{"year": "1997", "tier": "T1", "problem_label": "1", "problem_type": null, "exam": "USAMO", "problem": "Let $p_{1}, p_{2}, p_{3}, \\ldots$ be the prime numbers listed in increasing order, and let $0<x_{0}<1$ be a real number between 0 and 1 . For each positive integer $k$, define  $$ x_{k}= \\begin{cases}0 & \\text { if } x_{k-1}=0 \\\\ \\left\\{\\frac{p_{k}}{x_{k-1}}\\right\\} & \\text { if } x_{k-1} \\neq 0\\end{cases} $$  where $\\{x\\}$ denotes the fractional part of $x$. Find, with proof, all $x_{0}$ satisfying $0<x_{0}<1$ for which the sequence $x_{0}, x_{1}, x_{2}, \\ldots$ eventually becomes 0 .", "solution": "  ## The answer is $x_{0}$ rational.  If $x_{0}$ is irrational, then all $x_{i}$ are irrational by induction. So the sequence cannot become zero.  If $x_{0}$ is rational, then all are. Now one simply observes that the denominators of $x_{n}$ are strictly decreasing, until we reach $0=\\frac{0}{1}$. This concludes the proof.  Remark. The sequence $p_{k}$ could have been any sequence of integers.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1997-notes.jsonl"}}
{"year": "1997", "tier": "T1", "problem_label": "2", "problem_type": null, "exam": "USAMO", "problem": "Let $A B C$ be a triangle. Take noncollinear points $D, E, F$ on the perpendicular bisectors of $B C, C A, A B$ respectively. Show that the lines through $A, B, C$ perpendicular to $E F, F D, D E$ respectively are concurrent.", "solution": "  The three lines are the radical axii of the three circles centered at $D, E, F$, so they concur.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1997-notes.jsonl"}}
{"year": "1997", "tier": "T1", "problem_label": "3", "problem_type": null, "exam": "USAMO", "problem": "Prove that for any integer $n$, there exists a unique polynomial $Q$ with coefficients in $\\{0,1, \\ldots, 9\\}$ such that $Q(-2)=Q(-5)=n$.", "solution": "  If we let  $$ Q(x)=\\sum_{k \\geq 0} a_{k} x^{k} $$  then $a_{k}$ is uniquely determined by $n\\left(\\bmod 2^{k}\\right)$ and $n\\left(\\bmod 5^{k}\\right)$. Indeed, we can extract the coefficients of $Q$ exactly by the following algorithm:  - Define $b_{0}=c_{0}=n$. - For $i \\geq 0$, let $a_{i}$ be the unique digit satisfying $a_{i} \\equiv b_{i}(\\bmod 2), a_{i} \\equiv c_{i}(\\bmod 5)$. Then, define  $$ b_{i+1}=\\frac{b_{i}-a_{i}}{-2}, \\quad c_{i+1}=\\frac{c_{i}-a_{i}}{-5} . $$   In fact, we will prove the following claim: Claim - Suppose $b_{0}$ and $c_{0}$ are any integers such that  $$ b_{0} \\equiv c_{0} \\quad(\\bmod 3) . $$  Then defining $b_{i}$ and $c_{i}$ as above, we have $b_{i} \\equiv c_{i}(\\bmod 3)$ for all $i$, and $b_{N}=c_{N}=0$ for large enough $N$.   $$ (b, c) \\mapsto\\left(\\frac{b-a}{-2}, \\frac{c-a}{-5}\\right) $$  for some $0 \\leq a \\leq 9$ (a function in $b$ and $c$ ). The $b \\equiv c(\\bmod 3)$ is clearly preserved. Also, examining the size,  - If $|c|>2$, we have $\\left|\\frac{c-a}{-5}\\right| \\leq \\frac{|c|+9}{5}<|c|$. Thus, we eventually reach a pair with $|c| \\leq 2$. - Similarly, if $|b|>9$, we have $\\left|\\frac{b-a}{-2}\\right| \\leq \\frac{|b|+9}{2}<|b|$, so we eventually reach a pair with $|b| \\leq 9$. this leaves us with $5 \\cdot 19=95$ ordered pairs to check (though only about one third have $b \\equiv c(\\bmod 3))$. This can be done by the following code:  ``` import functools @functools.lru_cache() def f(x0, yO):     if x0 == 0 and y0 == 0:         return 0     if x0 % 2 == (y0 % 5) % 2:         d = y0 % 5     else:         d = (y0 % 5) + 5     x1 = (x0 - d) // (-2)     y1 = (y0 - d) // (-5)     return 1 + f(x1, y1) for x in range(-9, 10): for y in range(-2, 3):     if (x % 3 == y % 3):         print(f\"({x:2d}, {y:2d}) finished in {f(x,y)} moves\") ```  As this gives the output ![](https://cdn.mathpix.com/cropped/2024_11_19_35408d86cf32c99e435fg-06.jpg?height=1317&width=664&top_left_y=1169&top_left_x=296) we are done.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1997-notes.jsonl"}}
{"year": "1997", "tier": "T1", "problem_label": "4", "problem_type": null, "exam": "USAMO", "problem": "To clip a convex $n$-gon means to choose a pair of consecutive sides $A B, B C$ and to replace them by the three segments $A M, M N$, and $N C$, where $M$ is the midpoint of $A B$ and $N$ is the midpoint of $B C$. In other words, one cuts off the triangle $M B N$ to obtain a convex $(n+1)$-gon. A regular hexagon $\\mathcal{P}_{6}$ of area 1 is clipped to obtain a heptagon $\\mathcal{P}_{7}$. Then $\\mathcal{P}_{7}$ is clipped (in one of the seven possible ways) to obtain an octagon $\\mathcal{P}_{8}$, and so on. Prove that no matter how the clippings are done, the area of $\\mathcal{P}_{n}$ is greater than $\\frac{1}{3}$, for all $n \\geq 6$.", "solution": "  Call the original hexagon $A B C D E F$. We show the area common to triangles $A C E$ and $B D F$ is in every $\\mathcal{P}_{n}$; this solves the problem since the area is $1 / 3$.  For every side of a clipped polygon, we define its foundation recursively as follows:  - $A B, B C, C D, D E, E F, F A$ are each their own foundation (we also call these original sides). - When a new clipped edge is added, its foundation is the union of the foundations of the two edges it touches.  Hence, any foundations are nonempty subsets of original sides. Claim - All foundations are in fact at most two-element sets of adjacent original sides.   Now, if a side has foundation contained in $\\{A B, B C\\}$, say, then the side should be contained within triangle $A B C$. Hence the side does not touch $A C$. This proves the problem.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1997-notes.jsonl"}}
{"year": "1997", "tier": "T1", "problem_label": "5", "problem_type": null, "exam": "USAMO", "problem": "If $a, b, c>0$ prove that  $$ \\frac{1}{a^{3}+b^{3}+a b c}+\\frac{1}{b^{3}+c^{3}+a b c}+\\frac{1}{c^{3}+a^{3}+a b c} \\leq \\frac{1}{a b c} . $$", "solution": "  From $a^{3}+b^{3} \\geq a b(a+b)$, the left-hand side becomes  $$ \\sum_{\\mathrm{cyc}} \\frac{1}{a^{3}+b^{3}+a b c} \\leq \\sum_{\\mathrm{cyc}} \\frac{1}{a b(a+b+c)}=\\frac{1}{a b c} \\sum_{\\mathrm{cyc}} \\frac{c}{a+b+c}=\\frac{1}{a b c} $$", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1997-notes.jsonl"}}
{"year": "1997", "tier": "T1", "problem_label": "6", "problem_type": null, "exam": "USAMO", "problem": "Suppose the sequence of nonnegative integers $a_{1}, a_{2}, \\ldots, a_{1997}$ satisfies  $$ a_{i}+a_{j} \\leq a_{i+j} \\leq a_{i}+a_{j}+1 $$  for all $i, j \\geq 1$ with $i+j \\leq 1997$. Show that there exists a real number $x$ such that $a_{n}=\\lfloor n x\\rfloor$ for all $1 \\leq n \\leq 1997$.", "solution": "  We are trying to show there exists an $x \\in \\mathbb{R}$ such that  $$ \\frac{a_{n}}{n} \\leq x<\\frac{a_{n}+1}{n} \\quad \\forall n $$  This means we need to show  $$ \\max _{i} \\frac{a_{i}}{i}<\\min _{j} \\frac{a_{j}+1}{j} . $$  Replace 1997 by $N$. We will prove this by induction, but we will need some extra hypotheses on the indices $i, j$ which are used above.  ## Claim - Suppose that  - Integers $a_{1}, a_{2}, \\ldots, a_{N}$ satisfy the given conditions. - Let $i=\\operatorname{argmax}_{n} \\frac{a_{n}}{n}$; if there are ties, pick the smallest $i$. - Let $j=\\operatorname{argmin}_{n} \\frac{a_{n}+1}{n}$; if there are ties, pick the smallest $j$.  Then  $$ \\frac{a_{i}}{i}<\\frac{a_{j}+1}{j} $$  Moreover, these two fractions are in lowest terms, and are adjacent in the Farey sequence of order $\\max (i, j)$.   Now, consider the new number $a_{N}$. We have two cases:  - Suppose $i+j>N$. Then, no fraction with denominator $N$ can lie strictly inside the interval; so we may write for some integer $b$  $$ \\frac{b}{N} \\leq \\frac{a_{i}}{i}<\\frac{a_{j}+1}{j} \\leq \\frac{b+1}{N} $$  For purely algebraic reasons we have  $$ \\frac{b-a_{i}}{N-i} \\leq \\frac{b}{N} \\leq \\frac{a_{i}}{i}<\\frac{a_{j}+1}{j} \\leq \\frac{b+1}{N} \\leq \\frac{b-a_{j}}{N-j} . $$  Now,  $$ \\begin{aligned} a_{N} & \\geq a_{i}+a_{N-i} \\geq a_{i}+(N-i) \\cdot \\frac{a_{i}}{i} \\\\ & \\geq a_{i}+\\left(b-a_{i}\\right)=b \\\\ a_{N} & \\leq a_{j}+a_{N-j}+1 \\leq\\left(a_{j}+1\\right)+(N-j) \\cdot \\frac{a_{j}+1}{j} \\\\ & =\\left(a_{j}+1\\right)+\\left(b-a_{j}\\right)=b+1 \\end{aligned} $$  Thus $a_{N} \\in\\{b, b+1\\}$. This proves that $\\frac{a_{N}}{N} \\leq \\frac{a_{i}}{i}$ while $\\frac{a_{N}+1}{N} \\geq \\frac{a_{j}+1}{j}$. Moreover, the pair $(i, j)$ does not change, so all inductive hypotheses carry over.  - On the other hand, suppose $i+j=N$. Then we have  $$ \\frac{a_{i}}{i}<\\frac{a_{i}+a_{j}+1}{N}<\\frac{a_{j}+1}{j} . $$  Now, we know $a_{N}$ could be either $a_{i}+a_{j}$ or $a_{i}+a_{j}+1$. If it's the former, then $(i, j)$ becomes $(i, N)$. If it's the latter, then $(i, j)$ becomes $(N, j)$. The properties of Farey sequences ensure that the $\\frac{a_{i}+a_{j}+1}{N}$ is reduced, either way.", "metadata": {"resource_path": "USAMO/segmented/en-USAMO-1997-notes.jsonl"}}