}^{16}\\right)=1-2^{16} \\mathrm{x}^{16}+\\mathrm{O}\\left(\\mathrm{x}^{48}\\right)$. The $\\mathrm{x}^{16}$ term can only come from $\\left(1-\\mathrm{x}^{16}\\right)^{\\mathrm{n}}$, so $\\mathrm{n}=2^{16}$. Now there is no $\\mathrm{x}^{32}$ term, so putting $\\mathrm{N}=2^{16}$ we have $\\mathrm{NC} 2=16 \\mathrm{k}$, were $\\mathrm{NC} 2$ is the binomial coefficient $\\mathrm{N}(\\mathrm{N}-$ $1) / 2=2^{31}-2^{15}$. Hence $\\mathrm{k}=2^{27}-2^{11}$.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl", "problem_match": "# Problem 5", "solution_match": "# Solution"}}
{"year": "1990", "tier": "T1", "problem_label": "1", "problem_type": null, "exam": "USAMO", "problem": "A license plate has six digits from 0 to 9 and may have leading zeros. If two plates must always differ in at least two places, what is the largest number of plates that is possible?", "solution": "Answer: $10^{2}$.\n\nWe show by induction that we can find a set of $10^{\\mathrm{n}-1}$ plates for $\\mathrm{n}>1$ digits. It is true for $\\mathrm{n}=2$ : take the plates $00,11,22, \\ldots, 99$. Suppose it is true for $\\mathrm{n}$. If $\\mathrm{d}$ is a digit from 0 to 9 and $\\mathrm{s}$ is a plate of $\\mathrm{n}$ digits, let $[\\mathrm{d}, \\mathrm{s}]$ be the plate of $\\mathrm{n}+1$ digits which has a as its first digit, and the remaining digits the same as those of $\\mathrm{s}$, except that the last digit is that for $\\mathrm{s}$ plus $\\mathrm{d}$ (reduced mod 10 if necessary). Let $\\mathrm{S}$ be a set of plates for $\\mathrm{n}$ digits. We claim that the set $\\mathrm{S}^{\\prime}=\\{[\\mathrm{d}, \\mathrm{s}]: \\mathrm{d}=0,1, \\ldots$ or 9 and $\\mathrm{s}$ belongs to $\\mathrm{S}\\}$ is a set of plates for $\\mathrm{n}+1$ digits. It obviously has 10 times as many members as $\\mathrm{S}$, so this claim is sufficient to establish the induction.\n\nWe have to show that $[\\mathrm{a}, \\mathrm{s}]$ and $[\\mathrm{b}, \\mathrm{t}]$ differ in at least two places. If $\\mathrm{a}=\\mathrm{b}$, then $\\mathrm{s} \\neq \\mathrm{t}$, so $\\mathrm{s}$ and $\\mathrm{t}$ differ in at least two places. The same change is made to their last digits, so $[\\mathrm{a}, \\mathrm{s}]$ and $[\\mathrm{a}, \\mathrm{t}]$ differ in at least two places. If $\\mathrm{a} \\neq \\mathrm{b}$ and $\\mathrm{s}=\\mathrm{t}$, then $[\\mathrm{a}, \\mathrm{s}]$ and $[\\mathrm{b}, \\mathrm{s}]$ differ in both their first and last places. If $\\mathrm{a} \\neq \\mathrm{b}$ and $\\mathrm{s} \\neq \\mathrm{t}$, then $\\mathrm{s}$ and $\\mathrm{t}$ differ in at least two places and so the modified $\\mathrm{s}$ and $\\mathrm{t}$, differ in at least one place. But $[\\mathrm{a}, \\mathrm{s}]$ and $[\\mathrm{b}, \\mathrm{t}]$ also differ in the first place, so they differ in at least two places.\n\nSo we have established that the largest number is at least $10^{n-1}$ for $\\mathrm{n}$ digits.\n\nBut any two plates which differ only in the last digit cannot both be chosen. So at most $1 / 10$ of the $10^{\\mathrm{n}}$ possible plates can be chosen. That shows that $10^{\\mathrm{n}-1}$ is best possible.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl", "problem_match": "# Problem 1", "solution_match": "# Solution"}}
{"year": "1990", "tier": "T1", "problem_label": "2", "problem_type": null, "exam": "USAMO", "problem": "Define $f_{1}(x)=\\sqrt{ }\\left(x^{2}+48\\right)$ and $f_{n}(x)=\\sqrt{ }\\left(x^{2}+6 f_{n-1}(x)\\right)$. Find all real solutions to $f_{n}(x)=2 x$.", "solution": "Answer: For each $\\mathrm{n}, \\mathrm{x}=4$ is the only solution.\n\nObviously $\\mathrm{x}=4$ is a solution. Since $\\mathrm{f}_{\\mathrm{n}}(\\mathrm{x})>=0$, any solution must be non-negative. So we restrict attention to $\\mathrm{x} \\geq 0$.\n\nSuppose $x<4$. We show by induction that $f_{n}(x)>2 x$. For $n=1$, the claim is equivalent to $4 x^{2}<x^{2}+48$, or $x^{2}<16$, which is true. So suppose the result is true for $n$. Then $x^{2}+6 f_{n}(x)>x^{2}+12 x$. But $x<4$, so $3 x^{2}<$ $12 \\mathrm{x}$, so $4 \\mathrm{x}^{2}<\\mathrm{x}^{2}+12 \\mathrm{x}$. Hence $\\mathrm{f}_{\\mathrm{n}+1}(\\mathrm{x})>2 \\mathrm{x}$, as required.\n\nAn exactly similar argument shows that $\\mathrm{f}_{\\mathrm{n}}(\\mathrm{x})<2 \\mathrm{x}$ for $\\mathrm{x}>4$. Hence $\\mathrm{x}=4$ is the only solution.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl", "problem_match": "# Problem 2", "solution_match": "# Solution"}}
{"year": "1990", "tier": "T1", "problem_label": "3", "problem_type": null, "exam": "USAMO", "problem": "Show that for any odd positive integer we can always divide the set $\\{\\mathrm{n}, \\mathrm{n}+1, \\mathrm{n}+2, \\ldots, \\mathrm{n}+32\\}$ into two parts, one with 14 numbers and one with 19 , so that the numbers in each part can be arranged in a circle, with each\nnumber relatively prime to its two neighbours. For example, for $\\mathrm{n}=1$, arranging the numbers as $1,2,3, \\ldots$, 14 and $15,16,17, \\ldots, 33$, does not work, because 15 and 33 are not relatively prime.", "solution": "Suppose we use $\\mathrm{n}, \\mathrm{n}+1, \\ldots, \\mathrm{n}+13$ for the first circle. That certainly works for $\\mathrm{n}$ not divisible by 13 , since consecutive numbers are always relatively prime and any common divisor of $n$ and $n+13$ must also divide their difference 13. We could then take the second circle to be $n+15, n+14, n+16, n+17, \\ldots, n+32$ for $n \\neq 2$ $\\bmod 17$ or $n+14, n+15, \\ldots, n+29, n+30, n+32, n+31$ if $n=2 \\bmod 17$. Note that any common factor of $n+14$ and $n+16$ must divide their difference 2 , but $n$ is odd, so they are relatively prime. Similarly, $n+30$ and $n+32$.\n\nIf $\\mathrm{n}$ is divisible by 13 , then $\\mathrm{n}+19$ is not, so we can take $n+19, n+20, \\ldots, n+32$ for the first circle. Then we can take the second circle to be $n+1, n, n+2, n+3, \\ldots, n+18$ for $n \\neq 16 \\bmod 17$ or $n, n+1, \\ldots, n+15, n+16, n+18$, $\\mathrm{n}+17$ if $\\mathrm{n}=16 \\bmod 17$.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl", "problem_match": "# Problem 3", "solution_match": "# Solution"}}
{"year": "1990", "tier": "T1", "problem_label": "4", "problem_type": null, "exam": "USAMO", "problem": "How many positive integers can be written in base $\\mathrm{n}$ so that (1) the integer has no two digits the same, and (2) each digit after the first differs by one from an earlier digit? For example, in base 3 , the possible numbers are $1,2,10,12,21,102,120,210$.", "solution": "Answer: $2^{n+1}-2 n-2$.\n\nWe use a more elaborate induction hypothesis. We claim that for base $n+1$, the following numbers of integers satisfy the two conditions: $2^{n+1}-2 n-2$ not using the digit $n ; 2^{n}-1$ with the digit $n$ is the last position; $2^{\\mathrm{n}-1}$ with the digit $\\mathrm{n}$ in the last but 1 position; $2^{\\mathrm{n}-2}$ with the digit $\\mathrm{n}$ having 2 following digits; $2^{\\mathrm{n}-3}$ with the digit $\\mathrm{n}$ having 3 following digits; $\\ldots ; 1$ with the digit $\\mathrm{n}$ having $\\mathrm{n}$ following digits.\n\nThus for $\\mathrm{n}=2$, the claim is that there are 2 numbers only involving 0 and $1(1,10), 3$ numbers with 2 as the last digit $(2,12,102), 2$ numbers with one digit after $2(21,120)$ and 1 number with two digits after $2(210)$. Suppose this holds for base $n+1$.\n\nIf we add up the various possibilities we get $2^{n+1}-2 n-2+\\left(2^{n}-1+2^{n-1}+2^{n-2}+\\ldots+1\\right)=2^{n+1}-2 n-2+2^{n+1}-$ $2=2^{n+2}-2(n+1)-2$. The number of base $n+2$ integers not involving the digit $n+1$ is the same as the number of base $n+1$ numbers, which is (by induction and the addition above) $2^{n+2}-2(n+1)-2$. If a base $n+2$ number has the digit $\\mathrm{n}+1$ in the last place, then either that is the only digit in the number or the earlier digits must form a base $n+1$ number with the digit $n$ in it. There are $\\left(2^{n}-1+2^{n-1}+\\ldots+1\\right)=2^{n+1}-2$ such numbers, so in total we have $2^{n+1}-1$ base $n+2$ numbers with $n+1$ in the last place. If a base $n+2$ number has the digit $n+1$ in the penultimate place, then either the number has $n+1$ as the first digit, in which case it must be $n+1 n$, or the other digits form a base $\\mathrm{n}+1$ number with the digit $\\mathrm{n}$ in the penultimate place or earlier. There are $2^{\\mathrm{n}-1}+\\ldots+$ $1=2^{\\mathrm{n}}-1$ such numbers. So $2^{\\mathrm{n}}$ in total. Similarly for the other possibilities.\n\nFinally, we need to check the answer for $n=2\\left(1,10\\right.$, so two numbers and $\\left.2^{2+1}-2 \\cdot 2-2=8-4-2=2\\right)$.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl", "problem_match": "# Problem 4", "solution_match": "# Solution"}}
{"year": "1990", "tier": "T1", "problem_label": "5", "problem_type": null, "exam": "USAMO", "problem": "$\\mathrm{ABC}$ is acute-angled. The circle diameter $\\mathrm{AB}$ meets the altitude from $\\mathrm{C}$ at $\\mathrm{P}$ and $\\mathrm{Q}$. The circle diameter $\\mathrm{AC}$ meets the altitude from B at R and S. Show that P, Q, R and S lie on a circle.", "solution": "Use vectors. Take $\\mathrm{A}$ as the origin. Let $\\mathrm{AB}=\\mathbf{b}, \\mathrm{AC}=\\mathbf{c}, \\mathrm{AR}=\\mathbf{r}$. $\\mathrm{AR}$ is perpendicular to $\\mathrm{RC}$, so $\\mathbf{r} .(\\mathbf{c}-\\mathbf{r})=0$. $\\mathrm{BR}$ is perpendicular to $\\mathrm{AC}$, so $(\\mathbf{b}-\\mathbf{r}) . \\mathbf{c}=0$. Hence $\\mathbf{r . r}=\\mathbf{r . c}=$ b.c. Thus $|\\mathrm{AR}|=\\sqrt{ }(|\\mathrm{AB}| \\cdot|\\mathrm{AC}| \\cos \\mathrm{A})$. But the identical argument gives the same value for $|\\mathrm{AS}|$. The situtation is symmetrical between $\\mathrm{B}$ and $\\mathrm{C}$, so we get the same result for $|\\mathrm{AP}|$ and $|\\mathrm{AQ}|$. Hence all four points lie on a circle center $\\mathrm{A}$.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl", "problem_match": "# Problem 5", "solution_match": "# Solution"}}
{"year": "1991", "tier": "T1", "problem_label": "1", "problem_type": null, "exam": "USAMO", "problem": "An obtuse angled triangle has integral sides and one acute angle is twice the other. Find the smallest possible perimeter.", "solution": "Answer: $77=16+28+33$.\n\nLet the triangle be $\\mathrm{ABC}$ with angle $\\mathrm{A}$ obtuse and angle $\\mathrm{B}=2$ angle $\\mathrm{C}$. Let the sides be $\\mathrm{a}, \\mathrm{b}, \\mathrm{c}$ as usual. Note that $\\mathrm{a}>\\mathrm{b}>\\mathrm{c}$. We have $\\mathrm{b} \\sin \\mathrm{C}=\\mathrm{c} \\sin 2 \\mathrm{C}$ and $\\mathrm{c}^{2}=\\mathrm{a}^{2}+\\mathrm{b}^{2}-2 \\mathrm{ab} \\sin \\mathrm{C}$. Hence, $\\mathrm{b} / 2 \\mathrm{c}=\\sin 2 \\mathrm{C} / \\sin \\mathrm{C}=\\cos \\mathrm{C}=$ $\\left(a^{2}+b^{2}-c^{2}\\right) / 2 a b$. So $a b^{2}=a^{2} c+b^{2} c-c^{3}$. Hence $b^{2}(a-c)=c\\left(a^{2}-c^{2}\\right)$. Dividing by $a-c$ we get $b^{2}=c(a+c)$.\n\nNow the triangle with smallest perimeter will have $\\mathrm{a}, \\mathrm{b}, \\mathrm{c}$ relatively prime (otherwise we could divide by the common factor). Hence $\\mathrm{c}$ must be a square. For if $\\mathrm{c}$ and $\\mathrm{a}+\\mathrm{c}$ have a common factor, then so do a and $\\mathrm{c}$ and hence $\\mathrm{a}, \\mathrm{b}$ and $\\mathrm{c}$, which means they cannot be the minimal set. Clearly $\\mathrm{c}$ is not 1 (or the triangle would have nil area). $\\mathrm{c}=4$ gives $\\mathrm{a}=5$, which is too short, or $\\mathrm{a} \\geq 21$, which is too long. $\\mathrm{c}=9$ gives $\\mathrm{a}=7$ (too short), 18 (3 divides $\\mathrm{a}, \\mathrm{b}, \\mathrm{c}$ ), or $\\geq 40$ (too long). $\\mathrm{c}=16$ gives $\\mathrm{a}=20$ (too short) or $\\mathrm{a}=33$ which works. Larger $\\mathrm{c}$ gives $\\mathrm{a}$ larger perimeter. $\\operatorname{Eg} \\mathrm{c}=25$ gives $\\mathrm{a}=56, \\mathrm{~b}=45$ (perimeter 126).", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl", "problem_match": "# Problem 1", "solution_match": "# Solution"}}
{"year": "1991", "tier": "T1", "problem_label": "2", "problem_type": null, "exam": "USAMO", "problem": "For each non-empty subset of $\\{1,2, \\ldots, n\\}$ take the sum of the elements divided by the product. Show that the sum of the resulting quantities is $n^{2}+2 n-(n+1) s_{n}$, where $s_{n}=1+1 / 2+1 / 3+\\ldots+1 / n$.", "solution": "This is a straightforward induction. For $\\mathrm{n}=1$ the only term in the sum is $1 / 1$ with sum 1 . The formula gives $1^{2}+2.1-2.1=1$. So it is true for $n=1$.\n\nNote first that for $n$ the sum of the inverses of the products, including 1 for the empty set, is $(1+1 / 1)(1+$ $1 / 2)(1+1 / 3) \\ldots(1+1 / n)=n+1$ (it telescopes). Now the sum for $n+1$ is the sum for $n$ plus the sum of terms involving $n+1$. But if a term involves $n+1$, then the sum is increased by $n+1$ and the product is increased by a factor $n+1$. So the sum of all the terms involving $n+1$ is $(1 / n+1 \\times$ sum for $n)+$ (sum of inverse products for $n$ ) $=\\left(n^{2}+2 n-(n+1) s_{n}\\right) /(n+1)+(n+1)=n+1-1 /(n+1)-s_{n}+(n+1)=2(n+1)-s_{n+1}$. Hence the sum for $n+1$ is $\\left(n^{2}+2 n\\right)-(n+1) s_{n}+2(n+1)-s_{n+1}=(n+1)^{2}+2(n+1)-1-(n+1) s_{n}-s_{n+1}=(n+1)^{2}+2(n+1)-(n+1) /(n+1)-$ $(n+1) s_{n}-s_{n+1}=(n+1)^{2}+2(n+1)-(n+2) s_{n+1}$.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl", "problem_match": "# Problem 2", "solution_match": "# Solution"}}
{"year": "1991", "tier": "T1", "problem_label": "3", "problem_type": null, "exam": "USAMO", "problem": "Define the function $\\mathrm{f}$ on the natural numbers by $\\mathrm{f}(1)=2, \\mathrm{f}(\\mathrm{n})=2^{\\mathrm{f}(\\mathrm{n}-1)}$. Show that $\\mathrm{f}(\\mathrm{n})$ has the same residue mod $m$ for all sufficiently large $n$.", "solution": "This is the so-called tower of exponents, but HTML is not up showing it! The trick is to use induction on $\\mathrm{m}$ (which is not at all obvious). The result is trivial for $\\mathrm{m}=1$.\n\nIf $\\mathrm{m}$ is even, then we can write $\\mathrm{m}=2^{\\mathrm{a}} \\mathrm{b}$, where $\\mathrm{b}$ is odd. Then $\\mathrm{f}(\\mathrm{n})$ is eventually constant mod $\\mathrm{b}$. Obviously $\\mathrm{f}(\\mathrm{n})$ is eventually $0 \\bmod 2^{\\mathrm{a}}$, so the residue mod $\\mathrm{m}$ is eventually constant.\n\nIf $\\mathrm{m}$ is odd, then $2^{\\varphi(\\mathrm{m})}=1 \\bmod \\mathrm{m}$, where $\\varphi(\\mathrm{m})<\\mathrm{m}$. So by induction $\\mathrm{f}(\\mathrm{n})$ is eventually constant $\\bmod \\varphi(\\mathrm{m})$. Hence $\\mathrm{f}(\\mathrm{n}+1)=2^{\\mathrm{f}(\\mathrm{n})}$ is eventually constant $\\bmod m$.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl", "problem_match": "# Problem 3", "solution_match": "# Solution"}}
{"year": "1991", "tier": "T1", "problem_label": "4", "problem_type": null, "exam": "USAMO", "problem": "$\\mathrm{a}$ and $\\mathrm{b}$ are positive integers and $\\mathrm{c}=\\left(\\mathrm{a}^{\\mathrm{a}+1}+\\mathrm{b}^{\\mathrm{b}+1}\\right) /\\left(\\mathrm{a}^{\\mathrm{a}}+\\mathrm{b}^{\\mathrm{b}}\\right)$. By considering $\\left(\\mathrm{x}^{\\mathrm{n}}-\\mathrm{n}^{\\mathrm{n}}\\right) /(\\mathrm{x}-\\mathrm{n})$ or otherwise, show that $c^{a}+c^{b} \\geq a^{a}+b^{b}$.", "solution": "$\\left(x^{n}-n^{n}\\right) /(x-n)=x^{n-1}+x^{n-2} n+\\ldots+n^{n-1}$. If $x>n$, then each term is $>n^{n-1}$, so $\\left(x^{n}-n^{n}\\right) /(x-n)>n^{n}$. If $x<n$, then each term is $<=n^{n}$ with equality only for the last term, so $\\left(x^{n}-n^{n}\\right) /(x-n)<n^{n}$. So multiplying by $x-n$, which is positive for $x>n$ and negative for $x<n$, we get $\\left(x^{n}-n^{n}\\right)>(x-n) n^{n}$ except for $x=n$, and hence $\\left(x^{n}-n^{n}\\right) \\geq$ $(\\mathrm{x}-\\mathrm{n}) \\mathrm{n}^{\\mathrm{n}}$ for all $\\mathrm{x} \\geq 0$.\n\nPutting $x=c, n=a$, we get $\\left(c^{a}-a^{a}\\right) \\geq(c-a) a^{a}$. Similarly, putting $x=c, n=b$, we get $\\left(c^{b}-b^{b}\\right) \\geq(c-b) b^{b}$. Adding $\\left(c^{a}+c^{b}\\right)-\\left(a^{a}+b^{b}\\right) \\geq(c-a) a^{a}+(c-b) b^{b}=c\\left(a^{a}+b^{b}\\right)-a^{a+1}-b^{b+1}=0$, which is the required result.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl", "problem_match": "# Problem 4", "solution_match": "# Solution"}}
{"year": "1991", "tier": "T1", "problem_label": "5", "problem_type": null, "exam": "USAMO", "problem": "$\\mathrm{X}$ is a point on the side $\\mathrm{BC}$ of the triangle $\\mathrm{ABC}$. Take the other common tangent (apart from $\\mathrm{BC}$ ) to the incircles of ABX and ACX which intersects the segments AB and AC. Let it meet AX at Y. Show that the locus of $\\mathrm{Y}$, as $\\mathrm{X}$ varies, is the arc of a circle.", "solution": "We show that $\\mathrm{AY}=(\\mathrm{AB}+\\mathrm{AC}-\\mathrm{BC}) / 2=$ constant.\n\nLet the common tangent meet the incircles of $\\mathrm{ABX}, \\mathrm{ACX}$ at $\\mathrm{R}, \\mathrm{S}$ respectively. Let $\\mathrm{AX}$ meet them at $\\mathrm{P}, \\mathrm{Q}$ respectively and let $\\mathrm{BC}$ meet them at $\\mathrm{U}, \\mathrm{V}$ respectively. Let $\\mathrm{AB}$ meet the incircle of $\\mathrm{ABX}$ at $\\mathrm{K}$ and let $\\mathrm{AC}$ meet the incircle of $\\mathrm{ACX}$ at $\\mathrm{L}$. We have $\\mathrm{AY}=\\mathrm{AQ}-\\mathrm{QY}$ and $\\mathrm{AY}=\\mathrm{AP}-\\mathrm{PY}$. So adding $2 \\mathrm{AY}=\\mathrm{AP}+\\mathrm{AQ}-$ $(\\mathrm{YQ}+\\mathrm{YP})=\\mathrm{AP}+\\mathrm{AQ}-(\\mathrm{YS}+\\mathrm{YR})=\\mathrm{AP}+\\mathrm{AQ}-\\mathrm{RS}$. If we reflect about the line of centers of the two incircles $R$ goes to $U$ and $S$ to $V$. Hence $R S=U V$. So $2 A Y=A P+A Q-U V$. We have $A P=A K=A B-B K$ $=A B-B U$ and $A Q=A L=A C-C L=A C-C V$. Hence $2 A Y=A B+A C-B U-U V-C V=A B+A C-B C$.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl", "problem_match": "# Problem 5", "solution_match": "# Solution"}}
{"year": "1992", "tier": "T1", "problem_label": "1", "problem_type": null, "exam": "USAMO", "problem": "Let $a_{n}$ be the number written with $2^{n}$ nines. For example, $a_{0}=9, a_{1}=99, a_{2}=9999$. Let $b_{n}=\\Pi_{0}{ }^{n} a_{i}$. Find the sum of the digits of $b_{n}$.", "solution": "Answer: $9 \\cdot 2^{\\text {n }}$.\n\nInduction on $\\mathrm{n}$. We have $\\mathrm{b}_{0}=9$, digit sum 9 , and $\\mathrm{b}_{1}=891$, digit sum 18 , so the result is true for $\\mathrm{n}=0$ and 1 . Assume it is true for n-1. Obviously $\\mathrm{a}_{\\mathrm{n}}<10$ to the power of $2^{\\mathrm{n}}$, so $\\mathrm{b}_{\\mathrm{n}-1}<10$ to the power of $\\left(1+2+2^{2}+\\ldots+\\right.$ $\\left.2^{n-1}\\right)<10$ to the power of $2^{n}$. Now $b_{n}=b_{n-1} 10^{N}-b_{n-1}$, where $N=2^{n}$. But $b_{n-1}<10^{N}$, so $b_{n}=\\left(b_{n-1}-1\\right) 10^{N}+$ $\\left(10^{N}-b_{n-1}\\right)$ and the digit sum of $b_{n}$ is just the digit sum of $\\left(b_{n-1}-1\\right) 10^{N}$ plus the digit sum of $\\left(10^{N}-b_{n-1}\\right)$.\n\nNow $b_{n-1}$ is odd and so its last digit is non-zero, so the digit sum of $b_{n-1}-1$ is one less than the digit sum of $b_{n-}$ 1, and hence is $9 \\cdot 2^{n-1}-1$. Multiplying by $10^{N}$ does not change the digit sum. $\\left(10^{N}-1\\right)-b_{n-1}$ has $2^{n}$ digits, each 9 minus the corresponding digit of $b_{n-1}$, so its digit sum is $9 \\cdot 2^{n}-9 \\cdot 2^{n-1} \\cdot b_{n-1}$ is odd, so its last digit is not 0 and hence the last digit of $\\left(10^{N}-1\\right)-b_{n-1}$ is not 9 . So the digit sum of $10^{N}-b_{n-1}$ is $9 \\cdot 2^{n}-9 \\cdot 2^{n-1}+1$. Hence $b_{n}$ has digit sum $\\left(9 \\cdot 2^{n-1}-1\\right)+\\left(9 \\cdot 2^{n}-9 \\cdot 2^{n-1}+1\\right)=9 \\cdot 2^{n}$.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl", "problem_match": "# Problem 1", "solution_match": "# Solution"}}
{"year": "1992", "tier": "T1", "problem_label": "2", "problem_type": null, "exam": "USAMO", "problem": "Let $\\mathrm{k}=1^{0}$. Show that $\\Sigma_{0}^{88} 1 /(\\cos \\mathrm{nk} \\cos (\\mathrm{n}+1) \\mathrm{k})=\\cos \\mathrm{k} / \\sin ^{2} \\mathrm{k}$.", "solution": "$\\tan (n+1) k-\\tan n k=(\\sin (n+1) k \\cos n k-\\sin n k \\cos (n+1) k) /(\\cos n k \\cos (n+1) k)=\\sin k /(\\cos n k \\cos (n+1) k$ ). Using this expression the sum telescopes and we get $\\sum_{0}^{88} 1 /(\\cos n k \\cos (\\mathrm{n}+1) \\mathrm{k})=(\\tan 89 \\mathrm{k}-\\tan 0) / \\sin \\mathrm{k}$. But $\\tan 0=0$ and $\\tan 89 \\mathrm{k}=\\cot (\\pi / 2-89 \\mathrm{k})=\\cot \\mathrm{k}$.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl", "problem_match": "# Problem 2", "solution_match": "# Solution"}}
{"year": "1992", "tier": "T1", "problem_label": "3", "problem_type": null, "exam": "USAMO", "problem": "A set of 11 distinct positive integers has the property that we can find a subset with sum $n$ for any $n$ between 1 and 1500 inclusive. What is the smallest possible value for the second largest element?", "solution": "Answer: 248.\n\nBy taking the integers to be $1,2,4,8, \\ldots, 1024$ we can generate all integers up to 2047 . But by taking some integers smaller, we can do better. For example, 1, 2, 4, .., 128, 247, 248, 750 gives all integers up to 1500 . We can obviously use the integers $1,2,4, \\ldots, 128$ to generate all integers up to 255 . Adding 248 gives all integers from 256 up to 503 . Then adding 247 gives all integers from 504 to 750 . So adding 750 gives all integers up to 1500 . We show that we cannot do better than this.\n\nLet the integers be $a_{1}<a_{2}<\\ldots<a_{11}$. Put $s_{n}=a_{1}+\\ldots+a_{n}$. Take $N$ such that $s_{N-1}<1500 \\leq s_{N}$. Note that 1500 must be a sum of some of the integers so certainly $1500 \\leq s_{11}$. Equally we obviously have $\\mathrm{a}_{1}=1, \\mathrm{a}_{2}=2$, so $\\mathrm{N}$ is well-defined and we have $1<\\mathrm{N} \\leq 11$. Now if $1<\\mathrm{k} \\leq \\mathrm{N}$, we have $\\mathrm{s}_{\\mathrm{k}-1}<1500$ and hence $\\mathrm{s}_{\\mathrm{k}-1}+1 \\leq 1500$. So some sum of distinct $\\mathrm{a}_{\\mathrm{i}}$ must equal $\\mathrm{s}_{\\mathrm{k}-1}+1$ and it must involve an $\\mathrm{a}_{\\mathrm{i}}$ with $\\mathrm{i}>\\mathrm{k}-1$ since $\\mathrm{s}_{\\mathrm{k}-1}<\\mathrm{s}_{\\mathrm{k}-1}+1$. Hence $\\mathrm{a}_{\\mathrm{k}} \\leq \\mathrm{s}_{\\mathrm{k}-1}+1$ for $1<\\mathrm{k} \\leq \\mathrm{N}$.\n\nNow an easy induction gives $\\mathrm{s}_{\\mathrm{k}} \\leq 2_{\\mathrm{k}}-1$. We must have $\\mathrm{a}_{1}=1$ and hence $\\mathrm{s}_{1}=1$, so it is true for $\\mathrm{k}=1$.\n\nSuppose it is true for $\\mathrm{k}-1$. Then $\\mathrm{a}_{\\mathrm{k}} \\leq \\mathrm{s}_{\\mathrm{k}-1}+1 \\leq 2^{\\mathrm{k}-1}$, so $\\mathrm{s}_{\\mathrm{k}}=\\mathrm{s}_{\\mathrm{k}-1}+\\mathrm{a}_{\\mathrm{k}} \\leq 2^{\\mathrm{k}-1}-1+2^{\\mathrm{k}-1}=2^{\\mathrm{k}}-1$, so it is true for $\\mathrm{k}$. Hence for all $\\mathrm{k} \\leq \\mathrm{N}$. But $2^{10}-1=1023<1500$, so if $\\mathrm{N}<11$, then $\\mathrm{s}_{\\mathrm{N}}<1500$. Contradiction. Hence $\\mathrm{N}=11$.\n\nWe have $\\mathrm{s}_{\\mathrm{k}}=\\mathrm{s}_{\\mathrm{k}-1}+\\mathrm{a}_{\\mathrm{k}} \\leq 2 \\mathrm{~s}_{\\mathrm{k}-1}+1$. Hence $\\mathrm{s}_{\\mathrm{k}-1} \\geq\\left(\\mathrm{s}_{\\mathrm{k}}-1\\right) / 2$. So $\\mathrm{s}_{11} \\geq 1500$ implies $\\mathrm{s}_{10} \\geq 750$. But $\\mathrm{s}_{8} \\leq 255$, so $\\mathrm{a}_{9}+$ $\\mathrm{a}_{10} \\geq 495$, so $\\mathrm{a}_{10} \\geq 248$.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl", "problem_match": "# Problem 3", "solution_match": "# Solution"}}
{"year": "1992", "tier": "T1", "problem_label": "4", "problem_type": null, "exam": "USAMO", "problem": "Three chords of a sphere are meet at a point $\\mathrm{X}$ inside the sphere but are not coplanar. A sphere through an endpoint of each chord and $\\mathrm{X}$ touches the sphere through the other endpoints and $\\mathrm{X}$. Show that the chords have equal length.", "solution": "Let two of the chords be $\\mathrm{AB}$ and $\\mathrm{CD}$. Take the plane containing them. In this plane we have a circle through A, B, C, D, and a circle through A, C, X which touches a circle through B, D, X at X. We show that AX = CX. Let the common tangent at X meet the larger circle at $\\mathrm{E}$ and $\\mathrm{F}$. [Assume the points are in the order $\\mathrm{A}, \\mathrm{C}$, $\\mathrm{F}, \\mathrm{B}, \\mathrm{D}, \\mathrm{E}$ as we go round the circle.] We have $\\angle \\mathrm{XAC}=\\angle \\mathrm{BAC}$ (same angle $)=\\angle \\mathrm{BDC}($ circle $\\mathrm{ABCD})=$ $\\angle \\mathrm{BXF}(\\mathrm{XF}$ tangent to $\\mathrm{BXD})=\\angle \\mathrm{EXA}$ (opposite angle) $=\\angle \\mathrm{XCA}$ (EX tangent to AXC). So XAC is isosceles. So $\\mathrm{XA}=\\mathrm{XC}$. Similarly $\\mathrm{XB}=\\mathrm{XD}$. Hence $\\mathrm{AB}=\\mathrm{CD}$.\n\nSimilarly, the other pairs of chords are equal.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl", "problem_match": "# Problem 4", "solution_match": "# Solution"}}
{"year": "1992", "tier": "T1", "problem_label": "5", "problem_type": null, "exam": "USAMO", "problem": "A complex polynomial has degree 1992 and distinct zeros. Show that we can find complex numbers $\\mathrm{z}_{\\mathrm{n}}$, such that if $\\mathrm{p}_{1}(\\mathrm{z})=\\mathrm{z}-\\mathrm{z}_{1}$ and $\\mathrm{p}_{\\mathrm{n}}(\\mathrm{z})=\\mathrm{p}_{\\mathrm{n}-1}(\\mathrm{z})^{2}-\\mathrm{z}_{\\mathrm{n}}$, then the polynomial divides $\\mathrm{p}_{1992}(\\mathrm{z})$.", "solution": "Let the polynomial of degree 1992 be $\\mathrm{q}(\\mathrm{z})$. Suppose its roots are $\\mathrm{w}_{1}, \\mathrm{w}_{2}, \\ldots, \\mathrm{w}_{1992}$. Let $\\mathrm{S}_{1}=\\left\\{\\mathrm{w}_{1}, \\ldots, \\mathrm{w}_{1992}\\right\\}$.\n\nWe now define $S_{2}$ as follows. Let $z_{1}=\\left(w_{1}+w_{2}\\right) / 2$ and take $S_{2}$ to be the set of all numbers $\\left(w-z_{1}\\right)^{2}$ with $w$ in $\\mathrm{S}_{1}$. Note that $\\mathrm{w}=\\mathrm{w}_{1}$ and $\\mathrm{w}=\\mathrm{w}_{2}$ give the same number. It is possible that other pairs may also give the same number. But certainly $\\left|\\mathrm{S}_{2}\\right|<=\\left|\\mathrm{S}_{1}\\right|-1$. We now repeat this process until we get a set with only one member. Thus if $\\mathrm{S}_{\\mathrm{i}}$ has more than one member, then we take we take $\\mathrm{z}_{\\mathrm{i}}$ to be an average of any two distinct members. Then we take $\\mathrm{S}_{\\mathrm{i}+1}$ to be the set of all $\\left(\\mathrm{w}-\\mathrm{z}_{\\mathrm{i}}\\right)^{2}$ with $\\mathrm{w}$ in $\\mathrm{S}_{\\mathrm{i}}$. So after picking at most 1991 elements $\\mathrm{z}_{\\mathrm{i}}$ we have a set $\\mathrm{S}_{\\mathrm{N}}$ with only one member. Take $\\mathrm{Z}_{\\mathrm{N}}$ to be that one member, so that $\\mathrm{S}_{\\mathrm{N}+1}=\\{0\\}$. Now if $\\mathrm{N}<1991$, take the remaining $\\mathrm{z}_{\\mathrm{i}}$ to be 0 until we reach $\\mathrm{z}_{1992}$.\n\nNow as we allow $z$ to take the values in $S$ :\n\n$\\mathrm{p}_{1}(\\mathrm{z})=\\left(\\mathrm{z}-\\mathrm{z}_{1}\\right)$ takes 1992 possible values;\n\n$\\mathrm{p}_{2}(\\mathrm{z})=\\left(\\mathrm{z}-\\mathrm{z}_{1}\\right)^{2}-\\mathrm{z}_{2}$ takes at most 1991 possible values;\n\n$\\mathrm{p}_{3}(\\mathrm{z})=\\left(\\left(\\mathrm{z}-\\mathrm{z}_{1}{ }^{2}-\\mathrm{z}_{2}\\right)^{2}-\\mathrm{z}_{3}\\right.$ takes at most 1990 possible values;\n\n$\\mathrm{p}_{4}(\\mathrm{z})=\\left(\\left(\\left(\\mathrm{z}-\\mathrm{z}_{1}\\right)^{2}-\\mathrm{z}_{2}\\right)^{2}-\\mathrm{z}_{3}\\right)^{2}-\\mathrm{z}_{4}$ takes at most 1989 possible values;\n\n...\n\n$\\mathrm{p}_{1992}(\\mathrm{z})$ takes only the value 0 .\n\nSo every root of $\\mathrm{q}(\\mathrm{z})$ is also a root of $\\mathrm{p}_{1992}(\\mathrm{z})$. Hence $\\mathrm{q}(\\mathrm{z})$ must divide $\\mathrm{p}_{1992}(\\mathrm{z})$.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl", "problem_match": "# Problem 5", "solution_match": "# Solution"}}
{"year": "1993", "tier": "T1", "problem_label": "1", "problem_type": null, "exam": "USAMO", "problem": "$\\mathrm{n}>1$, and $\\mathrm{a}$ and $\\mathrm{b}$ are positive real numbers such that $\\mathrm{a}^{\\mathrm{n}}-\\mathrm{a}-1=0$ and $\\mathrm{b}^{2 \\mathrm{n}}-\\mathrm{b}-3 \\mathrm{a}=0$. Which is larger?", "solution": "Answer: $\\mathrm{a}>\\mathrm{b}$.\n\nNote that $\\mathrm{a}^{\\mathrm{n}}=\\mathrm{a}+1>1$ (since $\\mathrm{a}$ is positive). Hence $\\mathrm{a}>1$. So $\\mathrm{a}^{2 \\mathrm{n}}=(\\mathrm{a}+1)^{2}=\\mathrm{a}^{2}+2 \\mathrm{a}+1$. Put $\\mathrm{a}=1+k$, then $a^{2}=1+2 k+k^{2}>1+2 k$, so $a^{2}+1>2+2 k=2 a$. Hence $a^{2 n}>4 a$. So $(b / a)^{2 n}=(b+3 a) / a^{2 n}<(b+3 a) / 4 a$. If $\\mathrm{b} / \\mathrm{a} \\geq 1$, then $(\\mathrm{b}+3 \\mathrm{a}) / 4 \\mathrm{a} \\leq(\\mathrm{b}+3 \\mathrm{~b}) / 4 \\mathrm{a}=\\mathrm{b} / \\mathrm{a}$, so $(\\mathrm{b} / \\mathrm{a})^{2 \\mathrm{n}}<\\mathrm{b} / \\mathrm{a}$. Contradiction. Hence $\\mathrm{b} / \\mathrm{a}<1$.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl", "problem_match": "# Problem 1", "solution_match": "# Solution"}}
{"year": "1993", "tier": "T1", "problem_label": "2", "problem_type": null, "exam": "USAMO", "problem": "The diagonals of a convex quadrilateral meet at right angles at $X$. Show that the four points obtained by reflecting $\\mathrm{X}$ in each of the sides are cyclic.", "solution": "If we shrink the reflected points by $1 / 2$ about the point $X$, then we get the feet of the perpendiculars from $X$ to the sides. So it is sufficient to show that these four points are cyclic. Let the quadrilateral be ABCD. Let the feet of the perpendiculars from $\\mathrm{X}$ to $\\mathrm{AB}, \\mathrm{BC}, \\mathrm{CD}, \\mathrm{DA}$ be $\\mathrm{P}, \\mathrm{Q}, \\mathrm{R}, \\mathrm{S}$ respectively.\n\n$\\mathrm{XPBQ}$ is cyclic, so $\\angle \\mathrm{XQP}=\\angle \\mathrm{XBP}$. Similarly, $\\mathrm{XPAS}$ is cyclic, so $\\angle \\mathrm{XSP}=\\angle \\mathrm{XAP}$. But $\\mathrm{XBP}$ and $\\mathrm{XAP}$ are two angles in the triangle $\\mathrm{XAB}$ and the third is $\\angle \\mathrm{AXB}=90^{\\circ}$. Hence $\\angle \\mathrm{XQP}+\\angle \\mathrm{XSP}=90^{\\circ}$. Similarly $\\angle$ $\\mathrm{XQR}+\\angle \\mathrm{XSR}=90^{\\circ}$. Adding, $\\angle \\mathrm{PQR}+\\angle \\mathrm{PSR}=180^{\\circ}$, so $\\mathrm{PQRS}$ is cyclic.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl", "problem_match": "# Problem 2", "solution_match": "# Solution"}}
{"year": "1993", "tier": "T1", "problem_label": "3", "problem_type": null, "exam": "USAMO", "problem": "Let $\\mathrm{S}$ be the set of functions $\\mathrm{f}$ defined on reals in the closed interval $[0,1]$ with non-negative real values such that $\\mathrm{f}(1)=1$ and $\\mathrm{f}(\\mathrm{x})+\\mathrm{f}(\\mathrm{y}) \\leq \\mathrm{f}(\\mathrm{x}+\\mathrm{y})$ for all $\\mathrm{x}, \\mathrm{y}$ such that $\\mathrm{x}+\\mathrm{y} \\leq 1$. What is the smallest $\\mathrm{k}$ such that $\\mathrm{f}(\\mathrm{x}) \\leq$ $\\mathrm{kx}$ for all $\\mathrm{f}$ in $\\mathrm{S}$ and all $\\mathrm{x}$ ?", "solution": "Answer: $\\mathrm{k}=2$.\n\nConsider the function $f(x)=0$ for $0 \\leq x \\leq 1 / 2,1$ for $1 / 2<x \\leq 1$. If $x+y \\leq 1$, then at least one of $x, y$ is $\\leq 1 / 2$, so at least one of $f(x), f(y)$ is 0 . But $f$ is obviously increasing, so the other of $f(x), f(y)$ is $\\leq f(x+y)$. Thus $f$ satisfies the conditions. But $\\mathrm{f}(1 / 2+\\varepsilon)=1$, so $\\mathrm{k}$ cannot be smaller than 2 .\n\nSo now let $\\mathrm{f}$ be any function satisfying the conditions. We wish to show that $\\mathrm{f}(\\mathrm{x}) \\leq 2 \\mathrm{x}$. Putting $\\mathrm{y}=1$ - $\\mathrm{x}$, we get $\\mathrm{f}(\\mathrm{x})+\\mathrm{f}(\\mathrm{y})<=\\mathrm{f}(1)=1$. But $\\mathrm{f}(\\mathrm{y})$ is non-negative, so $\\mathrm{f}(\\mathrm{x}) \\leq 1$. Put $\\mathrm{y}=\\mathrm{x} \\leq 1 / 2$. Then $2 \\mathrm{f}(\\mathrm{x}) \\leq \\mathrm{f}(2 \\mathrm{x})$. A simple induction gives $2^{n} f(x) \\leq f\\left(2^{n} x\\right)$ for $x \\leq 1 / 2^{n}$. Now take any $x$ in $[0,1]$. If $x>1 / 2$, then $2 x>1$, so $f(x)<2 x$. If $x$ $\\leq 1 / 2$, choose $\\mathrm{n} \\geq 1$, so that $1 / 2^{n+1}<\\mathrm{x} \\leq 1 / 2^{n}$. Then $2^{\\mathrm{n}} \\mathrm{f}(\\mathrm{x}) \\leq \\mathrm{f}\\left(2^{n^{n}} \\mathrm{x}\\right) \\leq 1$, so $\\mathrm{f}(\\mathrm{x}) \\leq 1 / 2^{\\mathrm{n}} \\leq 2 \\mathrm{x}$.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl", "problem_match": "# Problem 3", "solution_match": "# Solution"}}
{"year": "1993", "tier": "T1", "problem_label": "4", "problem_type": null, "exam": "USAMO", "problem": "The sequence $a_{n}$ of odd positive integers is defined as follows: $a_{1}=r, a_{2}=s$, and $a_{n}$ is the greatest odd divisor of $a_{n-1}+a_{n-2}$. Show that, for sufficiently large $n, a_{n}$ is constant and find this constant (in terms of $r$ and $s$ ).", "solution": "This is awkward to get started. Note that if $a_{n-1}=a_{n-2}$, then $a_{n-1}+a_{n-2}=2 a_{n-1}$, whose greatest odd divisor is just $a_{n-1}$, so $a_{n}=a_{n-1}$. So once two consecutive terms are constant the following terms are constant.\n\nBoth $a_{n-1}$ and $a_{n-2}$ are odd, so $a_{n-1}+a_{n-2}$ is even and hence $a_{n} \\leq\\left(a_{n-1}+a_{n-2}\\right) / 2$. So if $a_{n-1}$ and $a_{n-2}$ are unequal, then $a_{n}<\\max \\left(a_{n-1}, a_{n-2}\\right)$. As already noted, if they are equal, then $a_{n}=\\max \\left(a_{n-1}, a_{n-2}\\right)$.\n\nPut $b_{n}=\\max \\left(a_{n}, a_{n-1}\\right)$. If $a_{n}<a_{n-1}$, then $a_{n+1}<\\max \\left(a_{n}, a_{n-1}\\right)=a_{n-1}$. Also $a_{n+2} \\leq \\max \\left(a_{n+1}, a_{n}\\right)<a_{n-1}$. Hence $\\max \\left(a_{n+2}, a_{n+1}\\right)<\\max \\left(a_{n}, a_{n-1}\\right)$ or $b_{n+2}<b_{n}$. If $a_{n}>a_{n-1}$, then $a_{n+1}<\\max \\left(a_{n}, a_{n-1}\\right)=a_{n}$, and $a_{n+2}<\\max \\left(a_{n+1}, a_{n}\\right)=$\n$a_{n}$. Hence $b_{n+2}<b_{n}$. Thus if $a_{n}$ and $a_{n-1}$ are unequal, then $b_{n+2}<b_{n}$. But obviously $a_{n}>0$ for all $n$, and so $b_{n}>0$ for all $n$. Also it is integral, and $b_{1}=\\max (r, s)$, so we can only have $b_{2 n+1}<b_{2 n-1}$ for at most $\\max (r, s)$ values of $n$. Hence for some $n$ we must have $a_{n}=a_{n-1}$ and then $a_{n}$ is constant from that point on.\n\nWe show that the constant is the greatest common divisor of $\\mathrm{r}$ and $\\mathrm{s}$. Use parentheses to denote the greatest common divisor, so the greatest common divisor of $r$ and $s$ is $(r, s)$. We have $\\left(a_{n-1}, a_{n-2}\\right)=\\left(a_{n-1}, a_{n-1}+a_{n-2}\\right)=$ $\\left(a_{n}, a_{n-1}\\right)$. So if $a_{n}$ is constant for $n \\geq N$, we have $(r, s)=\\left(a_{1}, a_{2}\\right)=\\left(a_{2}, a_{3}\\right)=\\ldots=\\left(a_{N}, a_{N+1}\\right)=a_{N}$. We claim that $\\mathrm{a}_{\\mathrm{n}}=\\mathrm{d}$ for $\\mathrm{n}>3$. Clearly $\\mathrm{d}$ divides $\\mathrm{r}+\\mathrm{s}$ and hence $\\mathrm{d}$ divides $\\mathrm{a}_{3}$. But $\\mathrm{d}$ divides $\\mathrm{a}_{2}=\\mathrm{s}$, so $\\mathrm{d}$ also divides $\\mathrm{a}_{2}+$ $\\mathrm{a}_{3}$ and hence d divides $\\mathrm{a}_{4}$. Now suppose some odd $\\mathrm{k}$ divides $\\mathrm{r}+\\mathrm{s}$, but does not divide $\\mathrm{s}$. So $\\mathrm{k}$ divides $\\mathrm{a}_{3}$, but not $\\mathrm{a}_{2}$. Hence it does not divide $\\mathrm{a}_{4}$.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl", "problem_match": "# Problem 4", "solution_match": "# Solution"}}
{"year": "1993", "tier": "T1", "problem_label": "5", "problem_type": null, "exam": "USAMO", "problem": "A sequence $\\mathrm{x}_{\\mathrm{n}}$ of positive reals satisfies $\\mathrm{x}_{n-1} \\mathrm{x}_{\\mathrm{n}+1} \\leq \\mathrm{x}_{\\mathrm{n}}{ }^{2}$. Let $\\mathrm{a}_{\\mathrm{n}}$ be the average of the terms $\\mathrm{x}_{0}, \\mathrm{x}_{1}, \\ldots, \\mathrm{x}_{\\mathrm{n}}$ and $b_{n}$ be the average of the terms $x_{1}, x_{2}, \\ldots, x_{n}$. Show that $a_{n} b_{n-1} \\geq a_{n-1} b_{n}$.", "solution": "Put $\\mathrm{k}=\\mathrm{x}_{1}+\\mathrm{x}_{2}+\\ldots+\\mathrm{x}_{\\mathrm{n}-1}$. We have to show that $\\left(\\mathrm{x}_{0}+\\mathrm{k}+\\mathrm{x}_{\\mathrm{n}}\\right) /(\\mathrm{n}+1) \\mathrm{k} /(\\mathrm{n}-1) \\geq\\left(\\mathrm{x}_{0}+\\mathrm{k}\\right) / \\mathrm{n} \\quad\\left(\\mathrm{k}+\\mathrm{x}_{\\mathrm{n}}\\right) / \\mathrm{n}$ or $\\mathrm{n}^{2}(\\mathrm{k}+$ $\\left.x_{0}+x_{n}\\right) \\geq\\left(n^{2}-1\\right)\\left(k+x_{0}\\right)\\left(k+x_{n}\\right)$. Simplifying slightly, this is equivalent to $k\\left(k+x_{0}+x_{n}\\right) \\geq\\left(n^{2}-1\\right) x_{0} x_{n}$. So it is evidently sufficient to show that $k \\geq(n-1)\\left(x_{0} x_{n}\\right)^{1 / 2}$. [Then AM/GM gives $x_{0}+x_{n} \\geq 2\\left(x_{0} x_{n}\\right)^{1 / 2}$, so $\\left(k+x_{0}+\\right.$ $\\left.\\left.\\mathrm{x}_{\\mathrm{n}}\\right) \\geq(\\mathrm{n}+1)\\left(\\mathrm{x}_{0} \\mathrm{x}_{\\mathrm{n}}\\right)^{1 / 2}.\\right]$\n\nWe have $\\mathrm{x}_{0} / \\mathrm{x}_{1} \\leq \\mathrm{x}_{1} / \\mathrm{x}_{2} \\leq \\ldots \\leq \\mathrm{x}_{\\mathrm{n}-1} / \\mathrm{x}_{\\mathrm{n}}$. Hence (apparently weakening) $\\mathrm{x}_{0} \\mathrm{x}_{\\mathrm{n}} \\leq \\mathrm{x}_{1} \\mathrm{x}_{\\mathrm{n}-1} \\leq \\mathrm{x}_{2} \\mathrm{x}_{\\mathrm{n}-2} \\leq \\ldots$. So $\\mathrm{k}=\\left(\\mathrm{x}_{1}+\\mathrm{x}_{\\mathrm{n}-}\\right.$ $\\left.{ }_{1}\\right)+\\left(x_{2}+x_{n-2}\\right)+\\ldots \\geq(n-1)\\left(x_{0} x_{n}\\right)^{1 / 2}$, using first AM/GM, then the relation just established.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl", "problem_match": "# Problem 5", "solution_match": "# Solution"}}
{"year": "1994", "tier": "T1", "problem_label": "1", "problem_type": null, "exam": "USAMO", "problem": "$a_{1}, a_{2}, a_{3}, \\ldots$ are positive integers such that $a_{n}>a_{n-1}+1$. Put $b_{n}=a_{1}+a_{2}+\\ldots+a_{n}$. Show that there is always a square in the range $b_{n}, b_{n}+1, b_{n}+2, \\ldots, b_{n+1}-1$.", "solution": "If the result fails then for some $m$ we have $b_{n}>m^{2}$ and $b_{n+1} \\leq(m+1)^{2}$. So $b_{n+1}{ }^{1 / 2}-b_{n}{ }^{1 / 2}<1$. Thus it is sufficient to prove that $b_{n+1}{ }^{1 / 2}-b_{n}{ }^{1 / 2} \\geq 1$. Squaring, that is equivalent to $a_{n+1} \\geq 2 b_{n}{ }^{1 / 2}+1$ or $b_{n}{ }^{1 / 2} \\leq\\left(a_{n+1}-1\\right) / 2$.\n\nWe have $a_{n-1}<=a_{n}-2$. So $b_{n} \\leq a_{n}+\\left(a_{n}-2\\right)+\\left(a_{n}-4\\right)-\\ldots-\\left(a_{n}-2(n-1)\\right)$. Adding extra terms to the rhs if necessary we get $b_{n}=1+3+5+\\ldots+a_{n}$ for $a_{n}$ odd, or $2+4+6+\\ldots+a_{n}$ for $a_{n}$ even. In the odd case there are $\\left(a_{n}+1\\right) / 2$ terms of average size $\\left(a_{n}+1\\right) / 2$ (group them in pairs working from the outside in), so the sum is $\\left(a_{n}+1\\right)^{2} / 4$. In the even case there are $a_{n} / 2$ terms of average size $\\left(a_{n}+2\\right) / 2$, so the sum is $a_{n}\\left(a_{n}+2\\right) / 4$. So in either case we have $b_{n} \\leq\\left(a_{n}+1\\right)^{2} / 4$ and hence $b_{n}{ }^{1 / 2} \\leq\\left(a_{n}+1\\right) / 2 \\leq\\left(a_{n+1}-1\\right) / 2$.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl", "problem_match": "# Problem 1", "solution_match": "# Solution"}}
{"year": "1994", "tier": "T1", "problem_label": "2", "problem_type": null, "exam": "USAMO", "problem": "The sequence $a_{1}, a_{2}, \\ldots, a_{99}$ has $a_{1}=a_{3}=a_{5}=\\ldots=a_{97}=1, a_{2}=a_{4}=a_{6}=\\ldots=a_{98}=2$, and $a_{99}=3$. We interpret subscripts greater than 99 by subtracting 99 , so that $\\mathbf{a}_{100}$ means $\\mathrm{a}_{1}$ etc. An allowed move is to change the value of any one of the $a_{n}$ to another member of $\\{1,2,3\\}$ different from its two neighbors, $a_{n-1}$ and $a_{n+1}$. Is there a sequence of allowed moves which results in $\\mathrm{a}_{\\mathrm{m}}=\\mathrm{a}_{\\mathrm{m}+2}=\\ldots=\\mathrm{a}_{\\mathrm{m}+96}=1, \\mathrm{a}_{\\mathrm{m}+1}=\\mathrm{a}_{\\mathrm{m}+3}=\\ldots=\\mathrm{a}_{\\mathrm{m}+95}=2$, $\\mathrm{a}_{\\mathrm{m}+97}=3, \\mathrm{a}_{\\mathrm{n}+98}=2$ for some $\\mathrm{m}$ ? [So if $\\mathrm{m}=1$, we have just interchanged the values of $\\mathrm{a}_{98}$ and $\\mathrm{a}_{99}$.]", "solution": "This is a classic invariant problem. We strongly suspect that there is no sequence of moves (otherwise it would be too easy to find), so that we must prove there is no sequence. The standard approach is to look for some invariant which is not changed by the allowed moves, but which is different for the initial and desired final positions.\n\nFor each member $a_{i}$ of the sequence let $b_{i}=$\n\n0 if $\\mathrm{a}_{\\mathrm{i}}=\\mathrm{a}_{\\mathrm{i}+1}$,\n\n1 if $\\left(\\mathrm{a}_{\\mathrm{i}}, \\mathrm{a}_{\\mathrm{i}+1}\\right)=(1,2),(2,3)$ or $(3,1)$\n\n-1 if $\\left(\\mathrm{a}_{\\mathrm{i}}, \\mathrm{a}_{\\mathrm{i}+1}\\right)=(1,3),(2,1)$ or $(3,2)$.\n\nWe hope that $b_{1}+b_{2}+\\ldots+b_{9 g}$ will be a suitable invariant. Suppose we make an allowed move by changing $a_{i}$. That has the effect of changing just $b_{i-1}$ and $b_{i}$. If $a_{i-1}=a_{i+1}$, then $b_{i-1}+b_{i}=0$, so the total of the $b_{j}$ does not change. If $a_{i-1}$ does not equal $a_{i+1}$, then we cannot change $a_{i}$ since it must be different from both. Thus allowed moves do not change $b_{1}+\\ldots+b_{g 9}$. So it is an invariant. We now check it is suitable. The initial value is +3 (there are 49 pairs $(1,2), 48$ pairs $(2,1), 1$ pair $(2,3)$ and 1 pair $(3,1)$ total $49-48+1+1=3$. But the desired final position has value -3 (it has 48 pairs $(1,2), 49$ pairs $(2,1), 1$ pair $(1,3)$ and 1 pair $(3,2)$, total 48 - 49 - 1 $-1=-3)$. So we cannot get from one to the other by allowed moves.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl", "problem_match": "# Problem 2", "solution_match": "# Solution"}}
{"year": "1994", "tier": "T1", "problem_label": "3", "problem_type": null, "exam": "USAMO", "problem": "The hexagon $\\mathrm{ABCDEF}$ has the following properties: (1) its vertices lie on a circle; (2) $\\mathrm{AB}=\\mathrm{CD}=\\mathrm{EF}$; and (3) the diagonals $\\mathrm{AD}, \\mathrm{BE}, \\mathrm{CF}$ meet at a point. Let $\\mathrm{X}$ be the intersection of $\\mathrm{AD}$ and $\\mathrm{CE}$. Show that $\\mathrm{CX} / \\mathrm{XE}=$ $(\\mathrm{AC} / \\mathrm{CE})^{2}$.", "solution": "Let the diagonals $\\mathrm{AD}, \\mathrm{BE}, \\mathrm{CF}$ meet at $\\mathrm{Y}$. We show first that the triangles $\\mathrm{AEC}$, $\\mathrm{YED}$ are similar. $\\angle \\mathrm{ACE}=$ $\\angle \\mathrm{ADE}(\\mathrm{ACDE}$ circle) $=\\mathrm{YDE}$ (same angle). $\\angle \\mathrm{AEB}=\\angle \\mathrm{CED}(\\mathrm{AB}=\\mathrm{CD}$ ), so $\\angle \\mathrm{AEB}+\\angle \\mathrm{BEC}=\\angle \\mathrm{CED}+$ $\\angle \\mathrm{BEC}$ or $\\angle \\mathrm{AEC}=\\angle \\mathrm{YED}$. So $\\mathrm{AEC}$ and $\\mathrm{YED}$ are similar, so $\\mathrm{AC} / \\mathrm{CE}=\\mathrm{YD} / \\mathrm{DE}$.\n\nWe next show that $\\mathrm{AEC}$ and $\\mathrm{CDY}$ are similar. $\\angle \\mathrm{AEC}=\\angle \\mathrm{ADC}$ (circle) $=\\angle \\mathrm{CDY}$ (same angle). $\\angle \\mathrm{EAC}=$ $\\angle \\mathrm{EAD}+\\angle \\mathrm{DAC}=\\angle \\mathrm{ECD}$ (circle) $+\\angle \\mathrm{ECF}(\\mathrm{EF}=\\mathrm{CD})=\\angle \\mathrm{DCY}$. So $\\mathrm{AEC}$ and $\\mathrm{CDY}$ are similar. So $\\mathrm{AC} / \\mathrm{CE}=\\mathrm{CY} / \\mathrm{YD}$.\n\nHence $(\\mathrm{AC} / \\mathrm{CE})^{2}=\\mathrm{CY} / \\mathrm{DE}$. Finally we show that $\\mathrm{CXY}$ and EXD are similar. That is almost obvious because $\\mathrm{CF}$ is parallel to $\\mathrm{DE}$ since $\\mathrm{CD}=\\mathrm{EF}$. Hence $\\mathrm{CY} / \\mathrm{DE}=\\mathrm{CX} / \\mathrm{XE}$, which gives the required result.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl", "problem_match": "# Problem 3", "solution_match": "# Solution"}}
{"year": "1994", "tier": "T1", "problem_label": "4", "problem_type": null, "exam": "USAMO", "problem": "$\\mathrm{x}_{\\mathrm{i}}$ is a infinite sequence of positive reals such that for all $\\mathrm{n}, \\mathrm{x}_{1}+\\mathrm{x}_{2}+\\ldots+\\mathrm{x}_{\\mathrm{n}} \\geq \\sqrt{ }$. Show that $\\mathrm{x}_{1}{ }^{2}+\\mathrm{x}_{2}{ }^{2}+\\ldots+$ $\\mathrm{x}_{\\mathrm{n}}^{2}>(1+1 / 2+1 / 3+\\ldots+1 / \\mathrm{n}) / 4$ for all $\\mathrm{n}$.", "solution": "Note that this is rather a weak inequality. Taking $\\mathrm{n}=1$, we get $\\mathrm{x}_{1}>=1$, but $(1+1 / 2+1 / 3+\\ldots+1 / 30)<4$, so it is only for $\\mathrm{n}>30$ that we need to consider $\\mathrm{x}_{2}$ ! Of course, weak inequalities can be awkward to prove.\n\nFor $\\mathrm{a}+\\mathrm{b}$ constant, we minimise $\\mathrm{a}^{2}+\\mathrm{b}^{2}$ by taking $|\\mathrm{a}-\\mathrm{b}|$ as small as possible. So we suspect that the minimum value of $x_{1}{ }^{2}+\\ldots+x_{n}{ }^{2}$ is when $x_{1}=1, x_{2}=\\sqrt{2}-1, x_{3}=\\sqrt{3}-\\sqrt{2}, x_{4}=\\sqrt{4}-\\sqrt{3}, \\ldots(*)$. Note that $\\sqrt{n}-$ $\\sqrt{ }(n-1)=1 /\\left(\\sqrt{n}+\\sqrt{ }(n-1)>1 /(2 \\sqrt{n})\\right.$. So for the values $(*)$ we have $x_{1}{ }^{2}+\\ldots+x_{n}{ }^{2}>(1+1 / 2+1 / 3+\\ldots+1 / n) / 4$. So it remains to show that the values $\\left(^{*}\\right)$ do indeed give the minimum.\n\nWe use Abel's partial summation formula (whose proof is trivial). Put $y_{n}=V_{n}-\\sqrt{ }(n-1), s_{n}=x_{1}+x_{2}+\\ldots+x_{n}$, $t_{n}=y_{1}+\\ldots+t_{n}$. So we assume $s_{n} \\geq t_{n}$. Note also that $y_{1}>y_{2}>y_{3}>\\ldots$. Then the partial summation formula is: $\\sum x_{i} y_{i}=\\sum s_{i}\\left(y_{i}-y_{i+1}\\right)+s_{n} y_{n+1}$ (where the sum is taken from 1 to $n$ ). We also have $\\sum y_{i}^{2}=\\sum t_{i}\\left(y_{i}-y_{i+1}\\right)+$ $\\mathrm{t}_{\\mathrm{n}} \\mathrm{y}_{\\mathrm{n}+1}$. But each term is not more than the corresponding term in the first equality, so $\\sum \\mathrm{y}_{\\mathrm{i}}{ }^{2} \\leq \\sum \\mathrm{x}_{\\mathrm{i}} \\mathrm{y}_{\\mathrm{i}}$. Now Cauchy-Schwartz gives $\\left(\\sum \\mathrm{x}_{\\mathrm{i}} \\mathrm{y}_{\\mathrm{i}}\\right)^{2} \\leq \\sum \\mathrm{x}_{\\mathrm{i}}^{2} \\sum \\mathrm{y}_{\\mathrm{i}}^{2} \\cdot$ Hence $\\sum \\mathrm{y}_{\\mathrm{i}}^{2} \\leq \\sum \\mathrm{x}_{\\mathrm{i}}^{2}$, as required.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl", "problem_match": "# Problem 4", "solution_match": "# Solution"}}
{"year": "1994", "tier": "T1", "problem_label": "5", "problem_type": null, "exam": "USAMO", "problem": "$\\mathrm{X}$ is a set of $\\mathrm{n}$ positive integers with sum $\\mathrm{s}$ and product $\\mathrm{p}$. Show for any integer $\\mathrm{N}>=\\mathrm{s}, \\Sigma$ ( parity( $\\mathrm{Y}$ ) ( $\\mathrm{N}-$ sum(Y))Cs $)=\\mathrm{p}$, where $\\mathrm{aCb}$ is the binomial coefficient $\\mathrm{a}!/(\\mathrm{b}!(\\mathrm{a}-\\mathrm{b})$ ! ), the sum is taken over all subsets $\\mathrm{Y}$ of $\\mathrm{X}$, parity $(\\mathrm{Y})=1$ if $\\mathrm{Y}$ is empty or has an even number of elements, -1 if $\\mathrm{Y}$ has an odd number of elements, and $\\operatorname{sum}(\\mathrm{Y})$ is the sum of the elements in $\\mathrm{Y}$.", "solution": "$\\left(\\sum(-1)^{n}\\right.$ binomial) with the sum over all subsets $\\mathrm{Y}$ of a set $\\mathrm{X}$ strongly suggests the principle of inclusion and exclusion.\n\nConsider all sequences of length $\\mathrm{N} \\geq \\mathrm{s}$ of $0 \\mathrm{~s}$ and $1 \\mathrm{~s}$ with a total of $\\mathrm{s} 1 \\mathrm{~s}$. We can regard positions $1,2, \\ldots$, $a_{1}$ as block $a_{1}$, positions $a_{1}+1, a_{2}+2, \\ldots, a_{1}+a_{2}$ as block $a_{2}$, positions $a_{1}+a_{2}+1, a_{1}+a_{2}+2, \\ldots, a_{1}+a_{2}+a_{3}$ as block $a_{3}$, and so on up to $a_{1}+\\ldots+a_{n-1}+1, \\ldots a_{1}+\\ldots+a_{n}$ as block $a_{n}$. We are interested in sequences which have just one 1 in each block. Counting directly there are obviously just p such sequences.\n\nBut we can also take $\\mathrm{N}^{\\prime}$ to be the total number of sequences of $0 \\mathrm{~s}$ and $1 \\mathrm{~s}$ with $\\mathrm{s} 1 \\mathrm{~s}$ and $\\mathrm{N}_{\\mathrm{Y}}$ to be the number with no 1s in block $\\mathrm{k}$ for $\\mathrm{k}$ in $\\mathrm{Y}$, where $\\mathrm{Y}$ is a subset of $\\mathrm{X}$. Requiring one 1 in each block is equivalent to requiring no block to have no 1s. So the PIE gives that $\\mathrm{p}=\\mathrm{N}^{\\prime}-\\mathrm{N}_{\\mathrm{Y}}$ for $\\mathrm{Y}$ with one element $+\\mathrm{N}_{\\mathrm{Y}}$ for $\\mathrm{Y}$ with 2 elements and so on. $\\mathrm{N}^{\\prime}=\\mathrm{NCs}$ and $\\mathrm{N}_{\\mathrm{Y}}=(\\mathrm{N}-\\operatorname{sum}(\\mathrm{Y}))$ Cs, so we have $\\mathrm{p}=\\Sigma(\\operatorname{parity}(\\mathrm{Y})(\\mathrm{N}-\\operatorname{sum}(\\mathrm{Y}))$ Cs $)$, which is the required relation.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl", "problem_match": "# Problem 5", "solution_match": "# Solution"}}
{"year": "1995", "tier": "T1", "problem_label": "1", "problem_type": null, "exam": "USAMO", "problem": "The sequence $\\mathrm{a}_{0} \\mathrm{a}_{1}, \\mathrm{a}_{2}, \\ldots$ of non-negative integers is defined as follows. The first $\\mathrm{p}-1$ terms are $0,1,2,3, \\ldots$, $p-2$. Then $a_{n}$ is the least positive integer so that there is no arithmetic progression of length $p$ in the first $\\mathrm{n}+1$ terms. If $\\mathrm{p}$ is an odd prime, show that $\\mathrm{a}_{\\mathrm{n}}$ is the number obtained by writing $\\mathrm{n}$ in base $\\mathrm{p}-1$, then treating the result as a number in base p. For example, if p is 5 , to get the 5 th term one writes 5 as 11 in base 4 , then treats this as a base 5 number to get 6 .", "solution": "Let $b_{n}$ be the number obtained by writing $n$ in base $p-1$ and then treating the result as a number in base $p$. The resulting sequence $\\mathrm{b}_{\\mathrm{n}}$ is all those non-negative integers whose base $\\mathrm{p}$ representation does not have a digit $\\mathrm{p}-1$. We show that $\\mathrm{b}_{\\mathrm{n}}$ cannot contain an arithmetic progression of length $\\mathrm{p}$. For suppose there was such a progression with difference $\\mathrm{d}$. Suppose the last non-zero digit of $\\mathrm{d}$ in base $\\mathrm{p}$ is $\\mathrm{k}$. Suppose the first term of the progression has digit $\\mathrm{h}$ in that position, then the terms of the progression have digit $\\mathrm{h}, \\mathrm{h}+\\mathrm{k}, \\mathrm{h}+2 \\mathrm{k}, \\ldots \\mathrm{h}+(\\mathrm{p}-1) \\mathrm{k}$ $\\bmod p$ in that position. But these must be a complete set of residues $\\bmod p$, so one of them must be $p-1$ mod p. So the corresponding term has a digit p-1 in this position. Contradiction.\n\nNow to show that $\\mathrm{a}_{\\mathrm{n}}=\\mathrm{b}_{\\mathrm{n}}$ we use induction on $\\mathrm{n}$. Evidently, it is true for $\\mathrm{n}<\\mathrm{p}-1$. Suppose it is true for all $\\mathrm{m}<$ $\\mathrm{n}$. It is sufficient to show that if $\\mathrm{b}_{\\mathrm{n}}<\\mathrm{m}<\\mathrm{b}_{\\mathrm{n}+1}$, then $\\left\\{\\mathrm{b}_{1}, \\mathrm{~b}_{2}, \\ldots, \\mathrm{b}_{\\mathrm{n}}, \\mathrm{m}\\right\\}$ contains an arithmetic progression. $\\mathrm{m}$ must contain a digit p-1, for otherwise it would be $\\mathrm{b}_{\\mathrm{k}}$ for some $\\mathrm{k}>\\mathrm{n}$. Let $\\mathrm{m}_{1}$ be the number obtained from $\\mathrm{m}$ by reducing every digit $\\mathrm{p}-1$ in $\\mathrm{m}$ by 1 . Then $\\mathrm{m}$ has no digit $\\mathrm{p}-1 \\mathrm{~s}$, so it must be some $\\mathrm{b}_{\\mathrm{k}}$ and hence one of $\\mathrm{b}_{1}, \\ldots$ , $b_{n}$. Now take $m_{2}$ to be the number obtained by reducing the same digits by another 1 . Similarly, define $m_{3}$, $\\ldots, \\mathrm{m}_{p-1}$. Then each $\\mathrm{m}_{\\mathrm{i}}$ is in $\\left\\{\\mathrm{b}_{1}, \\ldots, \\mathrm{b}_{\\mathrm{n}}\\right\\}$ and $\\mathrm{m}_{\\mathrm{p}-1}, \\ldots, \\mathrm{m}_{1}, \\mathrm{~m}$ is a progression of length $\\mathrm{p}$.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl", "problem_match": "# Problem 1", "solution_match": "# Solution"}}
{"year": "1995", "tier": "T1", "problem_label": "2", "problem_type": null, "exam": "USAMO", "problem": "A trigonometric map is any one of sin, cos, tan, arcsin, arccos and arctan. Show that given any positive rational number $\\mathrm{x}$, one can find a finite sequence of trigonometric maps which take 0 to $\\mathrm{x}$. [So we need to show that we can always find a sequence of trigonometric maps $\\mathrm{t}_{\\mathrm{i}}$ so that: $\\mathrm{x}_{1}=\\mathrm{t}_{0}(0), \\mathrm{x}_{2}=\\mathrm{t}_{1}\\left(\\mathrm{x}_{1}\\right), \\ldots, \\mathrm{x}_{\\mathrm{n}}=\\mathrm{t}_{\\mathrm{n}}$. $\\left.1\\left(\\mathrm{x}_{\\mathrm{n}-1}\\right), \\mathrm{x}=\\mathrm{t}_{\\mathrm{n}}\\left(\\mathrm{x}_{\\mathrm{n}}\\right) \\cdot\\right]$", "solution": "We have $\\cos ^{2} t+\\sin ^{2} t=1$. Hence $\\cos t=\\cos t /\\left(\\sqrt{ }\\left(\\cos ^{2} t+\\sin ^{2} t\\right)\\right)=1 /\\left(\\sqrt{ }\\left(1+\\tan ^{2} t\\right)\\right)$. So if we put $\\tan t=\\sqrt{x}$, then $\\cos \\tan ^{-1} \\sqrt{x}=1 / \\sqrt{ }(1+x)$. We also have $\\cos (\\pi / 2-x)=\\sin x$ and $\\tan (\\pi / 2-x)=1 / \\tan x$. So $\\tan ^{-1} \\cos ^{-1} \\sin ^{-1} \\tan ^{-}$ ${ }^{1} x=1 / x(1)$. Hence also $\\tan ^{-1} \\cos ^{-1} \\sin ^{-1} \\tan ^{-1} \\cos \\tan ^{-1} \\sqrt{x}=\\sqrt{ }(x+1)$ (2). These two relations solve the problem.\n\nUsing (2) and iterating we can get $\\sqrt{ }$ for any positive integer $n$. Hence in particular we can get $\\mathrm{n}$ for any positive integer $\\mathrm{n}$. Now suppose we want $\\mathrm{m} / \\mathrm{n}$ with $\\mathrm{m}$ and $\\mathrm{n}$ relatively prime. We show that $\\mathrm{m} / \\mathrm{n}$ can be achieved by induction on $n$. We have just dealt with the case $\\mathrm{n}=1$. Suppose we have dealt with all $\\mathrm{a} / \\mathrm{b}$ with $\\mathrm{b}$ $<\\mathrm{n}$. If $\\mathrm{m}>\\mathrm{n}$, then we can write $\\mathrm{m}=\\mathrm{qn}+\\mathrm{r}$ with $0<\\mathrm{r}<\\mathrm{n}$ and use (2) to reduce the problem to getting $\\mathrm{r} / \\mathrm{n}$. If $\\mathrm{m}<\\mathrm{n}$, then put $\\mathrm{r}=\\mathrm{m}$. Now use (1) to reduce the problem to $\\mathrm{n} / \\mathrm{r}$, which is solved by induction.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl", "problem_match": "# Problem 2", "solution_match": "# Solution"}}
{"year": "1995", "tier": "T1", "problem_label": "3", "problem_type": null, "exam": "USAMO", "problem": "The circumcenter $\\mathrm{O}$ of the triangle $\\mathrm{ABC}$ does not lie on any side or median. Let the midpoints of $\\mathrm{BC}, \\mathrm{CA}$, $\\mathrm{AB}$ be $\\mathrm{L}, \\mathrm{M}, \\mathrm{N}$ respectively. Take $\\mathrm{P}, \\mathrm{Q}, \\mathrm{R}$ on the rays $\\mathrm{OL}, \\mathrm{OM}, \\mathrm{ON}$ respectively so that $\\angle \\mathrm{OPA}=\\angle \\mathrm{OAL}$, $\\angle \\mathrm{OQB}=\\angle \\mathrm{OBM}$ and $\\angle \\mathrm{ORC}=\\angle \\mathrm{OCN}$. Show that $\\mathrm{AP}, \\mathrm{BQ}$ and $\\mathrm{CR}$ meet at a point.", "solution": "We show that the circumcircle $A B C$ is the incircle of $P Q R$. Then $(A R / A Q)(C Q / C P)(B P / B R)=1$ since $A R$ $=\\mathrm{BR}, \\mathrm{AQ}=\\mathrm{CQ}, \\mathrm{BP}=\\mathrm{CP}$ (equal tangents) and hence $\\mathrm{PA}, \\mathrm{QB}, \\mathrm{RC}$ are concurrent by Ceva's theorem.\n\nSo let the tangents to the circumcircle at $\\mathrm{B}$ and $\\mathrm{C}$ meet at $\\mathrm{P}^{\\prime}$. It is sufficient to show that $\\angle \\mathrm{OP} \\mathrm{A}=\\angle \\mathrm{OAL}$, for then it follows that $\\mathrm{P}^{\\prime}=\\mathrm{P}$ (since there is obviously a unique point on the ray OL at which the segment OA subtends the $\\angle \\mathrm{OAL})$.\n\nThis is curiously difficult to prove. Let $\\mathrm{LP}^{\\prime}$ meet the circle at $\\mathrm{K}$. Then $\\angle \\mathrm{KCP}^{\\prime}=\\angle \\mathrm{KBC}$ ( $\\mathrm{P}^{\\prime} \\mathrm{C}$ tangent) $=\\angle$ $\\mathrm{KCB}$ (KO perpendicular to $\\mathrm{BC}$, since $\\mathrm{L}$ midpoint) $=\\angle \\mathrm{KCL}$ (same angle). So $\\mathrm{KC}$ bisects $\\angle \\mathrm{P}^{\\prime} \\mathrm{CL}$. Hence $\\mathrm{KP}^{\\prime} / \\mathrm{KL}=\\mathrm{CP}^{\\prime} / \\mathrm{CL}$. But obviously $\\mathrm{CP}^{\\prime} / \\mathrm{CL}=\\mathrm{BP}^{\\prime} / \\mathrm{BL}$. So $\\mathrm{K}, \\mathrm{C}$ and $\\mathrm{B}$ lie on the circle of Apollonius, the points $\\mathrm{X}$ such that $\\mathrm{XP}^{\\prime} / \\mathrm{XL}$ is constant. But A also lies on the circle of Apollonius. Hence KA bisects $\\angle \\mathrm{P}^{\\prime} \\mathrm{AL}$. (See Canada 71/9 if you are not familiar with the circle of Apollonius.) But $\\angle \\mathrm{OP}^{\\prime} \\mathrm{A}+\\angle \\mathrm{KAP}^{\\prime}=\\angle \\mathrm{OKA}=\\angle$ $\\mathrm{OAK}(\\mathrm{OA}$ and $\\mathrm{OK}$ radii $)=\\angle \\mathrm{OAL}+\\angle \\mathrm{KAL}$. So $\\angle \\mathrm{OP}^{\\prime} \\mathrm{A}=\\angle \\mathrm{OAL}$.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl", "problem_match": "# Problem 3", "solution_match": "# Solution"}}
{"year": "1995", "tier": "T1", "problem_label": "4", "problem_type": null, "exam": "USAMO", "problem": "$a_{0}, a_{1}, a_{2}, \\ldots$ is an infinite sequence of integers such that $a_{n}-a_{m}$ is divisible by $n-m$ for all (unequal) $n$ and $m$. For some polynomial $\\mathrm{p}(\\mathrm{x})$ we have $\\mathrm{p}(\\mathrm{n})>\\left|\\mathrm{a}_{\\mathrm{n}}\\right|$ for all $\\mathrm{n}$. Show that there is a polynomial $\\mathrm{q}(\\mathrm{x})$ such that $\\mathrm{q}(\\mathrm{n})=$ $a_{n}$ for all $n$.", "solution": "Clearly for any finite $N$, we can find a polynomial $q(n)$ of degree $N$ such that $q(n)=a_{n}$ for $n=0, \\ldots, N$. Also, once $\\mathrm{a}_{0}, \\mathrm{a}_{1}, \\ldots, \\mathrm{a}_{\\mathrm{N}}$ are fixed, $\\mathrm{a}_{\\mathrm{m}}$ is somewhat constrained for $\\mathrm{m}>\\mathrm{N}$, because we require $\\mathrm{a}_{\\mathrm{m}}$ to be $\\mathrm{a}_{0}$ mod $\\mathrm{m}$, $a_{1} \\bmod m-1, \\ldots, a_{N} \\bmod m-N$. Put $M=1 c m(m, m-1, \\ldots, m-N)$. Then $a_{m}$ is determined mod M.\n\nWe would like to argue that $\\mathrm{q}(\\mathrm{m})$ is known to satisfy the congruences (because $m-n$ divides $q(m)-q(n)$ ), that $\\mathrm{q}(\\mathrm{m})$ and $\\mathrm{a}_{\\mathrm{m}}$ are bounded so that they cannot differ by as much as $\\mathrm{M}$ and hence must be equal. The snag is that $\\mathrm{q}(\\mathrm{x})$ does not necessarily have integer coeffcients, so it is not necessarily true that $\\mathrm{m}$ - $\\mathrm{n}$ divides $\\mathrm{q}(\\mathrm{m})$ $\\mathrm{q}(\\mathrm{n})$. [The argument is that $\\mathrm{m}-\\mathrm{n}$ divides $\\mathrm{m}^{\\mathrm{k}}-\\mathrm{n}^{\\mathrm{k}}$ for any integer $\\mathrm{k}$ and hence divides $\\mathrm{q}(\\mathrm{m})-\\mathrm{q}(\\mathrm{n})$ for any polynomial with integer coefficients.]\n\nHowever, the coefficients of $\\mathrm{q}(\\mathrm{x})$ are rational. So let $\\mathrm{Q}$ be the $1 \\mathrm{~cm}$ of the denominators. Then $\\mathrm{Q} q(\\mathrm{x})$ does have integer coefficients and $m-n$ does divide $Q(q(m)-q(n))$. So $Q q(m)=Q q(0)=Q a_{0}=Q a_{m} m o d m$, and similarly $Q \\mathrm{q}(\\mathrm{m})=Q \\mathrm{a}_{\\mathrm{m}}$ mod $\\mathrm{m}-1$ and so on. Hence $\\mathrm{Q} q(\\mathrm{~m})=Q \\mathrm{a}_{\\mathrm{m}}$ mod $M$.\n\nBut we know that $|\\mathrm{q}(\\mathrm{m})|<\\mathrm{c}^{\\mathrm{N}}$ for some constant $\\mathrm{c}$, since $\\mathrm{q}(\\mathrm{x})$ has degree N. Similarly, we know that $\\left|\\mathrm{a}_{\\mathrm{m}}\\right|$ is bounded by a polynomial of degree $\\mathrm{N}$ and hence $\\left|\\mathrm{a}_{\\mathrm{m}}\\right|<\\mathrm{c}^{\\prime} \\mathrm{m}^{\\mathrm{N}}$ for some constant $\\mathrm{c}^{\\prime}$. Now $\\mathrm{M}$ certainly exceeds the product of the $\\mathrm{N}+1$ numbers $\\mathrm{m}, \\mathrm{m}-1, \\ldots, \\mathrm{m}-\\mathrm{N}$ divided by the greatest common divisor for each of the $\\mathrm{N}(\\mathrm{N}+1) / 2$ pairs ( $\\mathrm{m}-\\mathrm{i}, \\mathrm{m}-\\mathrm{j})$. But each gcd is at most $\\mathrm{N}$, so $\\mathrm{M}$ is more than $\\mathrm{c}^{\\prime \\prime} \\mathrm{m}^{\\mathrm{N}+1}$ for some (small) constant $c^{\\prime \\prime}$ which does not depend on $m$. Hence for $m \\geq$ some $N^{\\prime}$ we have $M>|Q q(m)|+\\left|Q a_{m}\\right|$ and hence $q(m)=a_{m}$.\n\nSo we have $\\mathrm{q}(\\mathrm{m})=\\mathrm{a}_{\\mathrm{m}}$ for $\\mathrm{m} \\leq \\mathrm{N}$ and for $\\mathrm{m} \\geq \\mathrm{N}^{\\prime}$. Suppose $\\mathrm{N}<\\mathrm{m}<\\mathrm{N}^{\\prime}$. Then for any $\\mathrm{m}^{\\prime}>\\mathrm{N}^{\\prime}$ we have that ( $\\mathrm{m}^{\\prime}$ $-m)$ divides $Q q\\left(m^{\\prime}\\right)-Q q(m)$ and $Q a_{m^{\\prime}}-Q a_{m}$. So it must divide their difference. But $q\\left(m^{\\prime}\\right)=a_{m^{\\prime}}$, so it divides $Q\\left(q(m)-a_{m}\\right)$. But we can choose $m^{\\prime}$ so that $\\left(m^{\\prime}-m\\right)$ is prime to $Q$ and larger than $\\left(q(m)-a_{m}\\right)$. Hence $q(m)=a_{m}$. So we have established that $q(m)=a_{m}$ for all $m$.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl", "problem_match": "# Problem 4", "solution_match": "# Solution"}}
{"year": "1995", "tier": "T1", "problem_label": "5", "problem_type": null, "exam": "USAMO", "problem": "A graph with $\\mathrm{n}$ points and $\\mathrm{k}$ edges has no triangles. Show that it has a point $\\mathrm{P}$ such that there are at most $\\mathrm{k}(1$ $-4 k / \\mathrm{n}^{2}$ ) edges between points not joined to P (by an edge).", "solution": "Given a point $\\mathrm{P}$, the edges of the graph can be divided into three categories: (1) edges $\\mathrm{PQ}$, (2) edges $\\mathrm{QQ}^{\\prime}$, where Q is joined to $\\mathrm{P}$, and (3) edges $\\mathrm{Q}^{\\prime} \\mathrm{Q}^{\\prime \\prime}$, where $\\mathrm{Q}^{\\prime}$ and $\\mathrm{Q}^{\\prime \\prime}$ are not joined to $\\mathrm{P}$. Note that if $\\mathrm{Q}$ is joined to $\\mathrm{P}$ and there is an edge $\\mathrm{QQ}^{\\prime}$, then $\\mathrm{Q}^{\\prime}$ cannot be joined to $\\mathrm{P}$, or $\\mathrm{PQQ}$ ' would be a triangle. So the total number of edges in categories (1) and (2) is $\\sum^{\\prime} \\operatorname{deg} Q$, where $\\sum^{\\prime}$ indicates that the sum is taken over points $\\mathrm{Q}$ which are joined to $\\mathrm{P}$ (by an edge). Thus the total number of edges in category (3) is $\\mathrm{k}-\\Sigma^{\\prime}$ deg $\\mathrm{Q}$. So we have to show that for some $\\mathrm{P}, \\Sigma^{\\prime} \\operatorname{deg} \\mathrm{Q} \\geq 4 \\mathrm{k}^{2} / \\mathrm{n}^{2}$.\n\nThe total for all points $\\mathrm{P}$ is $\\sum_{\\mathrm{P}} \\sum^{\\prime} \\operatorname{deg} \\mathrm{Q}$. If we invert the order of the summations, this becomes $\\sum \\operatorname{deg}^{2} \\mathrm{Q}$, where the sum is over all points $Q$. Now by Cauchy, $\\left(\\sum \\operatorname{deg} Q\\right)^{2} \\leq\\left(\\sum 1^{2}\\right)\\left(\\sum \\operatorname{deg}^{2} Q\\right)=n \\sum \\operatorname{deg}^{2} Q$. But $\\sum$ deg $\\mathrm{Q}=2 \\mathrm{k}$, so $\\sum \\operatorname{deg}^{2} \\mathrm{Q} \\geq 4 \\mathrm{k}^{2} / \\mathrm{n}$ and hence $\\sum_{\\mathrm{P}} \\sum^{\\prime} \\operatorname{deg} \\mathrm{Q} \\geq 4 \\mathrm{k}^{2} / \\mathrm{n}$. But if a sum of $\\mathrm{n}$ terms is at least $4 \\mathrm{k}^{2} / \\mathrm{n}$, then at least one for the terms must be at least $4 \\mathrm{k}^{2} / \\mathrm{n}^{2}$. In other words, there is a point $P$ such that $\\sum^{\\prime} \\operatorname{deg} Q \\geq 4 \\mathrm{k}^{2} / \\mathrm{n}^{2}$, as required.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1972-2003.jsonl", "problem_match": "# Problem 5", "solution_match": "# Solution"}}