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: 509

2008 Tournament Of Towns, 2

A line parallel to the side $AC$ of triangle $ABC$ cuts the side $AB$ at $K$ and the side $BC$ at $M$. $O$ is the intersection point of $AM$ and $CK$. If $AK = AO$ and $KM = MC$, prove that $AM = KB$.

2019 Regional Olympiad of Mexico Center Zone, 3

Let $ABC$ be an acute triangle and $D$ a point on the side $BC$ such that $\angle BAD = \angle DAC$. The circumcircles of the triangles $ABD$ and $ACD$ intersect the segments $AC$ and $AB$ at $E$ and $F$, respectively. The internal bisectors of $\angle BDF$ and $\angle CDE$ intersect the sides $AB$ and $AC$ at $P$ and $Q$, respectively. Points $X$ and $Y$ are chosen on the side $BC$ such that $PX$ is parallel to $AC$ and $QY$ is parallel to $AB$. Finally, let $Z$ be the point of intersection of $BE$ and $CF$. Prove that $ZX = ZY$.

1957 Moscow Mathematical Olympiad, 354

In a quadrilateral $ABCD$ points $M$ and $N$ are the midpoints of the diagonals $AC$ and $BD$, respectively. The line through $M$ and $N$ meets $AB$ and $CD$ at $M'$ and $N'$, respectively. Prove that if $MM' = NN'$, then $AD // BC$.

2020 Puerto Rico Team Selection Test, 3

The side $BC$ of the triangle $ABC$ is extended beyond $C$ to $D$, such that $CD=BC$. The side $CA$ is extended beyond $A$ to $E$, such that $AE=2CA$. Prove that if $AD=BE$, then the triangle $ABC$ is right.

2013 Argentina National Olympiad Level 2, 2

Let $ABC$ be a right triangle. It is known that there are points $D$ on the side $AC$ and $E$ on the side $BC$ such that $AB = AD = BE$ and $BD$ is perpendicular to $DE$. Calculate the ratios $\frac{AB}{BC}$ and $\frac{BC}{CA}$.

1973 All Soviet Union Mathematical Olympiad, 177

Given an angle with the vertex $O$ and a circle touching its sides in the points $A$ and $B$. A ray is drawn from the point $A$ parallel to $[OB)$. It intersects with the circumference in the point $C$. The segment $[OC]$ intersects the circumference in the point $E$. The straight lines $(AE)$ and $(OB)$ intersect in the point $K$. Prove that $|OK| = |KB|$.

Swiss NMO - geometry, 2006.7

Let $ABCD$ be a cyclic quadrilateral with $\angle ABC = 60^o$ and $| BC | = | CD |$. Prove that $|CD| + |DA| = |AB|$

Champions Tournament Seniors - geometry, 2001.4

Given a convex pentagon $ABCDE$ in which $\angle ABC = \angle AED = 90^o$, $\angle BAC= \angle DAE$. Let $K$ be the midpoint of the side $CD$, and $P$ the intersection point of lines $AD$ and $BK$, $Q$ be the intersection point of lines $AC$ and $EK$. Prove that $BQ = PE$.

2001 Rioplatense Mathematical Olympiad, Level 3, 2

Let $ABC$ be an acute triangle and $A_1, B_1$ and $C_1$, points on the sides $BC, CA$ and $AB$, respectively, such that $CB_1 = A_1B_1$ and $BC_1 = A_1C_1$. Let $D$ be the symmetric of $A_1$ with respect to $B_1C_1, O$ and $O_1$ are the circumcenters of triangles $ABC$ and $A_1B_1C_1$, respectively. If $A \ne D, O \ne O_1$ and $AD$ is perpendicular to $OO_1$, prove that $AB = AC$.

Kyiv City MO Juniors Round2 2010+ geometry, 2020.8.2

Given a convex quadrilateral $ABCD$, in which $\angle CBD = 90^o$, $\angle BCD =\angle CAD$ and $AD= 2BC$. Prove that $CA =CD$. (Anton Trygub)

2014 Oral Moscow Geometry Olympiad, 2

Let $ABCD$ be a parallelogram. On side $AB$, point $M$ is taken so that $AD = DM$. On side $AD$ point $N$ is taken so that $AB = BN$. Prove that $CM = CN$.

1953 Kurschak Competition, 3

$ABCDEF$ is a convex hexagon with all its sides equal. Also $\angle A + \angle C + \angle E = \angle B + \angle D + \angle F$. Show that $\angle A = \angle D$, $\angle B = \angle E$ and $\angle C = \angle F$.

Indonesia MO Shortlist - geometry, g9

It is known that $ABCD$ is a parallelogram. The point $E$ is taken so that $BCED$ is a cyclic quadrilateral. Let $\ell$ be a line that passes through $A$, intersects the segment $DC$ at point $F$ and intersects the extension of the line $BC$ at $G$. Given $EF = EG = EC$. Prove that $\ell$ is the bisector of the angle $\angle BAD$.

2008 Oral Moscow Geometry Olympiad, 3

Given a quadrilateral $ABCD$. $A ', B', C'$ and $D'$ are the midpoints of the sides $BC, CB, BA$ and $AB$, respectively. It is known that $AA'= CC'$, $BB'= DD'$. Is it true that $ABCD$ is a parallelogram? (M. Volchkevich)

Novosibirsk Oral Geo Oly VII, 2020.5

Point $P$ is chosen inside triangle $ABC$ so that $\angle APC+\angle ABC=180^o$ and $BC=AP.$ On the side $AB$, a point $K$ is chosen such that $AK = KB + PC$. Prove that $CK \perp AB$.

2015 Indonesia MO Shortlist, G6

Let $ABC$ be an acute angled triangle with circumcircle $O$. Line $AO$ intersects the circumcircle of triangle $ABC$ again at point $D$. Let $P$ be a point on the side $BC$. Line passing through $P$ perpendicular to $AP$ intersects lines $DB$ and $DC$ at $E$ and $F$ respectively . Line passing through $D$ perpendicular to $BC$ intersects $EF$ at point $Q$. Prove that $EQ = FQ$ if and only if $BP = CP$.

Indonesia MO Shortlist - geometry, g2

Let $ABC$ be an isosceles triangle right at $C$ and $P$ any point on $CB$. Let also $Q$ be the midpoint of $AB$ and $R, S$ be the points on $AP$ such that $CR$ is perpendicular to $AP$ and $|AS|=|CR|$. Prove that the $|RS| = \sqrt2 |SQ|$.

2011 Sharygin Geometry Olympiad, 7

Points $P$ and $Q$ on sides $AB$ and $AC$ of triangle $ABC$ are such that $PB = QC$. Prove that $PQ < BC$.

2003 Singapore Team Selection Test, 2

Let $M$ be a point on the diameter $AB$ of a semicircle $\Gamma$. The perpendicular at $M$ meets the semicircle $\Gamma$ at $P$. A circle inside $\Gamma$. touches $\Gamma$. and is tangent to $PM$ at $Q$ and $AM$ at $R$. Prove that $P B = RB$.

Kyiv City MO Seniors Round2 2010+ geometry, 2018.10.3

In the acute triangle $ABC$ the orthocenter $H$ and the center of the circumscribed circle $O$ were noted. The line $AO$ intersects the side $BC$ at the point $D$. A perpendicular drawn to the side $BC$ at the point $D$ intersects the heights from the vertices $B$ and $C$ of the triangle $ABC$ at the points $X$ and $Y$ respectively. Prove that the center of the circumscribed circle $\Delta HXY$ is equidistant from the points $B$ and $C$. (Danilo Hilko)

Estonia Open Senior - geometry, 1994.2.2

The two sides $BC$ and $CD$ of an inscribed quadrangle $ABCD$ are of equal length. Prove that the area of this quadrangle is equal to $S =\frac12 \cdot AC^2 \cdot \sin \angle A$

2015 Latvia Baltic Way TST, 16

Points $X$ , $Y$, $Z$ lie on a line $k$ in this order. Let $\omega_1$, $\omega_2$, $\omega_3$ be three circles of diameters $XZ$, $XY$ , $YZ$ , respectively. Line $\ell$ passing through point $Y$ intersects $\omega_1$ at points $A$ and $D$, $\omega_2$ at $B$ and $\omega_3$ at $C$ in such manner that points $A, B, Y, X, D$ lie on $\ell$ in this order. Prove that $AB =CD$.

2018 Peru Iberoamerican Team Selection Test, P2

Let $ABC$ be a triangle with $AB = AC$ and let $D$ be the foot of the height drawn from $A$ to $BC$. Let $P$ be a point inside the triangle $ADC$ such that $\angle APB> 90^o$ and $\angle PAD + \angle PBD = \angle PCD$. The $CP$ and $AD$ lines are cut at $Q$ and the $BP$ and $AD$ lines cut into $R$. Let $T$ be a point in segment $AB$ such that $\angle TRB = \angle DQC$ and let S be a point in the extension of the segment $AP$ (on the $P$ side) such that $\angle PSR = 2 \angle PAR$. Prove that $RS = RT$.

2010 Sharygin Geometry Olympiad, 3

Let $ABCD$ be a convex quadrilateral and $K$ be the common point of rays $AB$ and $DC$. There exists a point $P$ on the bisectrix of angle $AKD$ such that lines $BP$ and $CP$ bisect segments $AC$ and $BD$ respectively. Prove that $AB = CD$.

Kyiv City MO Juniors Round2 2010+ geometry, 2012.7.3

In the triangle $ABC $ the median $BD$ is drawn, which is divided into three equal parts by the points $E $ and $F$ ($BE = EF = FD$). It is known that $AD = AF$ and $AB = 1$. Find the length of the segment $CE$.