This website contains problems from math contests. Problems and corresponding tags were obtained from the Art of Problem Solving website.

Tags were heavily modified to better represent problems.

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Found problems: 701

2002 IberoAmerican, 3

Let $P$ be a point in the interior of the equilateral triangle $\triangle ABC$ such that $\sphericalangle{APC}=120^\circ$. Let $M$ be the intersection of $CP$ with $AB$, and $N$ the intersection of $AP$ and $BC$. Find the locus of the circumcentre of the triangle $MBN$ as $P$ varies.

1985 IberoAmerican, 1

Find all the triples of integers $ (a, b,c)$ such that: \[ \begin{array}{ccc}a\plus{}b\plus{}c &\equal{}& 24\\ a^{2}\plus{}b^{2}\plus{}c^{2}&\equal{}& 210\\ abc &\equal{}& 440\end{array}\]

2008 IMO Shortlist, 6

There is given a convex quadrilateral $ ABCD$. Prove that there exists a point $ P$ inside the quadrilateral such that \[ \angle PAB \plus{} \angle PDC \equal{} \angle PBC \plus{} \angle PAD \equal{} \angle PCD \plus{} \angle PBA \equal{} \angle PDA \plus{} \angle PCB = 90^{\circ} \] if and only if the diagonals $ AC$ and $ BD$ are perpendicular. [i]Proposed by Dusan Djukic, Serbia[/i]

2013 Sharygin Geometry Olympiad, 23

Tags: symmetry , geometry
Two convex polytopes $A$ and $B$ do not intersect. The polytope $A$ has exactly $2012$ planes of symmetry. What is the maximal number of symmetry planes of the union of $A$ and $B$, if $B$ has a) $2012$, b) $2013$ symmetry planes? c) What is the answer to the question of p.b), if the symmetry planes are replaced by the symmetry axes?

2006 Bulgaria Team Selection Test, 1

[b]Problem 1.[/b] Points $D$ and $E$ are chosen on the sides $AB$ and $AC$, respectively, of a triangle $\triangle ABC$ such that $DE\parallel BC$. The circumcircle $k$ of triangle $\triangle ADE$ intersects the lines $BE$ and $CD$ at the points $M$ and $N$ (different from $E$ and $D$). The lines $AM$ and $AN$ intersect the side $BC$ at points $P$ and $Q$ such that $BC=2\cdot PQ$ and the point $P$ lies between $B$ and $Q$. Prove that the circle $k$ passes through the point of intersection of the side $BC$ and the angle bisector of $\angle BAC$. [i]Nikolai Nikolov[/i]

2010 IFYM, Sozopol, 3

Tags: geometry , symmetry
Through vertex $C$ of $\Delta ABC$ are constructed lines $l_1$ and $l_2$ which are symmetrical about the angle bisector $CL_c$. Prove that the projections of $A$ and $B$ on lines $l_1$ and $l_2$ lie on one circle.

1978 IMO Shortlist, 2

Two identically oriented equilateral triangles, $ABC$ with center $S$ and $A'B'C$, are given in the plane. We also have $A' \neq S$ and $B' \neq S$. If $M$ is the midpoint of $A'B$ and $N$ the midpoint of $AB'$, prove that the triangles $SB'M$ and $SA'N$ are similar.

2009 Purple Comet Problems, 22

The diagram shows a parabola, a line perpendicular to the parabola's axis of symmetry, and three similar isosceles triangles each with a base on the line and vertex on the parabola. The two smaller triangles are congruent and each have one base vertex on the parabola and one base vertex shared with the larger triangle. The ratio of the height of the larger triangle to the height of the smaller triangles is $\tfrac{a+\sqrt{b}}{c}$ where $a$, $b$, and $c$ are positive integers, and $a$ and $c$ are relatively prime. Find $a + b + c$. [asy] size(200); real f(real x) {return 1.2*exp(2/3*log(16-x^2));} path Q=graph(f,-3.99999,3.99999); path [] P={(-4,0)--(-2,0)--(-3,f(-3))--cycle,(-2,0)--(2,0)--(0,f(0))--cycle,(4,0)--(2,0)--(3,f(3))--cycle}; for(int k=0;k<3;++k) { fill(P[k],grey); draw(P[k]); } draw((-6,0)--(6,0),linewidth(1)); draw(Q,linewidth(1));[/asy]

Kyiv City MO 1984-93 - geometry, 1992.9.3

Tags: symmetry , geometry
Prove that a bounded figure cannot have more than one center of symmetry.

2013 ELMO Shortlist, 9

Let $ABCD$ be a cyclic quadrilateral inscribed in circle $\omega$ whose diagonals meet at $F$. Lines $AB$ and $CD$ meet at $E$. Segment $EF$ intersects $\omega$ at $X$. Lines $BX$ and $CD$ meet at $M$, and lines $CX$ and $AB$ meet at $N$. Prove that $MN$ and $BC$ concur with the tangent to $\omega$ at $X$. [i]Proposed by Allen Liu[/i]

1968 IMO Shortlist, 14

A line in the plane of a triangle $ABC$ intersects the sides $AB$ and $AC$ respectively at points $X$ and $Y$ such that $BX = CY$ . Find the locus of the center of the circumcircle of triangle $XAY .$

1998 AIME Problems, 2

Find the number of ordered pairs $(x,y)$ of positive integers that satisfy $x\le 2y\le 60$ and $y\le 2x\le 60.$

2007 AIME Problems, 3

Square $ABCD$ has side length $13$, and points $E$ and $F$ are exterior to the square such that $BE=DF=5$ and $AE=CF=12$. Find $EF^{2}$. [asy] size(200); defaultpen(fontsize(10)); real x=22.61986495; pair A=(0,26), B=(26,26), C=(26,0), D=origin, E=A+24*dir(x), F=C+24*dir(180+x); draw(B--C--F--D--C^^D--A--E--B--A, linewidth(0.7)); dot(A^^B^^C^^D^^E^^F); pair point=(13,13); label("$A$", A, dir(point--A)); label("$B$", B, dir(point--B)); label("$C$", C, dir(point--C)); label("$D$", D, dir(point--D)); label("$E$", E, dir(point--E)); label("$F$", F, dir(point--F));[/asy]

2004 AMC 10, 23

Circles $A$, $B$, and $C$ are externally tangent to each other and internally tangent to circle $D$. Circles $B$ and $C$ are congruent. Circle $A$ has radius 1 and passes through the center of $D$. What is the radius of circle $B$? [asy] size(200); defaultpen(linewidth(0.8));defaultpen(fontsize(8)); draw(Circle(origin, 2)); draw(Circle((-1,0), 1)); draw(Circle((6/9, 8/9), 8/9)); draw(Circle((6/9, -8/9), 8/9)); label("$A$", (-1.2, -0.2), NE); label("$B$", (6/9, 7/9), N); label("$C$", (6/9, -7/9), S); label("$D$", 2*dir(110), dir(110));[/asy] $ \textbf{(A)}\; \frac23\qquad \textbf{(B)}\; \frac{\sqrt{3}}2\qquad \textbf{(C)}\; \frac78\qquad \textbf{(D)}\; \frac89\qquad \textbf{(E)}\; \frac{1+\sqrt3}3 $

2015 AMC 10, 19

The isosceles right triangle $ABC$ has right angle at $C$ and area $12.5$. The rays trisecting $\angle{ACB}$ intersect $AB$ at $D$ and $E$. What is the area of $\triangle{CDE}$? $\textbf{(A) }\frac{5\sqrt{2}}{3}\qquad\textbf{(B) }\frac{50\sqrt{3}-75}{4}\qquad\textbf{(C) }\frac{15\sqrt{3}}{8}\qquad\textbf{(D) }\frac{50-25\sqrt{3}}{2}\qquad\textbf{(E) }\frac{25}{6}$

2014 Baltic Way, 12

Triangle $ABC$ is given. Let $M$ be the midpoint of the segment $AB$ and $T$ be the midpoint of the arc $BC$ not containing $A$ of the circumcircle of $ABC.$ The point $K$ inside the triangle $ABC$ is such that $MATK$ is an isosceles trapezoid with $AT\parallel MK.$ Show that $AK = KC.$

1998 IMO Shortlist, 4

For any two nonnegative integers $n$ and $k$ satisfying $n\geq k$, we define the number $c(n,k)$ as follows: - $c\left(n,0\right)=c\left(n,n\right)=1$ for all $n\geq 0$; - $c\left(n+1,k\right)=2^{k}c\left(n,k\right)+c\left(n,k-1\right)$ for $n\geq k\geq 1$. Prove that $c\left(n,k\right)=c\left(n,n-k\right)$ for all $n\geq k\geq 0$.

2012 AMC 10, 25

A bug travels from $A$ to $B$ along the segments in the hexagonal lattice pictured below. The segments marked with an arrow can be traveled only in the direction of the arrow, and the bug never travels the same segment more than once. How many different paths are there? [asy] size(10cm); draw((0.0,0.0)--(1.0,1.7320508075688772)--(3.0,1.7320508075688772)--(4.0,3.4641016151377544)--(6.0,3.4641016151377544)--(7.0,5.196152422706632)--(9.0,5.196152422706632)--(10.0,6.928203230275509)--(12.0,6.928203230275509)); draw((0.0,0.0)--(1.0,1.7320508075688772)--(3.0,1.7320508075688772)--(4.0,3.4641016151377544)--(6.0,3.4641016151377544)--(7.0,5.196152422706632)--(9.0,5.196152422706632)--(10.0,6.928203230275509)--(12.0,6.928203230275509)); draw((3.0,-1.7320508075688772)--(4.0,0.0)--(6.0,0.0)--(7.0,1.7320508075688772)--(9.0,1.7320508075688772)--(10.0,3.4641016151377544)--(12.0,3.464101615137755)--(13.0,5.196152422706632)--(15.0,5.196152422706632)); draw((6.0,-3.4641016151377544)--(7.0,-1.7320508075688772)--(9.0,-1.7320508075688772)--(10.0,0.0)--(12.0,0.0)--(13.0,1.7320508075688772)--(15.0,1.7320508075688776)--(16.0,3.464101615137755)--(18.0,3.4641016151377544)); draw((9.0,-5.196152422706632)--(10.0,-3.464101615137755)--(12.0,-3.464101615137755)--(13.0,-1.7320508075688776)--(15.0,-1.7320508075688776)--(16.0,0)--(18.0,0.0)--(19.0,1.7320508075688772)--(21.0,1.7320508075688767)); draw((12.0,-6.928203230275509)--(13.0,-5.196152422706632)--(15.0,-5.196152422706632)--(16.0,-3.464101615137755)--(18.0,-3.4641016151377544)--(19.0,-1.7320508075688772)--(21.0,-1.7320508075688767)--(22.0,0)); draw((0.0,-0.0)--(1.0,-1.7320508075688772)--(3.0,-1.7320508075688772)--(4.0,-3.4641016151377544)--(6.0,-3.4641016151377544)--(7.0,-5.196152422706632)--(9.0,-5.196152422706632)--(10.0,-6.928203230275509)--(12.0,-6.928203230275509)); draw((3.0,1.7320508075688772)--(4.0,-0.0)--(6.0,-0.0)--(7.0,-1.7320508075688772)--(9.0,-1.7320508075688772)--(10.0,-3.4641016151377544)--(12.0,-3.464101615137755)--(13.0,-5.196152422706632)--(15.0,-5.196152422706632)); draw((6.0,3.4641016151377544)--(7.0,1.7320508075688772)--(9.0,1.7320508075688772)--(10.0,-0.0)--(12.0,-0.0)--(13.0,-1.7320508075688772)--(15.0,-1.7320508075688776)--(16.0,-3.464101615137755)--(18.0,-3.4641016151377544)); draw((9.0,5.1961524)--(10.0,3.464101)--(12.0,3.46410)--(13.0,1.73205)--(15.0,1.732050)--(16.0,0)--(18.0,-0.0)--(19.0,-1.7320)--(21.0,-1.73205080)); draw((12.0,6.928203)--(13.0,5.1961524)--(15.0,5.1961524)--(16.0,3.464101615)--(18.0,3.4641016)--(19.0,1.7320508)--(21.0,1.732050)--(22.0,0)); dot((0,0)); dot((22,0)); label("$A$",(0,0),WNW); label("$B$",(22,0),E); filldraw((2.0,1.7320508075688772)--(1.6,1.2320508075688772)--(1.75,1.7320508075688772)--(1.6,2.232050807568877)--cycle,black); filldraw((5.0,3.4641016151377544)--(4.6,2.9641016151377544)--(4.75,3.4641016151377544)--(4.6,3.9641016151377544)--cycle,black); filldraw((8.0,5.196152422706632)--(7.6,4.696152422706632)--(7.75,5.196152422706632)--(7.6,5.696152422706632)--cycle,black); filldraw((11.0,6.928203230275509)--(10.6,6.428203230275509)--(10.75,6.928203230275509)--(10.6,7.428203230275509)--cycle,black); filldraw((4.6,0.0)--(5.0,-0.5)--(4.85,0.0)--(5.0,0.5)--cycle,white); filldraw((8.0,1.732050)--(7.6,1.2320)--(7.75,1.73205)--(7.6,2.2320)--cycle,black); filldraw((11.0,3.4641016)--(10.6,2.9641016)--(10.75,3.46410161)--(10.6,3.964101)--cycle,black); filldraw((14.0,5.196152422706632)--(13.6,4.696152422706632)--(13.75,5.196152422706632)--(13.6,5.696152422706632)--cycle,black); filldraw((8.0,-1.732050)--(7.6,-2.232050)--(7.75,-1.7320508)--(7.6,-1.2320)--cycle,black); filldraw((10.6,0.0)--(11,-0.5)--(10.85,0.0)--(11,0.5)--cycle,white); filldraw((14.0,1.7320508075688772)--(13.6,1.2320508075688772)--(13.75,1.7320508075688772)--(13.6,2.232050807568877)--cycle,black); filldraw((17.0,3.464101615137755)--(16.6,2.964101615137755)--(16.75,3.464101615137755)--(16.6,3.964101615137755)--cycle,black); filldraw((11.0,-3.464101615137755)--(10.6,-3.964101615137755)--(10.75,-3.464101615137755)--(10.6,-2.964101615137755)--cycle,black); filldraw((14.0,-1.7320508075688776)--(13.6,-2.2320508075688776)--(13.75,-1.7320508075688776)--(13.6,-1.2320508075688776)--cycle,black); filldraw((16.6,0)--(17,-0.5)--(16.85,0)--(17,0.5)--cycle,white); filldraw((20.0,1.7320508075688772)--(19.6,1.2320508075688772)--(19.75,1.7320508075688772)--(19.6,2.232050807568877)--cycle,black); filldraw((14.0,-5.196152422706632)--(13.6,-5.696152422706632)--(13.75,-5.196152422706632)--(13.6,-4.696152422706632)--cycle,black); filldraw((17.0,-3.464101615137755)--(16.6,-3.964101615137755)--(16.75,-3.464101615137755)--(16.6,-2.964101615137755)--cycle,black); filldraw((20.0,-1.7320508075688772)--(19.6,-2.232050807568877)--(19.75,-1.7320508075688772)--(19.6,-1.2320508075688772)--cycle,black); filldraw((2.0,-1.7320508075688772)--(1.6,-1.2320508075688772)--(1.75,-1.7320508075688772)--(1.6,-2.232050807568877)--cycle,black); filldraw((5.0,-3.4641016)--(4.6,-2.964101)--(4.75,-3.4641)--(4.6,-3.9641016)--cycle,black); filldraw((8.0,-5.1961524)--(7.6,-4.6961524)--(7.75,-5.19615242)--(7.6,-5.696152422)--cycle,black); filldraw((11.0,-6.9282032)--(10.6,-6.4282032)--(10.75,-6.928203)--(10.6,-7.428203)--cycle,black);[/asy] $ \textbf{(A)}\ 2112\qquad\textbf{(B)}\ 2304\qquad\textbf{(C)}\ 2368\qquad\textbf{(D)}\ 2384\qquad\textbf{(E)}\ 2400 $

1996 AMC 12/AHSME, 30

A hexagon inscribed in a circle has three consecutive sides each of length $3$ and three consecutive sides each of length $5$. The chord of the circle that divides the hexagon into two trapezoids, one with three sides each of length $3$ and the other with three sides each of length $5$, has length equal to $\frac mn$, where $m$ and $n$ are relatively prime positive integers. Find $m + n$. $\text{(A)}\ 309 \qquad \text{(B)}\ 349 \qquad \text{(C)}\ 369 \qquad \text{(D)}\ 389\qquad \text{(E)}\ 409$

1991 Arnold's Trivium, 11

Investigate the convergence of the integral \[\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\frac{dxdy}{1+x^4y^4}\]

2010 Germany Team Selection Test, 1

Let $ABC$ be a triangle. The incircle of $ABC$ touches the sides $AB$ and $AC$ at the points $Z$ and $Y$, respectively. Let $G$ be the point where the lines $BY$ and $CZ$ meet, and let $R$ and $S$ be points such that the two quadrilaterals $BCYR$ and $BCSZ$ are parallelogram. Prove that $GR=GS$. [i]Proposed by Hossein Karke Abadi, Iran[/i]

2012 AMC 10, 11

Tags: symmetry , rotation
A dessert chef prepares the dessert for every day of a week starting with Sunday. The dessert each day is either cake, pie, ice cream, or pudding. The same dessert may not be served two days in a row. There must be cake on Friday because of a birthday. How many different dessert menus for the week are possible? $ \textbf{(A)}\ 729\qquad\textbf{(B)}\ 972\qquad\textbf{(C)}\ 1024\qquad\textbf{(D)}\ 2187\qquad\textbf{(E)}\ 2304 $

2011 Macedonia National Olympiad, 2

Acute-angled $~$ $\triangle{ABC}$ $~$ is given. A line $~$ $l$ $~$ parallel to side $~$ $AB$ $~$ passing through vertex $~$ $C$ $~$ is drawn. Let the angle bisectors of $~$ $\angle{BAC}$ $~$ and $~$ $\angle{ABC}$ $~$ intersect the sides $~$ $BC$ and $~$ $AC$ at points $~$ $D$ $~$ and $~$ $F$, and line $~$ $l$ $~$ at points $~$ $E$ $~$ and $~$ $G$ $~$ respectively. Prove that if $~$ $\overline{DE}=\overline{GF}$ $~$ then $~$ $\overline{AC}=\overline{BC}\, .$

1997 Dutch Mathematical Olympiad, 5

Given is a triangle $ABC$ and a point $K$ within the triangle. The point $K$ is mirrored in the sides of the triangle: $P , Q$ and $R$ are the mirrorings of $K$ in $AB , BC$ and $CA$, respectively . $M$ is the center of the circle passing through the vertices of triangle $PQR$. $M$ is mirrored again in the sides of triangle $ABC$: $P', Q'$ and $R'$ are the mirror of $M$ in $AB$ respectively, $BC$ and $CA$. a. Prove that $K$ is the center of the circle passing through the vertices of triangle $P'Q'R'$ . b. Where should you choose $K$ within triangle $ABC$ so that $M$ and $K$ coincide? Prove your answer.

2020 Ukrainian Geometry Olympiad - April, 2

Let $\Gamma$ be a circle and $P$ be a point outside, $PA$ and $PB$ be tangents to $\Gamma$ , $A, B \in \Gamma$ . Point $K$ is an arbitrary point on the segment $AB$. The circumscirbed circle of $\vartriangle PKB$ intersects $\Gamma$ for the second time at point $T$, point $P'$ is symmetric to point $P$ wrt point $A$. Prove that $\angle PBT = \angle P'KA$.