{"year": "2014", "tier": "T2", "problem_label": "1", "problem_type": null, "exam": "EGMO", "problem": "Determine all real constants $t$ such that whenever $a, b, c$ are the lengths of the sides of a triangle, then so are $a^{2}+b c t, b^{2}+c a t, c^{2}+a b t$.\n\nProposed by S. Khan, UNK\nThe answer is the interval $[2 / 3,2]$.", "solution": "If $t<2 / 3$, take a triangle with sides $c=b=1$ and $a=2-\\epsilon$. Then $b^{2}+c a t+c^{2}+$ $a b t-a^{2}-b c t=3 t-2+\\epsilon(4-2 t-\\epsilon) \\leq 0$ for small positive $\\epsilon$; for instance, for any $0<\\epsilon<(2-3 t) /(4-2 t)$.\n\nOn the other hand, if $t>2$, then take a triangle with sides $b=c=1$ and $a=\\epsilon$. Then $b^{2}+c a t+c^{2}+a b t-a^{2}-b c t=2-t+\\epsilon(2 t-\\epsilon) \\leq 0$ for small positive $\\epsilon$; for instance, for any $0<\\epsilon<(t-2) /(2 t)$.\nNow assume that $2 / 3 \\leq t \\leq 2$ and $b+c>a$. Then using $(b+c)^{2} \\geq 4 b c$ we obtain\n\n$$\n\\begin{aligned}\nb^{2}+c a t+c^{2}+a b t-a^{2}-b c t & =(b+c)^{2}+a t(b+c)-(2+t) b c-a^{2} \\\\\n& \\geq(b+c)^{2}+a t(b+c)-\\frac{1}{4}(2+t)(b+c)^{2}-a^{2} \\\\\n& \\geq \\frac{1}{4}(2-t)(b+c)^{2}+a t(b+c)-a^{2}\n\\end{aligned}\n$$\n\nAs $2-t \\geq 0$ and $t>0$, this last expression is an increasing function of $b+c$, and hence using $b+c>a$ we obtain\n\n$$\nb^{2}+c a t+c^{2}+a b t-a^{2}-b c t>\\frac{1}{4}(2-t) a^{2}+t a^{2}-a^{2}=\\frac{3}{4}\\left(t-\\frac{2}{3}\\right) a^{2} \\geq 0\n$$\n\nas $t \\geq 2 / 3$. The other two inequalities follow by symmetry.", "metadata": {"resource_path": "EGMO/segmented/en-2014-solutions-day1.jsonl", "problem_match": "\n1.", "solution_match": "# Solution 1."}}
{"year": "2014", "tier": "T2", "problem_label": "1", "problem_type": null, "exam": "EGMO", "problem": "Determine all real constants $t$ such that whenever $a, b, c$ are the lengths of the sides of a triangle, then so are $a^{2}+b c t, b^{2}+c a t, c^{2}+a b t$.\n\nProposed by S. Khan, UNK\nThe answer is the interval $[2 / 3,2]$.", "solution": "After showing that $t$ must be in the interval $[2 / 3,2]$ as in Solution 1, we let $x=$ $(c+a-b) / 2, y=(a+b-c) / 2$ and $z=(b+c-a) / 2$ so that $a=x+y, b=y+z$, $c=z+x$. Then we have:\n$b^{2}+c a t+c^{2}+a b t-a^{2}-b c t=\\left(x^{2}+y^{2}-z^{2}+x y+x z+y z\\right) t+2\\left(z^{2}+x z+y z-x y\\right)$\nSince this linear function of $t$ is positive both at $t=2 / 3$ where\n$\\frac{2}{3}\\left(x^{2}+y^{2}-z^{2}+x y+x z+y z\\right)+2\\left(z^{2}+x z+y z-x y\\right)=\\frac{2}{3}\\left((x-y)^{2}+4(x+y) z+2 z^{2}\\right)>0$\nand at $t=2$ where\n$2\\left(x^{2}+y^{2}-z^{2}+x y+x z-y z\\right)+2\\left(z^{2}+x z+y z+x y\\right)=2\\left(x^{2}+y^{2}\\right)+4(x+y) z>0$,\nit is positive on the entire interval $[2 / 3,2]$.", "metadata": {"resource_path": "EGMO/segmented/en-2014-solutions-day1.jsonl", "problem_match": "\n1.", "solution_match": "# Solution 2."}}
{"year": "2014", "tier": "T2", "problem_label": "1", "problem_type": null, "exam": "EGMO", "problem": "Determine all real constants $t$ such that whenever $a, b, c$ are the lengths of the sides of a triangle, then so are $a^{2}+b c t, b^{2}+c a t, c^{2}+a b t$.\n\nProposed by S. Khan, UNK\nThe answer is the interval $[2 / 3,2]$.", "solution": "After the point in Solution 2 where we obtain\n$b^{2}+c a t+c^{2}+a b t-a^{2}-b c t=\\left(x^{2}+y^{2}-z^{2}+x y+x z+y z\\right) t+2\\left(z^{2}+x z+y z-x y\\right)$\nwe observe that the right hand side can be rewritten as\n\n$$\n(2-t) z^{2}+(x-y)^{2} t+(3 t-2) x y+z(x+y)(2+t)\n$$\n\nAs the first three terms are non-negative and the last term is positive, the result follows.", "metadata": {"resource_path": "EGMO/segmented/en-2014-solutions-day1.jsonl", "problem_match": "\n1.", "solution_match": "# Solution 3."}}
{"year": "2014", "tier": "T2", "problem_label": "1", "problem_type": null, "exam": "EGMO", "problem": "Determine all real constants $t$ such that whenever $a, b, c$ are the lengths of the sides of a triangle, then so are $a^{2}+b c t, b^{2}+c a t, c^{2}+a b t$.\n\nProposed by S. Khan, UNK\nThe answer is the interval $[2 / 3,2]$.", "solution": "First we show that $t$ must be in the interval $[2 / 3,2]$ as in Solution 1. Then:\nCase 1: If $a \\geq b, c$, then $a b+a c-b c>0,2\\left(b^{2}+c^{2}\\right) \\geq(b+c)^{2}>a^{2}$ and $t \\geq 2 / 3$ implies:\n\n$$\n\\begin{aligned}\nb^{2}+c a t+c^{2}+a b t-a^{2}-b c t & =b^{2}+c^{2}-a^{2}+(a b+a c-b c) t \\\\\n& \\geq\\left(b^{2}+c^{2}-a^{2}\\right)+\\frac{2}{3}(a b+a c-b c) \\\\\n& \\geq \\frac{1}{3}\\left(3 b^{2}+3 c^{2}-3 a^{2}+2 a b+2 a c-2 b c\\right) \\\\\n& \\geq \\frac{1}{3}\\left[\\left(2 b^{2}+2 c^{2}-a^{2}\\right)+(b-c)^{2}+2 a(b+c-a)\\right] \\\\\n& >0\n\\end{aligned}\n$$\n\nCase 2: If $b \\geq a, c$, then $b^{2}+c^{2}-a^{2}>0$. If also $a b+a c-b c \\geq 0$, then $b^{2}+c a t+$ $c^{2}+a b t-a^{2}-b c t>0$. If, on the other hand, $a b+a c-b c \\leq 0$, then since $t \\leq 2$, we have:\n\n$$\n\\begin{aligned}\nb^{2}+c a t+c^{2}+a b t-a^{2}-b c t & \\geq b^{2}+c^{2}-a^{2}+2(a b+a c-b c) \\\\\n& \\geq(b-c)^{2}+a(b+c-a)+a(b+c) \\\\\n& >0\n\\end{aligned}\n$$\n\nBy symmetry, we are done.", "metadata": {"resource_path": "EGMO/segmented/en-2014-solutions-day1.jsonl", "problem_match": "\n1.", "solution_match": "# Solution 4."}}
{"year": "2014", "tier": "T2", "problem_label": "2", "problem_type": null, "exam": "EGMO", "problem": "Let $D$ and $E$ be two points on the sides $A B$ and $A C$, respectively, of a triangle $A B C$, such that $D B=B C=C E$, and let $F$ be the point of intersection of the lines $C D$ and $B E$. Prove that the incenter $I$ of the triangle $A B C$, the orthocenter $H$ of the triangle $D E F$ and the midpoint $M$ of the $\\operatorname{arc} B A C$ of the circumcircle of the triangle $A B C$ are collinear.\n\nProposed by Danylo Khilko, UKR", "solution": "As $D B=B C=C E$ we have $B I \\perp C D$ and $C I \\perp B E$. Hence $I$ is orthocenter of triangle $B F C$. Let $K$ be the point of intersection of the lines $B I$ and $C D$, and let $L$ be the point of intersection of the lines $C I$ and $B E$. Then we have the power relation $I B \\cdot I K=I C \\cdot I L$. Let $U$ and $V$ be the feet of the perpendiculars from $D$ to $E F$ and $E$ to $D F$, respectively. Now we have the power relation $D H \\cdot H U=E H \\cdot H V$.\n![](https://cdn.mathpix.com/cropped/2024_11_22_d396c315567a32ce5b15g-5.jpg?height=1058&width=1025&top_left_y=908&top_left_x=496)\n\nLet $\\omega_{1}$ and $\\omega_{2}$ be the circles with diameters $B D$ and $C E$, respectively. From the power relations above we conclude that $I H$ is the radical axis of the circles $\\omega_{1}$ and $\\omega_{2}$.\nLet $O_{1}$ and $O_{2}$ be centers of $\\omega_{1}$ and $\\omega_{2}$, respectively. Then $M B=M C, B O_{1}=C O_{2}$ and $\\angle M B O_{1}=\\angle M C O_{2}$, and the triangles $M B O_{1}$ and $M C O_{2}$ are congruent. Hence $M O_{1}=M O_{2}$. Since radii of $\\omega_{1}$ and $\\omega_{2}$ are equal, this implies that $M$ lies on the radical axis of $\\omega_{1}$ and $\\omega_{2}$ and $M, I, H$ are collinear.", "metadata": {"resource_path": "EGMO/segmented/en-2014-solutions-day1.jsonl", "problem_match": "\n2.", "solution_match": "# Solution 1. "}}
{"year": "2014", "tier": "T2", "problem_label": "2", "problem_type": null, "exam": "EGMO", "problem": "Let $D$ and $E$ be two points on the sides $A B$ and $A C$, respectively, of a triangle $A B C$, such that $D B=B C=C E$, and let $F$ be the point of intersection of the lines $C D$ and $B E$. Prove that the incenter $I$ of the triangle $A B C$, the orthocenter $H$ of the triangle $D E F$ and the midpoint $M$ of the $\\operatorname{arc} B A C$ of the circumcircle of the triangle $A B C$ are collinear.\n\nProposed by Danylo Khilko, UKR", "solution": "Let the points $K, L, U, V$ be as in Solution 1. Le $P$ be the point of intersection of $D U$ and $E I$, and let $Q$ be the point of intersection of $E V$ and $D I$.\n\nSince $D B=B C=C E$, the points $C I$ and $B I$ are perpendicular to $B E$ and $C D$, respectively. Hence the lines $B I$ and $E V$ are parallel and $\\angle I E B=\\angle I B E=$ $\\angle U E H$. Similarly, the lines $C I$ and $D U$ are parallel and $\\angle I D C=\\angle I C D=\\angle V D H$. Since $\\angle U E H=\\angle V D H$, the points $D, Q, F, P, E$ are concyclic. Hence $I P \\cdot I E=$ $I Q \\cdot I D$.\n\nLet $R$ be the second point intersection of the circumcircle of triangle $H E P$ and the line $H I$. As $I H \\cdot I R=I P \\cdot I E=I Q \\cdot I D$, the points $D, Q, H, R$ are also concyclic. We have $\\angle D Q H=\\angle E P H=\\angle D F E=\\angle B F C=180^{\\circ}-\\angle B I C=90^{\\circ}-\\angle B A C / 2$. Now using the concylicity of $D, Q, H, R$, and $E, P, H, R$ we obtain $\\angle D R H=$ $\\angle E R H=\\angle 180^{\\circ}-\\left(90^{\\circ}-\\angle B A C / 2\\right)=90^{\\circ}+\\angle B A C / 2$. Hence $R$ is inside the triangle $D E H$ and $\\angle D R E=360^{\\circ}-\\angle D R H-\\angle E R H=180^{\\circ}-\\angle B A C$ and it follows that the points $A, D, R, E$ are concyclic.\n![](https://cdn.mathpix.com/cropped/2024_11_22_d396c315567a32ce5b15g-6.jpg?height=1063&width=1014&top_left_y=1038&top_left_x=512)\n\nAs $M B=M C, B D=C E, \\angle M B D=\\angle M C E$, the triangles $M B D$ and $M C E$ are congruent and $\\angle M D A=\\angle M E A$. Hence the points $M, D, E, A$ are concylic. Therefore the points $M, D, R, E, A$ are concylic. Now we have $\\angle M R E=180^{\\circ}-$ $\\angle M A E=180^{\\circ}-\\left(90^{\\circ}+\\angle B A C / 2\\right)=90^{\\circ}-\\angle B A C / 2$ and since $\\angle E R H=90^{\\circ}+$ $\\angle B A C / 2$, we conclude that the points $I, H, R, M$ are collinear.", "metadata": {"resource_path": "EGMO/segmented/en-2014-solutions-day1.jsonl", "problem_match": "\n2.", "solution_match": "# Solution 2."}}
{"year": "2014", "tier": "T2", "problem_label": "2", "problem_type": null, "exam": "EGMO", "problem": "Let $D$ and $E$ be two points on the sides $A B$ and $A C$, respectively, of a triangle $A B C$, such that $D B=B C=C E$, and let $F$ be the point of intersection of the lines $C D$ and $B E$. Prove that the incenter $I$ of the triangle $A B C$, the orthocenter $H$ of the triangle $D E F$ and the midpoint $M$ of the $\\operatorname{arc} B A C$ of the circumcircle of the triangle $A B C$ are collinear.\n\nProposed by Danylo Khilko, UKR", "solution": "Suppose that we have a coordinate system and $\\left(b_{x}, b_{y}\\right),\\left(c_{x}, c_{y}\\right),\\left(d_{x}, d_{y}\\right),\\left(e_{x}, e_{y}\\right)$ are the coordinates of the points $B, C, D, E$, respectively. From $\\overrightarrow{B I} \\cdot \\overrightarrow{C D}=0, \\overrightarrow{C I} \\cdot \\overrightarrow{B E}=$ $0, \\overrightarrow{E H} \\cdot \\overrightarrow{C D}=0, \\overrightarrow{D H} \\cdot \\overrightarrow{B E}=0$ we obtain $\\overrightarrow{I H} \\cdot(\\vec{B}-\\vec{C}-\\vec{E}+\\vec{D})=0$. Hence the slope of the line $I H$ is $\\left(c_{x}+e_{x}-b_{x}-d_{x}\\right) /\\left(b_{y}+d_{y}-c_{y}-e_{y}\\right)$.\nAssume that the $x$-axis lies along the line $B C$, and let $\\alpha=\\angle B A C, \\beta=\\angle A B C$, $\\theta=\\angle A C B$. Since $D B=B C=C E$, we have $c_{x}-b_{x}=B C, e_{x}-d_{x}=B C-$ $B C \\cos \\beta-B C \\cos \\theta, b_{y}=c_{y}=0, d_{y}-e_{y}=B C \\sin \\beta-B C \\sin \\theta$. Therefore the slope of $I H$ is $(2-\\cos \\beta-\\cos \\theta) /(\\sin \\beta-\\sin \\theta)$.\n\nNow we will show that the slope of the line $M I$ is the same. Let $r$ and $R$ be the inradius and circumradius of the triangle $A B C$, respectively. As $\\angle B M C=$ $\\angle B A C=\\alpha$ and $B M=M C$, we have\n\n$$\nm_{y}-i_{y}=\\frac{B C}{2} \\cot \\left(\\frac{\\alpha}{2}\\right)-r \\text { and } m_{x}-i_{x}=\\frac{A C-A B}{2}\n$$\n\nwhere $\\left(m_{x}, m_{y}\\right)$ and $\\left(i_{x}, i_{y}\\right)$ are the coordinates of $M$ and $I$, respectively. Therefore the slope of $M I$ is $(B C \\cot (\\alpha / 2)-2 r) /(A C-A B)$.\n\nNow the equality of these slopes follows using\n\n$$\n\\frac{B C}{\\sin \\alpha}=\\frac{A C}{\\sin \\beta}=\\frac{A B}{\\sin \\theta}=2 R\n$$\n\nhence\n\n$$\nB C \\cot \\left(\\frac{\\alpha}{2}\\right)=4 R \\cos ^{2}\\left(\\frac{\\alpha}{2}\\right)=2 R(1+\\cos \\alpha)\n$$\n\nand\n\n$$\n\\frac{r}{R}=\\cos \\alpha+\\cos \\beta+\\cos \\theta-1\n$$\n\nas\n\n$$\n\\frac{B C \\cot (\\alpha / 2)-2 r}{A C-A B}=\\frac{2 R(1+\\cos \\alpha)-2 r}{2 R(\\sin \\beta-\\sin \\theta)}=\\frac{2-\\cos \\beta-\\cos \\theta}{\\sin \\beta-\\sin \\theta}\n$$\n\ngiving the collinearity of the points $I, H, M$.", "metadata": {"resource_path": "EGMO/segmented/en-2014-solutions-day1.jsonl", "problem_match": "\n2.", "solution_match": "# Solution 3."}}
{"year": "2014", "tier": "T2", "problem_label": "2", "problem_type": null, "exam": "EGMO", "problem": "Let $D$ and $E$ be two points on the sides $A B$ and $A C$, respectively, of a triangle $A B C$, such that $D B=B C=C E$, and let $F$ be the point of intersection of the lines $C D$ and $B E$. Prove that the incenter $I$ of the triangle $A B C$, the orthocenter $H$ of the triangle $D E F$ and the midpoint $M$ of the $\\operatorname{arc} B A C$ of the circumcircle of the triangle $A B C$ are collinear.\n\nProposed by Danylo Khilko, UKR", "solution": "Let the bisectors $B I$ and $C I$ meet the circumcircle of $A B C$ again at $P$ and $Q$, respectively. Let the altitude of $D E F$ belonging to $D$ meet $B I$ at $R$ and the one belonging to $E$ meet $C I$ at $S$.\n\nSince $B I$ is angle bisector of the iscosceles triangle $C B D, B I$ and $C D$ are perpendicular. Since $E H$ and $D F$ are also perpendicular, $H S$ and $R I$ are parallel. Similarly, $H R$ and $S I$ are parallel, and hence $H S I R$ is a parallelogram.\n\nOn the other hand, as $M$ is the midpoint of the $\\operatorname{arc} B A C$, we have $\\angle M P I=$ $\\angle M P B=\\angle M Q C=\\angle M Q I$, and $\\angle P I Q=(\\widehat{P A}+\\widehat{C B}+\\widehat{A Q}) / 2=(\\widehat{P C}+\\widehat{C B}+$ $\\widehat{B Q}) / 2=\\angle P M Q$. Therefore $M P I Q$ is a parallelogram.\n\nSince $C I$ is angle bisector of the iscosceles triangle $B C E$, the triangle $B S E$ is also isosceles. Hence $\\angle F B S=\\angle E B S=\\angle S E B=\\angle H E F=\\angle H D F=\\angle R D F=$ $\\angle F C S$ and $B, S, F, C$ are concyclic. Similarly, $B, F, R, C$ are concyclic. Therefore $B, S, R, C$ are concyclic. As $B, Q, P, C$ are also concyclic, $S R$ an $Q P$ are parallel.\n\nNow it follows that HSIR and MQIP are homothetic parallelograms, and therefore $M, H, I$ are collinear.\n![](https://cdn.mathpix.com/cropped/2024_11_22_d396c315567a32ce5b15g-8.jpg?height=1196&width=1075&top_left_y=1142&top_left_x=563)", "metadata": {"resource_path": "EGMO/segmented/en-2014-solutions-day1.jsonl", "problem_match": "\n2.", "solution_match": "# Solution 4."}}
{"year": "2014", "tier": "T2", "problem_label": "3", "problem_type": null, "exam": "EGMO", "problem": "We denote the number of positive divisors of a positive integer $m$ by $d(m)$ and the number of distinct prime divisors of $m$ by $\\omega(m)$. Let $k$ be a positive integer. Prove that there exist infinitely many positive integers $n$ such that $\\omega(n)=k$ and $d(n)$ does not divide $d\\left(a^{2}+b^{2}\\right)$ for any positive integers $a, b$ satisfying $a+b=n$.\n\nProposed by JPN", "solution": "We will show that any number of the form $n=2^{p-1} m$ where $m$ is a positive integer that has exactly $k-1$ prime factors all of which are greater than 3 and $p$ is a prime number such that $(5 / 4)^{(p-1) / 2}>m$ satisfies the given condition.\n\nSuppose that $a$ and $b$ are positive integers such that $a+b=n$ and $d(n) \\mid d\\left(a^{2}+b^{2}\\right)$. Then $p \\mid d\\left(a^{2}+b^{2}\\right)$. Hence $a^{2}+b^{2}=q^{c p-1} r$ where $q$ is a prime, $c$ is a positive integer and $r$ is a positive integer not divisible by $q$. If $q \\geq 5$, then\n\n$$\n2^{2 p-2} m^{2}=n^{2}=(a+b)^{2}>a^{2}+b^{2}=q^{c p-1} r \\geq q^{p-1} \\geq 5^{p-1}\n$$\n\ngives a contradiction. So $q$ is 2 or 3 .\nIf $q=3$, then $a^{2}+b^{2}$ is divisible by 3 and this implies that both $a$ and $b$ are divisible by 3 . This means $n=a+b$ is divisible by 3 , a contradiction. Hence $q=2$.\n\nNow we have $a+b=2^{p-1} m$ and $a^{2}+b^{2}=2^{c p-1} r$. If the highest powers of 2 dividing $a$ and $b$ are different, then $a+b=2^{p-1} m$ implies that the smaller one must be $2^{p-1}$ and this makes $2^{2 p-2}$ the highest power of 2 dividing $a^{2}+b^{2}=2^{c p-1} r$, or equivalently, $c p-1=2 p-2$, which is not possible. Therefore $a=2^{t} a_{0}$ and $b=2^{t} b_{0}$ for some positive integer $t<p-1$ and odd integers $a_{0}$ and $b_{0}$. Then $a_{0}^{2}+b_{0}^{2}=2^{c p-1-2 t} r$. The left side of this equality is congruent to 2 modulo 4 , therefore $c p-1-2 t$ must be 1. But then $t<p-1$ gives $(c / 2) p=t+1<p$, which is not possible either.", "metadata": {"resource_path": "EGMO/segmented/en-2014-solutions-day1.jsonl", "problem_match": "\n3.", "solution_match": "# Solution."}}