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

Points $D,E$ lie on segments $AB,AC$ of $\triangle ABC$ such that $DE\parallel BC$. Let $O_1,O_2$ be the circumcenters of $\triangle ABE, \triangle ACD$ respectively. Line $O_1O _2$ meets $AC$ at $P$, and $AB$ at $Q$. Let $O$ be the circumcenter of $\triangle APQ$, and $M$ be the intersection of $AO$ extended and $BC$. Prove that $M$ is the midpoint of $BC$.
Let a triangle $ABC$ be given. Consider the circle touching its circumcircle at $A$ and touching externally its incircle at some point $A_1$. Points $B_1$ and $C_1$ are defined similarly. a) Prove that lines $AA_1, BB_1$ and $CC1$ concur. b) Let $A_2$ be the touching point of the incircle with $BC$. Prove that lines $AA_1$ and $AA_2$ are symmetric about the bisector of angle $\angle A$.
Let $PA, PB$ be tangents to a circle centered at $O$, and $C$ a point on the minor arc $AB$. The perpendicular from $C$ to $PC$ intersects internal angle bisectors of $AOC,BOC$ at $D,E$. Show that $CD=CE$
Let $C_0$ be the midpoint of hypotenuse $AB$ of triangle $ABC$; $AA_1, BB_1$ the bisectors of this triangle; $I$ its incenter. Prove that the lines $C_0I$ and $A_1B_1$ meet on the altitude from $C$. [i]Proposed by A.Zaslavsky[/i]
Circles $Q_1$ and $Q_2$ are tangent to each other externally at $T$. Points $A$ and $E$ are on $Q_1$, lines $AB$ and $DE$ are tangent to $Q_2$ at $B$ and $D$, respectively, lines $AE$ and $BD$ meet at point $P$. Prove that (1) $\frac{AB}{AT}=\frac{ED}{ET}$; (2) $\angle ATP + \angle ETP = 180^{\circ}$. [asy]import graph; size(5.97cm); real lsf=0.5; pathpen=linewidth(0.7); pointpen=black; pen fp=fontsize(10); pointfontpen=fp; real xmin=-6,xmax=5.94,ymin=-3.19,ymax=3.43; pair Q_1=(-2.5,-0.5), T=(-1.5,-0.5), Q_2=(0.5,-0.5), A=(-2.09,0.41), B=(-0.42,1.28), D=(-0.2,-2.37), P=(-0.52,2.96); D(CR(Q_1,1)); D(CR(Q_2,2)); D(A--B); D((-3.13,-1.27)--D); D(P--(-3.13,-1.27)); D(P--D); D(T--(-3.13,-1.27)); D(T--A); D(T--P); D(Q_1); MP("Q_1",(-2.46,-0.44),NE*lsf); D(T); MP("T",(-1.46,-0.44),NE*lsf); D(Q_2); MP("Q_2",(0.54,-0.44),NE*lsf); D(A); MP("A",(-2.22,0.58),NE*lsf); D(B); MP("B",(-0.35,1.45),NE*lsf); D((-3.13,-1.27)); MP("E",(-3.52,-1.62),NE*lsf); D(D); MP("D",(-0.17,-2.31),NE*lsf); D(P); MP("P",(-0.47,3.02),NE*lsf); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy]
Let $\Delta ABC$ be a triangle with circumcenter $O$ and circumcircle $w$. Let $S$ be the center of the circle which is tangent with $AB$, $AC$, and $w$ (in the inside), and let the circle meet $w$ at point $K$. Let the circle with diameter $AS$ meet $w$ at $T$. If $M$ is the midpoint of $BC$, show that $K,T,M,O$ are concyclic.
Let $\triangle ABC$ have circumcenter $O$ and incenter $I$ with $\overline{IA}\perp\overline{OI}$, circumradius $13$, and inradius $6$. Find $AB\cdot AC$.
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 $ABCDE$ be a convex pentagon such that $\angle ABC = \angle AED = 90^\circ$. Suppose that the midpoint of $CD$ is the circumcenter of triangle $ABE$. Let $O$ be the circumcenter of triangle $ACD$. Prove that line $AO$ passes through the midpoint of segment $BE$.
In the triangle $ABC$ let $B'$ and $C'$ be the midpoints of the sides $AC$ and $AB$ respectively and $H$ the foot of the altitude passing through the vertex $A$. Prove that the circumcircles of the triangles $AB'C'$,$BC'H$, and $B'CH$ have a common point $I$ and that the line $HI$ passes through the midpoint of the segment $B'C'.$
Let $[AB]$ be a chord of the circle $\Gamma$ not passing through its center and let $M$ be the midpoint of $[AB].$ Let $C$ be a variable point on $\Gamma$ different from $A$ and $B$ and $P$ be the point of intersection of the tangent lines at $A$ of circumcircle of $CAM$ and at $B$ of circumcircle of $CBM.$ Show that all $CP$ lines pass through a fixed point.
The triangle $ABC$ is given. Let $P_1$ and $P_2$ be points on the side $AB$ such that $P_2$ lies on the segment $BP_1$ and $AP_1 = BP_2$. Similarly, $Q_1$ and $Q_2$ are points on the side $BC$ such that $Q_2$ lies on the segment $BQ_1$ and $BQ_1 = CQ_2$. The segments $P_1Q_2$ and $P_2Q_1$ intersect at the point $R{}$, and the circles $P_1P_2R$ and $Q_1Q_2R$ intersect a second time at the point $S{}$ lying inside the triangle $P_1Q_1R$. Let $M{}$ be the midpoint of the segment $AC$. Prove that the angles $P_1RS$ and $Q_1RM$ are equal.
In acute triangle $ABC$, $AB > AC$. $D$ and $E$ are the midpoints of $AB$, $AC$ respectively. The circumcircle of $ADE$ intersects the circumcircle of $BCE$ again at $P$. The circumcircle of $ADE$ intersects the circumcircle $BCD$ again at $Q$. Prove that $AP = AQ$.
Let $ABCD$ be a trapezoid with bases $ AB$ and $CD$ $(AB>CD)$. Diagonals $AC$ and $BD$ intersect in point $ N$ and lines $AD$ and $BC$ intersect in point $ M$. The circumscribed circles of $ADN$ and $BCN$ intersect in point $ P$, different from point $ N$. Prove that the angles $AMP$ and $BMN$ are equal.
Let $ABC$ be a triangle with $\angle{C} = 90^{\circ}$, and let $H$ be the foot of the altitude from $C$. A point $D$ is chosen inside the triangle $CBH$ so that $CH$ bisects $AD$. Let $P$ be the intersection point of the lines $BD$ and $CH$. Let $\omega$ be the semicircle with diameter $BD$ that meets the segment $CB$ at an interior point. A line through $P$ is tangent to $\omega$ at $Q$. Prove that the lines $CQ$ and $AD$ meet on $\omega$.
Let $P$ be a point interior to triangle $ABC$ (with $CA \neq CB$). The lines $AP$, $BP$ and $CP$ meet again its circumcircle $\Gamma$ at $K$, $L$, respectively $M$. The tangent line at $C$ to $\Gamma$ meets the line $AB$ at $S$. Show that from $SC = SP$ follows $MK = ML$. [i]Proposed by Marcin E. Kuczma, Poland[/i]
Construct a triangle, by straight edge and compass, if the three points where the extensions of the medians intersect the circumcircle of the triangle are given.
Given an acute triangle $ABC$, let $M$ be the midpoint of $BC$ and $H$ the orthocentre. Let $\Gamma$ be the circle with diameter $HM$, and let $X,Y$ be distinct points on $\Gamma$ such that $AX,AY$ are tangent to $\Gamma$. Prove that $BXYC$ is cyclic.
Let $ ABCD$ be a convex quadrilateral. Points $ P,Q$ and $ R$ are the feets of the perpendiculars from point $ D$ to lines $ BC, CA$ and $ AB$, respectively. Prove that $ PQ\equal{}QR$ if and only if the bisectors of the angles $ ABC$ and $ ADC$ meet on segment $ AC$.
In an acute-angled triangle $ABC$, the point $H{}$ is the orthocenter, $M{}$ is the midpoint of the side $BC$ and $\omega$ is the circumcircle. The lines $AH, BH$ and $CH{}$ intersect $\omega$ a second time at points $D, E$ and $F{}$ respectively. The ray $MH$ intersects $\omega$ at point $J{}$. The points $K{}$ and $L{}$ are the centers of the inscribed circles of the triangles $DEJ$ and $DFJ$ respectively. Prove that $KL\parallel BC$.
A scalene triangle $ABC$ is inscribed in a circle $\Gamma$. The bisector of angle $A$ meets $BC$ at $E$. Let $M$ be the midpoint of the arc $BAC$. The line $ME$ intersects $\Gamma$ again at $D$. Show that the circumcentre of triangle $AED$ coincides with the intersection point of the tangent to $\Gamma$ at $D$ and the line $BC$.
Given an acute scalene triangle $ABC$ inscribed in circle $(O)$. Let $H$ be its orthocenter and $M$ be the midpoint of $BC$. Let $D$ lie on the opposite rays of $HA$ so that $BC=2DM$. Let $D'$ be the reflection of $D$ through line $BC$ and $X$ be the intersection of $AO$ and $MD$. a) Show that $AM$ bisects $D'X$. b) Similarly, we define the points $E,F$ like $D$ and $Y,Z$ like $X$. Let $S$ be the intersection of tangent lines from $B,C$ with respect to $(O)$. Let $G$ be the projection of the midpoint of $AS$ to the line $AO$. Show that there exists a point with the same power to all the circles $(BEY),(CFZ),(SGO)$ and $(O)$.
In acute triangle $ABC$,$AC<BC$,$M$ is midpoint of $AB$ and $\Omega$ is it's circumcircle.Let $C^\prime$ be antipode of $C$ in $\Omega$. $AC^\prime$ and $BC^\prime$ intersect with $CM$ at $K,L$,respectively.The perpendicular drawn from $K$ to $AC^\prime$ and perpendicular drawn from $L$ to $BC^\prime$ intersect with $AB$ and each other and form a triangle $\Delta$.Prove that circumcircles of $\Delta$ and $\Omega$ are tangent.(M.Kungozhin)
$ABC$ is an obtuse triangle satisfying $\angle A>90^\circ$, and its circumcenter $O$ and circumcircle $\omega_1$. Let $\omega_2$ be a circle passing $C$ with center $B$. $\omega_2$ meets $BC$ at $D$. $\omega_1$ meets $AD$ and $\omega_2$ at $E$ and $F(\neq C)$, respectively. $AF$ meets $\omega_2$ at $G(\neq F)$. Prove that the intersection of $CE$ and $BG$ lies on the circumcircle of $AOB$.
Let $k$ be a circle with centre $O$. Let $AB$ be a chord of this circle with midpoint $M\neq O$. The tangents of $k$ at the points $A$ and $B$ intersect at $T$. A line goes through $T$ and intersects $k$ in $C$ and $D$ with $CT < DT$ and $BC = BM$. Prove that the circumcentre of the triangle $ADM$ is the reflection of $O$ across the line $AD$.