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: 3882

2003 China Team Selection Test, 1

Triangle $ABC$ is inscribed in circle $O$. Tangent $PD$ is drawn from $A$, $D$ is on ray $BC$, $P$ is on ray $DA$. Line $PU$ ($U \in BD$) intersects circle $O$ at $Q$, $T$, and intersect $AB$ and $AC$ at $R$ and $S$ respectively. Prove that if $QR=ST$, then $PQ=UT$.

2016 Croatia Team Selection Test, Problem 3

Let $ABC$ be an acute triangle with circumcenter $O$. Points $E$ and $F$ are chosen on segments $OB$ and $OC$ such that $BE = OF$. If $M$ is the midpoint of the arc $EOA$ and $N$ is the midpoint of the arc $AOF$, prove that $\sphericalangle ENO + \sphericalangle OMF = 2 \sphericalangle BAC$.

Swiss NMO - geometry, 2016.1

Let $ABC$ be a triangle with $\angle BAC = 60^o$. Let $E$ be the point on the side $BC$ , such that $2 \angle BAE = \angle ACB$ . Let $D$ be the second intersection of $AB$ and the circumcircle of the triangle $AEC$ and $P$ be the second intersection of $CD$ and the circumcircle of the triangle $DBE$. Calculate the angle $\angle BAP$.

1996 Brazil National Olympiad, 4

$ABC$ is acute-angled. $D$ s a variable point on the side BC. $O_1$ is the circumcenter of $ABD$, $O_2$ is the circumcenter of $ACD$, and $O$ is the circumcenter of $AO_1O_2$. Find the locus of $O$.

2001 Baltic Way, 9

Given a rhombus $ABCD$, find the locus of the points $P$ lying inside the rhombus and satisfying $\angle APD+\angle BPC=180^{\circ}$.

2015 Bosnia Herzegovina Team Selection Test, 2

Let $D$ be an arbitrary point on side $AB$ of triangle $ABC$. Circumcircles of triangles $BCD$ and $ACD$ intersect sides $AC$ and $BC$ at points $E$ and $F$, respectively. Perpendicular bisector of $EF$ cuts $AB$ at point $M$, and line perpendicular to $AB$ at $D$ at point $N$. Lines $AB$ and $EF$ intersect at point $T$, and the second point of intersection of circumcircle of triangle $CMD$ and line $TC$ is $U$. Prove that $NC=NU$

2012 Brazil Team Selection Test, 1

Let $ABC$ be an acute triangle. Let $\omega$ be a circle whose centre $L$ lies on the side $BC$. Suppose that $\omega$ is tangent to $AB$ at $B'$ and $AC$ at $C'$. Suppose also that the circumcentre $O$ of triangle $ABC$ lies on the shorter arc $B'C'$ of $\omega$. Prove that the circumcircle of $ABC$ and $\omega$ meet at two points. [i]Proposed by Härmel Nestra, Estonia[/i]

2009 Germany Team Selection Test, 3

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]

1992 India Regional Mathematical Olympiad, 8

The cyclic octagon $ABCDEFGH$ has sides $a,a,a,a,b,b,b,b$ respectively. Find the radius of the circle that circumscribes $ABCDEFGH.$

2011 China Western Mathematical Olympiad, 3

In triangle $ABC$ with $AB>AC$ and incenter $I$, the incircle touches $BC,CA,AB$ at $D,E,F$ respectively. $M$ is the midpoint of $BC$, and the altitude at $A$ meets $BC$ at $H$. Ray $AI$ meets lines $DE$ and $DF$ at $K$ and $L$, respectively. Prove that the points $M,L,H,K$ are concyclic.

2007 Korea National Olympiad, 2

$ ABC$ is a triangle which is not isosceles. Let the circumcenter and orthocenter of $ ABC$ be $ O$, $ H$, respectively, and the altitudes of $ ABC$ be $ AD$, $ BC$, $ CF$. Let $ K\neq A$ be the intersection of $ AD$ and circumcircle of triangle $ ABC$, $ L$ be the intersection of $ OK$ and $ BC$, $ M$ be the midpoint of $ BC$, $ P$ be the intersection of $ AM$ and the line that passes $ L$ and perpendicular to $ BC$, $ Q$ be the intersection of $ AD$ and the line that passes $ P$ and parallel to $ MH$, $ R$ be the intersection of line $ EQ$ and $ AB$, $ S$ be the intersection of $ FD$ and $ BE$. If $ OL \equal{} KL$, then prove that two lines $ OH$ and $ RS$ are orthogonal.

2009 Princeton University Math Competition, 3

A rectangular piece of paper $ABCD$ has sides of lengths $AB = 1$, $BC = 2$. The rectangle is folded in half such that $AD$ coincides with $BC$ and $EF$ is the folding line. Then fold the paper along a line $BM$ such that the corner $A$ falls on line $EF$. How large, in degrees, is $\angle ABM$? [asy] size(180); pathpen = rgb(0,0,0.6)+linewidth(1); pointpen = black+linewidth(3); pointfontpen = fontsize(10); pen dd = rgb(0,0,0.6) + linewidth(0.7) + linetype("4 4"), dr = rgb(0.8,0,0), dg = rgb(0,0.6,0), db = rgb(0,0,0.6)+linewidth(1); pair A=(0,1), B=(0,0), C=(2,0), D=(2,1), E=A/2, F=(2,.5), M=(1/3^.5,1), N=reflect(B,M)*A; D(B--M--D("N",N,NE)--B--D("C",C,SE)--D("D",D,NE)--M); D(D("M",M,plain.N)--D("A",A,NW)--D("B",B,SW),dd); D(D("E",E,W)--D("F",F,plain.E),dd); [/asy]

1999 Bulgaria National Olympiad, 2

The vertices A,B,C of an acute-angled triangle ABC lie on the sides B1C1, C1A1, A1B1 respectively of a triangle A1B1C1 similar to the triangle ABC (∠A = ∠A1, etc.). Prove that the orthocenters of triangles ABC and A1B1C1 are equidistant from the circumcenter of △ABC.

2018 Yasinsky Geometry Olympiad, 4

Let $I_a$ be the point of the center of an ex-circle of the triangle $ABC$, which touches the side $BC$ . Let $W$ be the intersection point of the bisector of the angle $\angle A$ of the triangle $ABC$ with the circumcircle of the triangle $ABC$. Perpendicular from the point $W$ on the straight line $AB$, intersects the circumcircle of $ABC$ at the point $P$. Prove, that if the points $B, P, I_a$ lie on the same line, then the triangle $ABC$ is isosceles. (Mykola Moroz)

1987 IberoAmerican, 3

Let $ABCD$ be a convex quadrilateral and let $P$ and $Q$ be the points on the sides $AD$ and $BC$ respectively such that $\frac{AP}{PD}=\frac{BQ}{QC}=\frac{AB}{CD}$. Prove that the line $PQ$ forms equal angles with the lines $AB$ and $CD$.

2006 Tuymaada Olympiad, 2

Let $ABC$ be a triangle, $G$ it`s centroid, $H$ it`s orthocenter, and $M$ the midpoint of the arc $\widehat{AC}$ (not containing $B$). It is known that $MG=R$, where $R$ is the radius of the circumcircle. Prove that $BG\geq BH$. [i]Proposed by F. Bakharev[/i]

2010 IMO, 4

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]

2013 Romania Team Selection Test, 2

Circles $\Omega $ and $\omega $ are tangent at a point $P$ ($\omega $ lies inside $\Omega $). A chord $AB$ of $\Omega $ is tangent to $\omega $ at $C;$ the line $PC$ meets $\Omega $ again at $Q.$ Chords $QR$ and $QS$ of $ \Omega $ are tangent to $\omega .$ Let $I,X,$ and $Y$ be the incenters of the triangles $APB,$ $ARB,$ and $ASB,$ respectively. Prove that $\angle PXI+\angle PYI=90^{\circ }.$

2012 Middle European Mathematical Olympiad, 5

Let $ K $ be the midpoint of the side $ AB $ of a given triangle $ ABC $. Let $ L $ and $ M$ be points on the sides $ AC $ and $ BC$, respectively, such that $ \angle CLK = \angle KMC $. Prove that the perpendiculars to the sides $ AB, AC, $ and $ BC $ passing through $ K,L, $ and $M$, respectively, are concurrent.

2009 Sharygin Geometry Olympiad, 20

Suppose $ H$ and $ O$ are the orthocenter and the circumcenter of acute triangle $ ABC$; $ AA_1$, $ BB_1$ and $ CC_1$ are the altitudes of the triangle. Point $ C_2$ is the reflection of $ C$ in $ A_1B_1$. Prove that $ H$, $ O$, $ C_1$ and $ C_2$ are concyclic.

2017 Taiwan TST Round 1, 1

Let ${\cal C}_1$ and ${\cal C}_2$ be concentric circles, with ${\cal C}_2$ in the interior of ${\cal C}_1$. From a point $A$ on ${\cal C}_1$ one draws the tangent $AB$ to ${\cal C}_2$ ($B\in {\cal C}_2$). Let $C$ be the second point of intersection of $AB$ and ${\cal C}_1$, and let $D$ be the midpoint of $AB$. A line passing through $A$ intersects ${\cal C}_2$ at $E$ and $F$ in such a way that the perpendicular bisectors of $DE$ and $CF$ intersect at a point $M$ on $AB$. Find, with proof, the ratio $AM/MC$.

2018-2019 Winter SDPC, 7

In triangle $ABC$, let $D$ be on side $BC$. The line through $D$ parallel to $AB,AC$ meet $AC,AB$ at $E,F$, respectively. (a) Show that if $D$ varies on line $BC$, the circumcircle of $AEF$ passes through a fixed point $T$. (b) Show that if $D$ lies on line $AT$, then the circumcircle of $AEF$ is tangent to the circumcircle of $BTC$.

2006 IMO Shortlist, 7

In a triangle $ ABC$, let $ M_{a}$, $ M_{b}$, $ M_{c}$ be the midpoints of the sides $ BC$, $ CA$, $ AB$, respectively, and $ T_{a}$, $ T_{b}$, $ T_{c}$ be the midpoints of the arcs $ BC$, $ CA$, $ AB$ of the circumcircle of $ ABC$, not containing the vertices $ A$, $ B$, $ C$, respectively. For $ i \in \left\{a, b, c\right\}$, let $ w_{i}$ be the circle with $ M_{i}T_{i}$ as diameter. Let $ p_{i}$ be the common external common tangent to the circles $ w_{j}$ and $ w_{k}$ (for all $ \left\{i, j, k\right\}= \left\{a, b, c\right\}$) such that $ w_{i}$ lies on the opposite side of $ p_{i}$ than $ w_{j}$ and $ w_{k}$ do. Prove that the lines $ p_{a}$, $ p_{b}$, $ p_{c}$ form a triangle similar to $ ABC$ and find the ratio of similitude. [i]Proposed by Tomas Jurik, Slovakia[/i]

2020 Bulgaria Team Selection Test, 1

In acute triangle $\triangle ABC$, $BC>AC$, $\Gamma$ is its circumcircle, $D$ is a point on segment $AC$ and $E$ is the intersection of the circle with diameter $CD$ and $\Gamma$. $M$ is the midpoint of $AB$ and $CM$ meets $\Gamma$ again at $Q$. The tangents to $\Gamma$ at $A,B$ meet at $P$, and $H$ is the foot of perpendicular from $P$ to $BQ$. $K$ is a point on line $HQ$ such that $Q$ lies between $H$ and $K$. Prove that $\angle HKP=\angle ACE$ if and only if $\frac{KQ}{QH}=\frac{CD}{DA}$.

1995 Italy TST, 4

In a triangle $ABC$, $P$ and $Q$ are the feet of the altitudes from $B$ and $A$ respectively. Find the locus of the circumcentre of triangle $PQC$, when point $C$ varies (with $A$ and $B$ fixed) in such a way that $\angle ACB$ is equal to $60^{\circ}$.