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: 847

Let $ABC$ be a triangle, and $M_a, M_b, M_c$ be the midpoints of $BC, CA, AB$, respectively. Extend $M_bM_c$ so that it intersects $\odot (ABC)$ at $P$. Let $AP$ and $BC$ intersect at $Q$. Prove that the tangent at $A$ to $\odot(ABC)$ and the tangent at $P$ to $\odot (P QM_a)$ intersect on line $BC$. (Li4)
Let $ABCD$ be an isosceles trapezoid with $AB\parallel CD$. Let $E$ be the midpoint of $AC$. Denote by $\omega$ and $\Omega$ the circumcircles of the triangles $ABE$ and $CDE$, respectively. Let $P$ be the crossing point of the tangent to $\omega$ at $A$ with the tangent to $\Omega$ at $D$. Prove that $PE$ is tangent to $\Omega$. [i]Jakob Jurij Snoj, Slovenia[/i]
Let $\triangle ABC$ be an acute-angled triangle with $AB>AC$, $O$ be its circumcenter and $D$ the midpoint of side $BC$. The circle with diameter $AD$ meets sides $AB,AC$ again at points $E,F$ respectively. The line passing through $D$ parallel to $AO$ meets $EF$ at $M$. Show that $EM=MF$.
Given a non-isosceles triangle $ABC$, let $D,E$, and $F$ denote the midpoints of the sides $BC,CA$, and $AB$ respectively. The circle $BCF$ and the line $BE$ meet again at $P$, and the circle $ABE$ and the line $AD$ meet again at $Q$. Finally, the lines $DP$ and $FQ$ meet at $R$. Prove that the centroid $G$ of the triangle $ABC$ lies on the circle $PQR$. [i](United Kingdom) David Monk[/i]
The midpoints of the sides $BC$, $CA$ and $AB$ of triangle $ABC$ are $M$, $N$ and $P$ respectively . $G$ is the intersection point of the medians. The circumscribed circle around $BGP$ intersects the line $MP$ at the point $K$ (different than $P$).The circle circumscribed around $CGN$ intersects the line $MN$ at point $L$ (different than $N$). Prove that $\angle BAK = \angle CAL$.
Let $ABCD$ be a cyclic quadrilateral with circumcircle $\odot O.$ The lines tangent to $\odot O$ at $A,B$ intersect at $L.$ $M$ is the midpoint of the segment $AB.$ The line passing through $D$ and parallel to $CM$ intersects $ \odot (CDL) $ at $F.$ Line $CF$ intersects $DM$ at $K,$ and intersects $\odot O$ at $E$ (different from point $C$). Prove that $EK=DK.$
Let $O$ be the circumcenter of an acute-angled triangle $ABC$, let $T$ be the circumcenter of the triangle $AOC$, and let $M$ be the midpoint of the segment $AC$. We take a point $D$ on the side $AB$ and a point $E$ on the side $BC$ that satisfy $\angle BDM = \angle BEM = \angle ABC$. Show that the straight lines $BT$ and $DE$ are perpendicular.
Two circles in the plane, $\gamma_1$ and $\gamma_2$, meet at points $M$ and $N$. Let $A$ be a point on $\gamma_1$, and let $D$ be a point on $\gamma_2$. The lines $AM$ and $AN$ meet again $\gamma_2$ at points $B$ and $C$, respectively, and the lines $DM$ and $DN$ meet again $\gamma_1$ at points $E$ and $F$, respectively. Assume the order $M$, $N$, $F$, $A$, $E$ is circular around $\gamma_1$, and the segments $AB$ and $DE$ are congruent. Prove that the points $A$, $F$, $C$ and $D$ lie on a circle whose centre does not depend on the position of the points $A$ and $D$ on the respective circles, subject to the assumptions above. [i]***[/i]
Let $ABC$ be an acute triangle with circumcenter $O$ such that $AB=4$, $AC=5$, and $BC=6$. Let $D$ be the foot of the altitude from $A$ to $BC$, and $E$ be the intersection of $AO$ with $BC$. Suppose that $X$ is on $BC$ between $D$ and $E$ such that there is a point $Y$ on $AD$ satisfying $XY\parallel AO$ and $YO\perp AX$. Determine the length of $BX$.
Let $ABC$ be an acute triangle with orthocenter $H$ and circumcircle $\Gamma$. Let $BH$ intersect $AC$ at $E$, and let $CH$ intersect $AB$ at $F$. Let $AH$ intersect $\Gamma$ again at $P \neq A$. Let $PE$ intersect $\Gamma$ again at $Q \neq P$. Prove that $BQ$ bisects segment $\overline{EF}$. [i]Proposed by Luke Robitaille[/i]
Let $\triangle ABC$ be an acute triangle with circumcenter $O$ such that $AB<AC$, let $Q$ be the intersection of the external bisector of $\angle A$ with $BC$, and let $P$ be a point in the interior of $\triangle ABC$ such that $\triangle BPA$ is similar to $\triangle APC$. Show that $\angle QPA + \angle OQB = 90^{\circ}$. [i]Alex Zhu.[/i]
Let $ABC$ be an acute triangle. Let $M$ be the midpoint of side $BC$, and let $E$ and $F$ be the feet of the altitudes from $B$ and $C$, respectively. Suppose that the common external tangents to the circumcircles of triangles $BME$ and $CMF$ intersect at a point $K$, and that $K$ lies on the circumcircle of $ABC$. Prove that line $AK$ is perpendicular to line $BC$. [i]Kevin Cong[/i]
Let $\omega$ be the incircle of triangle $ABC$ and $\omega$ touches $BC$ at $D$. $AD$ meets $\omega$ again at $L$. Let $K$ be $A$-excenter, and $M,N$ be the midpoint of $BC,KM$, respectively. Prove that $B,C,N,L$ are concyclic.
Let $ABC$ be a triangle ($\angle A\neq 90^\circ$). $BE,CF$ are the altitudes of the triangle. The bisector of $\angle A$ intersects $EF,BC$ at $M,N$. Let $P$ be a point such that $MP\perp EF$ and $NP\perp BC$. Prove that $AP$ passes through the midpoint of $BC$. [i]Proposed by Iman Maghsoudi, Hooman Fattahi[/i]
Let $\omega$ be the circumcircle of acute triangle $ABC$. Two tangents of $\omega$ from $B$ and $C$ intersect at $P$, $AP$ and $BC$ intersect at $D$. Point $E$, $F$ are on $AC$ and $AB$ such that $DE \parallel BA$ and $DF \parallel CA$. (1) Prove that $F,B,C,E$ are concyclic. (2) Denote $A_{1}$ the centre of the circle passing through $F,B,C,E$. $B_{1}$, $C_{1}$ are difined similarly. Prove that $AA_{1}$, $BB_{1}$, $CC_{1}$ are concurrent.
We are given an acute triangle $ABC$ with $AB\neq AC$. Let $D$ be a point of $BC$ such that $DA$ is tangent to the circumcircle of $ABC$. Let $E$ and $F$ be the circumcenters of triangles $ABD$ and $ACD$, respectively, and let $M$ be the midpoints $EF$. Prove that the line tangent to the circumcircle of $AMD$ through $D$ is also tangent to the circumcircle of $ABC$. [i]Proposed by Patrik Bak, Slovakia[/i]
Point $S$ is the midpoint of arc $ACB$ of the circumscribed circle $k$ around triangle $ABC$ with $AC>BC$. Let $I$ be the incenter of triangle $ABC$. Line $SI$ intersects $k$ again at point $T$. Let $D$ be the reflection of $I$ across $T$ and $M$ be the midpoint of side $AB$. Line $IM$ intersects the line through $D$, parallel to $AB$, at point $E$. Prove that $AE=BD$.
The midpoint $M$ of $[AC]$ of a triangle $\triangle ABC$ is between $C$ and the feet $H$ of the altitude from $B$. If $m(\widehat{ABH}) = m(\widehat{MBC})$, $m(\widehat{ACB}) = 15^{\circ}$, and $|HM|=2\sqrt{3}$, then $|AC|=?$ $ \textbf{(A)}\ 6 \qquad \textbf{(B)}\ 5 \sqrt 2 \qquad \textbf{(C)}\ 8 \qquad \textbf{(D)}\ \frac{16}{\sqrt3} \qquad \textbf{(E)}\ 10$
Let $ ABC$ be a triangle, let $ AB > AC$. Its incircle touches side $ BC$ at point $ E$. Point $ D$ is the second intersection of the incircle with segment $ AE$ (different from $ E$). Point $ F$ (different from $ E$) is taken on segment $ AE$ such that $ CE \equal{} CF$. The ray $ CF$ meets $ BD$ at point $ G$. Show that $ CF \equal{} FG$.
In a triangle $ABC$, with $AB \ne BC$, $E$ is a point on the line $AC$ such that $BE$ is perpendicular to $AC$. A circle passing through $A$ and touching the line $BE$ at a point $P \ne B$ intersects the line $AB$ for the second time at $X$. Let $Q$ be a point on the line $PB$ different from $P$ such that $BQ = BP$. Let $Y$ be the point of intersection of the lines $CP$ and $AQ$. Prove that the points $C, X, Y, A$ are concyclic if and only if $CX$ is perpendicular to $AB$.
Let $A$, $B$, and $C$ be three points such that $B$ is the midpoint of segment $AC$ and let $P$ be a point such that $<PBC=60$. Equilateral triangle $PCQ$ is constructed such that $B$ and $Q$ are on different half=planes with respect to $PC$, and the equilateral triangle $APR$ is constructed in such a way that $B$ and $R$ are in the same half-plane with respect to $AP$. Let $X$ be the point of intersection of the lines $BQ$ and $PC$, and let $Y$ be the point of intersection of the lines $BR$ and $AP$. Prove that $XY$ and $AC$ are parallel.
Let $\Gamma$ be a circle and $A$ a point outside $\Gamma$. The tangent lines to $\Gamma$ through $A$ touch $\Gamma$ at $B$ and $C$. Let $M$ be the midpoint of $AB$. The segment $MC$ meets $\Gamma$ again at $D$ and the line $AD$ meets $\Gamma$ again at $E$. Given that $AB=a$, $BC=b$, compute $CE$ in terms of $a$ and $b$.
Let $ABCD$ be a cyclic quadrilateral satisfying $AD^2 + BC^2 = AB^2$. The diagonals of $ABCD$ intersect at $E$. Let $P$ be a point on side $\overline{AB}$ satisfying $\angle APD = \angle BPC$. Show that line $PE$ bisects $\overline{CD}$. [i]Proposed by Ankan Bhattacharya[/i]
Triangle $ABC$ has sides $AB = 25$, $BC = 30$, and $CA=20$. Let $P,Q$ be the points on segments $AB,AC$, respectively, such that $AP=5$ and $AQ=4$. Suppose lines $BQ$ and $CP$ intersect at $R$ and the circumcircles of $\triangle{BPR}$ and $\triangle{CQR}$ intersect at a second point $S\ne R$. If the length of segment $SA$ can be expressed in the form $\frac{m}{\sqrt{n}}$ for positive integers $m,n$, where $n$ is not divisible by the square of any prime, find $m+n$. [i]Victor Wang[/i]
Let $ABC$ be an acute triangle, $B,C$ fixed, $A$ moves on the big arc $BC$ of $(ABC)$. Let $O$ be the circumcenter of $(ABC)$ $(B,O,C$ are not collinear, $AB \ne AC)$, $(I)$ is the incircle of triangle $ABC$. $(I)$ tangents to $BC$ at $D$. Let $I_a$ be the $A$-excenter of triangle $ABC$. $I_aD$ cuts $OI$ at $L$. Let $E$ lies on $(I)$ such that $DE \parallel AI$. a) $LE$ cuts $AI$ at $F$. Prove that $AF=AI$. b) Let $M$ lies on the circle $(J)$ go through $I_a,B,C$ such that $I_aM \parallel AD$. $MD$ cuts $(J)$ again at $N$. Prove that the midpoint $T$ of $MN$ lies on a fixed circle.