{"year": "2016", "tier": "T3", "problem_label": "1", "problem_type": null, "exam": "INMO", "problem": "Let $A B C$ be triangle in which $A B=A C$. Suppose the orthocentre of the triangle lies on the in-circle. Find the ratio $A B / B C$.", "solution": "Since the triangle is isosceles, the orthocentre lies on the perpendicular $A D$ from $A$ on to $B C$. Let it cut the in-circle at $H$. Now we are given that $H$ is the orthocentre of the triangle. Let $A B=A C=b$ and $B C=2 a$. Then $B D=a$. Observe that $b>a$ since $b$ is the hypotenuse and $a$ is a leg of a right-angled triangle. Let $B H$ meet $A C$ in $E$ and $C H$ meet $A B$ in $F$. By Pythagoras theorem applied to $\\triangle B D H$, we get\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_8eb8bd67d3f7fc489f33g-1.jpg?height=328&width=496&top_left_y=600&top_left_x=1313)\n\n$$\nB H^{2}=H D^{2}+B D^{2}=4 r^{2}+a^{2}\n$$\n\nwhere $r$ is the in-radius of $A B C$. We want to compute $B H$ in another way. Since $A, F, H, E$ are con-cyclic, we have\n\n$$\nB H \\cdot B E=B F \\cdot B A\n$$\n\nBut $B F \\cdot B A=B D \\cdot B C=2 a^{2}$, since $A, F, D, C$ are con-cyclic. Hence $B H^{2}=4 a^{4} / B E^{2}$. But\n\n$$\nB E^{2}=4 a^{2}-C E^{2}=4 a^{2}-B F^{2}=4 a^{2}-\\left(\\frac{2 a^{2}}{b}\\right)^{2}=\\frac{4 a^{2}\\left(b^{2}-a^{2}\\right)}{b^{2}}\n$$\n\nThis leads to\n\n$$\nB H^{2}=\\frac{a^{2} b^{2}}{b^{2}-a^{2}}\n$$\n\nThus we get\n\n$$\n\\frac{a^{2} b^{2}}{b^{2}-a^{2}}=a^{2}+4 r^{2}\n$$\n\nThis simplifies to $\\left(a^{4} /\\left(b^{2}-a^{2}\\right)\\right)=4 r^{2}$. Now we relate $a, b, r$ in another way using area. We know that $[A B C]=r s$, where $s$ is the semi-perimeter of $A B C$. We have $s=(b+b+2 a) / 2=b+a$. On the other hand area can be calculated using Heron's formula::\n\n$$\n[A B C]^{2}=s(s-2 a)(s-b)(s-b)=(b+a)(b-a) a^{2}=a^{2}\\left(b^{2}-a^{2}\\right)\n$$\n\nHence\n\n$$\nr^{2}=\\frac{[A B C]^{2}}{s^{2}}=\\frac{a^{2}\\left(b^{2}-a^{2}\\right)}{(b+a)^{2}}\n$$\n\nUsing this we get\n\n$$\n\\frac{a^{4}}{b^{2}-a^{2}}=4\\left(\\frac{a^{2}\\left(b^{2}-a^{2}\\right)}{(b+a)^{2}}\\right)\n$$\n\nTherefore $a^{2}=4(b-a)^{2}$, which gives $a=2(b-a)$ or $2 b=3 a$. Finally,\n\n$$\n\\frac{A B}{B C}=\\frac{b}{2 a}=\\frac{3}{4}\n$$\n\n## Alternate Solution 1:\n\nWe use the known facts $B H=2 R \\cos B$ and $r=4 R \\sin (A / 2) \\sin (B / 2) \\sin (C / 2)$, where $R$ is the circumradius of $\\triangle A B C$ and $r$ its in-radius. Therefore\n\n$$\nH D=B H \\sin \\angle H B D=2 R \\cos B \\sin \\left(\\frac{\\pi}{2}-C\\right)=2 R \\cos ^{2} B\n$$\n\nsince $\\angle C=\\angle B$. But $\\angle B=(\\pi-\\angle A) / 2$, since $A B C$ is isosceles. Thus we obtain\n\n$$\nH D=2 R \\cos ^{2}\\left(\\frac{\\pi}{2}-\\frac{A}{2}\\right)\n$$\n\nHowever $H D$ is also the diameter of the in circle. Therefore $H D=2 r$. Thus we get\n\n$$\n2 R \\cos ^{2}\\left(\\frac{\\pi}{2}-\\frac{A}{2}\\right)=2 r=8 R \\sin (A / 2) \\sin ^{2}((\\pi-A) / 4)\n$$\n\nThis reduces to\n\n$$\n\\sin (A / 2)=2(1-\\sin (A / 2))\n$$\n\nTherefore $\\sin (A / 2)=2 / 3$. We also observe that $\\sin (A / 2)=B D / A B$. Finally\n\n$$\n\\frac{A B}{B C}=\\frac{A B}{2 B D}=\\frac{1}{2 \\sin (A / 2)}=\\frac{3}{4}\n$$\n\n## Alternate Solution 2:\n\nLet $D$ be the mid-point of $B C$. Extend $A D$ to meet the circumcircle in $L$. Then we know that $H D=D L$. But $H D=2 r$. Thus $D L=2 r$. Therefore $I L=I D+D L=r+2 r=3 r$. We also know that $L B=L I$. Therefore $L B=3 r$. This gives\n\n$$\n\\frac{B L}{L D}=\\frac{3 r}{2 r}=\\frac{3}{2}\n$$\n\nBut $\\triangle B L D$ is similar to $\\triangle A B D$. So\n\n$$\n\\frac{A B}{B D}=\\frac{B L}{L D}=\\frac{3}{2}\n$$\n\nFinally,\n\n$$\n\\frac{A B}{B C}=\\frac{A B}{2 B D}=\\frac{3}{4}\n$$\n\n## Alternate Solution 3:\n\nLet $D$ be the mid-point of $B C$ and $E$ be the mid-point of $D C$. Since $D I=I H(=r)$ and $D E=E C$, the mid-point theorem implies that $I E \\| C H$. But $C H \\perp A B$. Therefore $E I \\perp A B$. Let $E I$ meet $A B$ in $F$. Then $F$ is the point of tangency of the incircle of $\\triangle A B C$ with $A B$. Since the incircle is also tangent to $B C$ at $D$, we have $B F=B D$. Observe that $\\triangle B F E$ is similar to $\\triangle B D A$. Hence\n\n$$\n\\frac{A B}{B D}=\\frac{B E}{B F}=\\frac{B E}{B D}=\\frac{B D+D E}{B D}=1+\\frac{D E}{B D}=\\frac{3}{2}\n$$\n\nThis gives\n\n$$\n\\frac{A B}{B C}=\\frac{3}{4}\n$$", "metadata": {"resource_path": "INMO/segmented/en-sol-inmo16.jsonl", "problem_match": "\n1.", "solution_match": "\nSolution:"}}
{"year": "2016", "tier": "T3", "problem_label": "2", "problem_type": null, "exam": "INMO", "problem": "For positive real numbers $a, b, c$, which of the following statements necessarily implies $a=b=c$ : (I) $a\\left(b^{3}+c^{3}\\right)=b\\left(c^{3}+a^{3}\\right)=c\\left(a^{3}+b^{3}\\right)$, (II) $a\\left(a^{3}+b^{3}\\right)=b\\left(b^{3}+c^{3}\\right)=c\\left(c^{3}+a^{3}\\right)$ ? Justify your answer.", "solution": "We show that (I) need not imply that $a=b=c$ where as (II) always implies $a=b=c$.\n\nObserve that $a\\left(b^{3}+c^{3}\\right)=b\\left(c^{3}+a^{3}\\right)$ gives $c^{3}(a-b)=a b\\left(a^{2}-b^{2}\\right)$. This gives either $a=b$ or $a b(a+b)=c^{3}$. Similarly, $b=c$ or $b c(b+c)=a^{3}$. If $a \\neq b$ and $b \\neq c$, we obtain\n\n$$\na b(a+b)=c^{3}, \\quad b c(b+c)=a^{3}\n$$\n\nTherefore\n\n$$\nb\\left(a^{2}-c^{2}\\right)+b^{2}(a-c)=c^{3}-a^{3}\n$$\n\nThis gives $(a-c)\\left(a^{2}+b^{2}+c^{2}+a b+b c+c a\\right)=0$. Since $a, b, c$ are positive, the only possibility is $a=c$. We have therefore 4 possibilities: $a=b=c ; a \\neq b, b \\neq c$ and $c=a ; b \\neq c, c \\neq a$ and $a=b$; $c \\neq a, a \\neq b$ and $b=c$.\n\nSuppose $a=b$ and $b, a \\neq c$. Then $b\\left(c^{3}+a^{3}\\right)=c\\left(a^{3}+b^{3}\\right)$ gives $a c^{3}+a^{4}=2 c a^{3}$. This implies that $a(a-c)\\left(a^{2}-a c-c^{2}\\right)=0$. Therefore $a^{2}-a c-c^{2}=0$. Putting $a / c=x$, we get the quadratic equation $x^{2}-x-1=0$. Hence $x=(1+\\sqrt{5}) / 2$. Thus we get\n\n$$\na=b=\\left(\\frac{1+\\sqrt{5}}{2}\\right) c, \\quad c \\text { arbitrary positive real number. }\n$$\n\nSimilarly, we get other two cases:\n\n$$\n\\begin{aligned}\n& b=c=\\left(\\frac{1+\\sqrt{5}}{2}\\right) a, \\quad a \\text { arbitrary positive real number } \\\\\n& c=a=\\left(\\frac{1+\\sqrt{5}}{2}\\right) b, \\quad b \\text { arbitrary positive real number. }\n\\end{aligned}\n$$\n\nAnd $a=b=c$ is the fourth possibility.\n\nConsider (II): $a\\left(a^{3}+b^{3}\\right)=b\\left(b^{3}+c^{3}\\right)=c\\left(c^{3}+a^{3}\\right)$. Suppose $a, b, c$ are mutually distinct. We may assume $a=\\max \\{a, b, c\\}$. Hence $a>b$ and $a>c$. Using $a>b$, we get from the first relation that $a^{3}+b^{3}<b^{3}+c^{3}$. Therefore $a^{3}<c^{3}$ forcing $a<c$. This contradicts $a>c$. We conclude that $a, b, c$ cannot be mutually distinct. This means some two must be equal. If $a=b$, the equality of the first two expressions give $a^{3}+b^{3}=b^{3}+c^{3}$ so that $a=c$. Similarly, we can show that $b=c$ implies $b=a$ and $c=a$ gives $c=b$.\n\nAlternate for (II) by a contestant: We can write\n\n$$\n\\begin{aligned}\n\\frac{a^{3}}{c}+\\frac{b^{3}}{c} & =\\frac{c^{3}}{a}+a^{2} \\\\\n\\frac{b^{3}}{a}+\\frac{c^{3}}{a} & =\\frac{a^{3}}{b}+b^{2} \\\\\n\\frac{c^{3}}{b}+\\frac{a^{3}}{b} & =\\frac{b^{3}}{c}+c^{2}\n\\end{aligned}\n$$\n\nAdding, we get\n\n$$\n\\frac{a^{3}}{c}+\\frac{b^{3}}{a}+\\frac{c^{3}}{b}=a^{2}+b^{2}+c^{2}\n$$\n\nUsing C-S inequality, we have\n\n$$\n\\begin{aligned}\n\\left(a^{2}+b^{2}+c^{2}\\right)^{2} & =\\left(\\frac{\\sqrt{a^{3}}}{\\sqrt{c}} \\cdot \\sqrt{a c}+\\frac{\\sqrt{b^{3}}}{\\sqrt{a}} \\cdot \\sqrt{b a}+\\frac{\\sqrt{c^{3}}}{\\sqrt{b}} \\cdot \\sqrt{c b}\\right)^{2} \\\\\n& \\leq\\left(\\frac{a^{3}}{c}+\\frac{b^{3}}{a}+\\frac{c^{3}}{b}\\right)(a c+b a+c b) \\\\\n& =\\left(a^{2}+b^{2}+c^{2}\\right)(a b+b c+c a)\n\\end{aligned}\n$$\n\nThus we obtain\n\n$$\na^{2}+b^{2}+c^{2} \\leq a b+b c+c a\n$$\n\nHowever this implies $(a-b)^{2}+(b-c)^{2}+(c-a)^{2} \\leq 0$ and hence $a=b=c$.", "metadata": {"resource_path": "INMO/segmented/en-sol-inmo16.jsonl", "problem_match": "\n2.", "solution_match": "\nSolution:"}}
{"year": "2016", "tier": "T3", "problem_label": "3", "problem_type": null, "exam": "INMO", "problem": "Let $\\mathbb{N}$ denote the set of all natural numbers. Define a function $T: \\mathbb{N} \\rightarrow \\mathbb{N}$ by $T(2 k)=k$ and $T(2 k+1)=2 k+2$. We write $T^{2}(n)=T(T(n))$ and in general $T^{k}(n)=T^{k-1}(T(n))$ for any $k>1$.\n\n(i) Show that for each $n \\in \\mathbb{N}$, there exists $k$ such that $T^{k}(n)=1$.\n\n(ii) For $k \\in \\mathbb{N}$, let $c_{k}$ denote the number of elements in the set $\\left\\{n: T^{k}(n)=1\\right\\}$. Prove that $c_{k+2}=c_{k+1}+c_{k}$, for $k \\geq 1$.", "solution": "(i) For $n=1$, we have $T(1)=2$ and $T^{2}(1)=T(2)=1$. Hence we may assume that $n>1$.\n\nSuppose $n>1$ is even. Then $T(n)=n / 2$. We observe that $(n / 2) \\leq n-1$ for $n>1$.\n\nSuppose $n>1$ is odd so that $n \\geq 3$. Then $T(n)=n+1$ and $T^{2}(n)=(n+1) / 2$. Again we see that $(n+1) / 2 \\leq(n-1)$ for $n \\geq 3$.\n\nThus we see that in at most $2(n-1)$ steps $T$ sends $n$ to 1 . Hence $k \\leq 2(n-1)$. (Here $2(n-1)$ is only a bound. In reality, less number of steps will do.)\n\n(ii) We show that $c_{n}=f_{n+1}$, where $f_{n}$ is the $n$-th Fibonacci number.\n\nLet $n \\in \\mathbb{N}$ and let $k \\in \\mathbb{N}$ be such that $T^{k}(n)=1$. Here $n$ can be odd or even. If $n$ is even, it can be either of the form $4 d+2$ or of the form $4 d$.\n\nIf $n$ is odd, then $1=T^{k}(n)=T^{k-1}(n+1)$. (Observe that $k>1$; otherwise we get $n+1=1$ which is impossible since $n \\in \\mathbb{N}$.) Here $n+1$ is even.\n\nIf $n=4 d+2$, then again $1=T^{k}(4 d+2)=T^{k-1}(2 d+1)$. Here $2 d+1=n / 2$ is odd.\n\nThus each solution of $T^{k-1}(m)=1$ produces exactly one solution of $T^{k}(n)=1$ and $n$ is either odd or of the form $4 d+2$.\n\nIf $n=4 d$, we see that $1=T^{k}(4 d)=T^{k-1}(2 d)=T^{k-2}(d)$. This shows that each solution of $T^{k-2}(m)=1$ produces exactly one solution of $T^{k}(n)=1$ of the form $4 d$.\n\nThus the number of solutions of $T^{k}(n)=1$ is equal to the number of solutions of $T^{k-1}(m)=1$ and the number of solutions of $T^{k-2}(l)=1$ for $k>2$. This shows that $c_{k}=c_{k-1}+c_{k-2}$ for $k>2$. We also observe that 2 is the only number which goes to 1 in one step and 4 is the only number which goes to 1 in two steps. Hence $c_{1}=1$ and $c_{2}=2$. This proves that $c_{n}=f_{n+1}$ for all $n \\in \\mathbb{N}$.", "metadata": {"resource_path": "INMO/segmented/en-sol-inmo16.jsonl", "problem_match": "\n3.", "solution_match": "## Solution:"}}
{"year": "2016", "tier": "T3", "problem_label": "4", "problem_type": null, "exam": "INMO", "problem": "Suppose 2016 points of the circumference of a circle are coloured red and the remaining points are coloured blue. Given any natural number $n \\geq 3$, prove that there is a regular $n$-sided polygon all of whose vertices are blue.", "solution": "Let $A_{1}, A_{2}, \\ldots, A_{2016}$ be 2016 points on the circle which are coloured red and the remain-\ning blue. Let $n \\geq 3$ and let $B_{1}, B_{2}, \\ldots, B_{n}$ be a regular $n$-sided polygon inscribed in this circle with the vertices chosen in anti-clock-wise direction. We place $B_{1}$ at $A_{1}$. (It is possible, in this position, some other $B$ 's also coincide with some other $A$ 's.) Rotate the polygon in anti-clock-wise direction gradually till some $B$ 's coincide with (an equal number of) $A$ 's second time. We again rotate the polygon in the same direction till some $B$ 's coincide with an equal number of $A$ 's third time, and so on until we return to the original position, i.e., $B_{1}$ at $A_{1}$. We see that the number of rotations will not be more than $2016 \\times n$, that is, at most these many times some $B$ 's would have coincided with an equal number of $A$ 's. Since the interval $\\left(0,360^{\\circ}\\right)$ has infinitely many points, we can find a value $\\alpha^{\\circ} \\in\\left(0,360^{\\circ}\\right)$ through which the polygon can be rotated from its initial position such that no $B$ coincides with any $A$. This gives a $n$-sided regular polygon having only blue vertices.\n\nAlternate Solution: Consider a regular $2017 \\times n$-gon on the circle; say, $A_{1} A_{2} A_{3} \\cdots A_{2017 n}$. For each $j, 1 \\leq j \\leq 2017$, consider the points $\\left\\{A_{k}: k \\equiv j(\\bmod 2017)\\right\\}$. These are the vertices of a regular $n$-gon, say $S_{j}$. We get 2017 regular $n$-gons; $S_{1}, S_{2}, \\ldots, S_{2017}$. Since there are only 2016 red points, by pigeon-hole principle there must be some $n$-gon among these 2017 which does not contain any red point. But then it is a blue $n$-gon.", "metadata": {"resource_path": "INMO/segmented/en-sol-inmo16.jsonl", "problem_match": "\n4.", "solution_match": "\nSolution:"}}
{"year": "2016", "tier": "T3", "problem_label": "5", "problem_type": null, "exam": "INMO", "problem": "Let $A B C$ be a right-angled triangle with $\\angle B=90^{\\circ}$. Let $D$ be a point on $A C$ such that the in-radii of the triangles $A B D$ and $C B D$ are equal. If this common value is $r^{\\prime}$ and if $r$ is the in-radius of triangle $A B C$, prove that\n\n$$\n\\frac{1}{r^{\\prime}}=\\frac{1}{r}+\\frac{1}{B D}\n$$", "solution": "Let $E$ and $F$ be the incentres of triangles $A B D$ and $C B D$ respectively. Let the incircles of triangles $A B D$ and $C B D$ touch $A C$ in $P$ and $Q$ respectively. If $\\angle B D A=\\theta$, we see that\n\n$$\nr^{\\prime}=P D \\tan (\\theta / 2)=Q D \\cot (\\theta / 2)\n$$\n\nHence\n\n$$\nP Q=P D+Q D=r^{\\prime}\\left(\\cot \\frac{\\theta}{2}+\\tan \\frac{\\theta}{2}\\right)=\\frac{2 r^{\\prime}}{\\sin \\theta}\n$$\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_8eb8bd67d3f7fc489f33g-5.jpg?height=336&width=613&top_left_y=1252&top_left_x=1271)\n\nBut we observe that\n\n$$\nD P=\\frac{B D+D A-A B}{2}, \\quad D Q=\\frac{B D+D C-B C}{2}\n$$\n\nThus $P Q=(b-c-a+2 B D) / 2$. We also have\n\n$$\n\\begin{aligned}\n\\frac{a c}{2}=[A B C]=[A B D]+[C B D]=r^{\\prime} \\frac{(A B+B D+D A)}{2} & +r^{\\prime} \\frac{(C B+B D+D C)}{2} \\\\\n= & r^{\\prime} \\frac{(c+a+b+2 B D)}{2}=r^{\\prime}(s+B D)\n\\end{aligned}\n$$\n\nBut\n\n$$\nr^{\\prime}=\\frac{P Q \\sin \\theta}{2}=\\frac{P Q \\cdot h}{2 B D}\n$$\n\nwhere $h$ is the altitude from $B$ on to $A C$. But we know that $h=a c / b$. Thus we get\n\n$$\na c=2 \\times r^{\\prime}(s+B D)=2 \\times \\frac{P Q \\cdot h}{2 \\times B D}(s+B D)=\\frac{(b-c-a+2 B D) c a(s+B D)}{2 \\times B D \\times b}\n$$\n\nThus we get\n\n$$\n2 \\times B D \\times b=2 \\times(B D-(s-b))(s+B D)\n$$\n\nThis gives $B D^{2}=s(s-b)$. Since $A B C$ is a right-angled triangle $r=s-b$. Thus we get $B D^{2}=r s$. On the other hand, we also have $[A B C]=r^{\\prime}(s+B D)$. Thus we get\n\n$$\nr s=[A B C]=r^{\\prime}(s+B D)\n$$\n\nHence\n\n$$\n\\frac{1}{r^{\\prime}}=\\frac{1}{r}+\\frac{B D}{r s}=\\frac{1}{r}+\\frac{1}{B D}\n$$\n\n## Alternate Solution 1: Observe that\n\n$$\n\\frac{r^{\\prime}}{r}=\\frac{A P}{A X}=\\frac{C Q}{C X}=\\frac{A P+C Q}{A C}\n$$\n\nwhere $X$ is the point at which the incircle of $A B C$ touches the side $A C$. If $s_{1}$ and $s_{2}$ are respectively the semi-perimeters of triangles $A B D$ and $C B D$, we know $A P=s_{1}-B D$ and $C Q=s_{2}-B D$. Therefore\n\n$$\n\\frac{r^{\\prime}}{r}=\\frac{\\left(s_{1}-B D\\right)+\\left(s_{2}-B D\\right)}{A C}=\\frac{s_{1}+s_{2}-2 B D}{b}\n$$\n\nBut\n\n$$\ns_{1}+s_{2}=\\frac{A D+B D+c}{2}+\\frac{C D+B D+a}{2}=\\frac{(a+b+c)+2 B D}{2}=\\frac{s+B D}{2}\n$$\n\nThis gives\n\n$$\n\\frac{r^{\\prime}}{r}=\\frac{s+B D-2 B D}{b}=\\frac{s-B D}{b}\n$$\n\nWe also have\n\n$$\nr^{\\prime}=\\frac{[A B D]}{s_{1}}=\\frac{[C B D]}{s_{2}}=\\frac{[A B D]+[C B D]}{s_{1}+s_{2}}=\\frac{[A B C]}{s+B D}=\\frac{r s}{s+B D}\n$$\n\nThis implies that\n\n$$\n\\frac{r^{\\prime}}{r}=\\frac{s}{s+B D}\n$$\n\nComparing the two expressions for $r^{\\prime} / r$, we see that\n\n$$\n\\frac{s-B D}{b}=\\frac{s}{s+B D}\n$$\n\nTherefore $s^{2}-B D^{2}=b s$, or $B D^{2}=s(s-b)$. Thus we get $B D=\\sqrt{s(s-b)}$.\n\nWe know now that\n\n$$\n\\frac{r^{\\prime}}{r}=\\frac{s}{s+B D}=\\frac{s-B D}{b}=\\frac{B D}{(s-b)+B D}=\\frac{\\sqrt{s(s-b)}}{(s-b)+\\sqrt{s(s-b)}}=\\frac{\\sqrt{s}}{\\sqrt{s-b}+\\sqrt{s}}\n$$\n\nTherefore\n\n$$\n\\frac{r}{r^{\\prime}}=1+\\sqrt{\\frac{s-b}{s}}\n$$\n\nThis gives\n\n$$\n\\frac{1}{r^{\\prime}}=\\frac{1}{r}+\\left(\\sqrt{\\frac{s-b}{s}}\\right) \\frac{1}{r}\n$$\n\nBut\n\n$$\n\\left(\\sqrt{\\frac{s-b}{s}}\\right) \\frac{1}{r}=\\left(\\frac{s-b}{\\sqrt{s(s-b)}}\\right) \\frac{1}{r}=\\left(\\frac{s-b}{B D}\\right) \\frac{1}{r}\n$$\n\nIf $\\angle B=90^{\\circ}$, we know that $r=s-b$. Therfore we get\n\n$$\n\\frac{1}{r^{\\prime}}=\\frac{1}{r}+\\left(\\frac{s-b}{B D}\\right) \\frac{1}{r}=\\frac{1}{r}+\\frac{1}{B D}\n$$\n\nAlternate Solution 2 by a contestant: Observe that $\\angle E D F=90^{\\circ}$. Hence $\\triangle E D P$ is similar to $\\triangle D F Q$. Therefore $D P \\cdot D Q=E P \\cdot F Q$. Taking $D P=y_{1}$ and $D Q=x_{1}$, we get $x_{1} y_{1}=\\left(r^{\\prime}\\right)^{2}$. We also observe that $B D=x_{1}+x_{2}=y_{1}+y_{2}$. Since $\\angle E B F=45^{\\circ}$, we get\n\n$$\n1=\\tan 45^{\\circ}=\\tan \\left(\\beta_{1}+\\beta_{2}\\right)=\\frac{\\tan \\beta_{1}+\\tan \\beta_{2}}{1-\\tan \\beta_{1} \\tan \\beta_{2}}\n$$\n\n![](https://cdn.mathpix.com/cropped/2024_06_05_8eb8bd67d3f7fc489f33g-7.jpg?height=342&width=569&top_left_y=661&top_left_x=1277)\n\nBut $\\tan \\beta_{1}=r^{\\prime} / y_{2}$ and $\\tan \\beta_{2}=r^{\\prime} / x_{2}$. Hence we obtain\n\n$$\n1=\\frac{\\left(r^{\\prime} / y_{2}\\right)+\\left(r^{\\prime} / x_{2}\\right)}{1-\\left(r^{\\prime}\\right)^{2} / x_{2} y_{2}}\n$$\n\nSolving for $r^{\\prime}$, we get\n\n$$\nr^{\\prime}=\\frac{x_{2} y_{2}-x_{1} y_{1}}{x_{2}+y_{2}}\n$$\n\nWe also know\n\n$$\nr=\\frac{A B+B C-A C}{2}=\\frac{x_{2}+y_{2}-\\left(x_{1}+y_{1}\\right)}{2}=\\frac{\\left(x_{2}-x_{1}\\right)+\\left(y_{2}-y_{1}\\right)}{2}\n$$\n\nFinally,\n\n$$\n\\begin{aligned}\n\\frac{1}{r}+\\frac{1}{B D} & =\\frac{2}{\\left(x_{2}-x_{1}\\right)+\\left(y_{2}-y_{1}\\right)}+\\frac{1}{x_{1}+x_{2}} \\\\\n& =\\frac{2 x_{1}+2 x_{2}+\\left(x_{2}-x_{1}\\right)+\\left(y_{2}-y_{1}\\right)}{\\left(x_{1}+x_{2}\\right)\\left(\\left(x_{2}-x_{1}\\right)+\\left(y_{2}-y_{1}\\right)\\right)}\n\\end{aligned}\n$$\n\nBut we can write\n\n$$\n2 x_{1}+2 x_{2}+\\left(x_{2}-x_{1}\\right)+\\left(y_{2}-y_{1}\\right)=\\left(x_{1}+x_{2}+x_{2}-x_{1}\\right)+\\left(y_{1}+y_{2}+y_{2}-y_{1}\\right)=2\\left(x_{2}+y_{2}\\right)\n$$\n\nand\n\n$$\n\\begin{aligned}\n\\left(x_{1}+x_{2}\\right)\\left(\\left(x_{2}-x_{1}\\right)+\\left(y_{2}-y_{1}\\right)\\right)=2\\left(x_{1}+x_{2}\\right) & \\left(x_{2}-y_{1}\\right) \\\\\n& =2\\left(x_{2}\\left(x_{2}+x_{1}-y_{1}\\right)-x_{1} y_{1}\\right)=2\\left(x_{2} y_{2}-x_{1} y_{1}\\right)\n\\end{aligned}\n$$\n\nTherefore\n\n$$\n\\frac{1}{r}+\\frac{1}{B D}=\\frac{2\\left(x_{2}+y_{2}\\right)}{2\\left(x_{2} y_{2}-x_{1} y_{1}\\right)}=\\frac{1}{r^{\\prime}}\n$$\n\nRemark: One can also choose $B=(0,0), A=(0, a)$ and $C=(1,0)$ and the coordinate geometry proof gets reduced considerbly.", "metadata": {"resource_path": "INMO/segmented/en-sol-inmo16.jsonl", "problem_match": "\n5.", "solution_match": "\nSolution:"}}
{"year": "2016", "tier": "T3", "problem_label": "6", "problem_type": null, "exam": "INMO", "problem": "Consider a non-constant arithmetic progression $a_{1}, a_{2}, \\ldots, a_{n}, \\ldots$ Suppose there exist relatively prime positive integers $p>1$ and $q>1$ such that $a_{1}^{2}, a_{p+1}^{2}$ and $a_{q+1}^{2}$ are also the terms of the same arithmetic progression. Prove that the terms of the arithmetic progression are all integers.", "solution": "Let us take $a_{1}=a$. We have\n\n$$\na^{2}=a+k d, \\quad(a+p d)^{2}=a+l d, \\quad(a+q d)^{2}=a+m d\n$$\n\nThus we have\n\n$$\na+l d=(a+p d)^{2}=a^{2}+2 p a d+p^{2} d^{2}=a+k d+2 p a d+p^{2} d^{2}\n$$\n\nSince we have non-constant AP, we see that $d \\neq 0$. Hence we obtain $2 p a+p^{2} d=l-k$. Similarly, we get $2 q a+q^{2} d=m-k$. Observe that $p^{2} q-p q^{2} \\neq 0$. Otherwise $p=q$ and $\\operatorname{gcd}(p, q)=p>1$ which is a contradiction to the given hypothesis that $\\operatorname{gcd}(p, q)=1$. Hence we can solve the two equations for $a, d$ :\n\n$$\na=\\frac{p^{2}(m-k)-q^{2}(l-k)}{2\\left(p^{2} q-p q^{2}\\right)}, \\quad d=\\frac{q(l-k)-p(m-k)}{p^{2} q-p q^{2}}\n$$\n\nIt follows that $a, d$ are rational numbers. We also have\n\n$$\np^{2} a^{2}=p^{2} a+k p^{2} d\n$$\n\nBut $p^{2} d=l-k-2 p a$. Thus we get\n\n$$\np^{2} a^{2}=p^{2} a+k(l-k-2 p a)=(p-2 k) p a+k(l-k)\n$$\n\nThis shows that pa satisfies the equation\n\n$$\nx^{2}-(p-2 k) x-k(l-k)=0\n$$\n\nSince $a$ is rational, we see that $p a$ is rational. Write $p a=w / z$, where $w$ is an integer and $z$ is a natural numbers such that $\\operatorname{gcd}(w, z)=1$. Substituting in the equation, we obtain\n\n$$\nw^{2}-(p-2 k) w z-k(l-k) z^{2}=0\n$$\n\nThis shows $z$ divides $w$. Since $\\operatorname{gcd}(w, z)=1$, it follows that $z=1$ and $p a=w$ an integer. (In fact any rational solution of a monic polynomial with integer coefficients is necessarily an integer.) Similarly, we can prove that $q a$ is an integer. Since $\\operatorname{gcd}(p, q)=1$, there are integers $u$ and $v$ such that $p u+q v=1$. Therefore $a=(p a) u+(q a) v$. It follows that $a$ is an integer.\n\nBut $p^{2} d=l-k-2 p a$. Hence $p^{2} d$ is an integer. Similarly, $q^{2} d$ is also an integer. Since $\\operatorname{gcd}\\left(p^{2}, q^{2}\\right)=1$, it follows that $d$ is an integer. Combining these two, we see that all the terms of the AP are integers.\n\nAlternatively, we can prove that $a$ and $d$ are integers in another way. We have seen that $a$ and $d$ are rationals; and we have three relations:\n\n$$\na^{2}=a+k d, \\quad p^{2} d+2 p a=n_{1}, \\quad q^{2} d+2 q a=n_{2}\n$$\n\nwhere $n_{1}=l-k$ and $n_{2}=m-k$. Let $a=u / v$ and $d=x / y$ where $u, x$ are integers and $v, y$ are natural numbers, and $\\operatorname{gcd}(u, v)=1, \\operatorname{gcd}(x, y)=1$. Putting this in these relations, we obtain\n\n$$\n\\begin{aligned}\nu^{2} y & =u v y+k x v^{2} \\\\\n2 p u y+p^{2} v x & =v y n_{1} \\\\\n2 q u y+q^{2} v x & =v y n_{2}\n\\end{aligned}\n$$\n\nNow (1) shows that $v \\mid u^{2} y$. Since $\\operatorname{gcd}(u, v)=1$, it follows that $v \\mid y$. Similarly (2) shows that $y \\mid p^{2} v x$. Using $\\operatorname{gcd}(y, x)=1$, we get that $y \\mid p^{2} v$. Similarly, (3) shows that $y \\mid q^{2} v$. Therefore $y$ divides $\\operatorname{gcd}\\left(p^{2} v, q^{2} v\\right)=v$. The two results $v \\mid y$ and $y \\mid v$ imply $v=y$, since both $v, y$ are positive.\n\nSubstitute this in (1) to get\n\n$$\nu^{2}=u v+k x v\n$$\n\nThis shows that $v \\mid u^{2}$. Since $\\operatorname{gcd}(u, v)=1$, it follows that $v=1$. This gives $v=y=1$. Finally $a=u$ and $d=x$ which are integers.", "metadata": {"resource_path": "INMO/segmented/en-sol-inmo16.jsonl", "problem_match": "\n6.", "solution_match": "\nSolution:"}}