Olympiad problems

2012 Switzerland - Final Round, 3

The circles $k_1$ and $k_2$ intersect at points $D$ and $P$. The common tangent of the two circles on the side of $D$ touches $k_1$ at $A$ and $k_2$ at $B$. The straight line $AD$ intersects $k_2$ for a second time at $C$. Let $M$ be the center of the segment $BC$. Show that $ \angle DPM = \angle BDC$ .

Durer Math Competition CD Finals - geometry, 2019.D3

Tags: geometry , equal segments , equal angles , angle

a) Does there exist a quadrilateral with (both of) the following properties: three of its edges are of the same length, but the fourth one is different, and three of its angles are equal, but the fourth one is different? b) Does there exist a pentagon with (both of) the following properties: four of its edges are of the same length, but the fifth one is different, and four of its angles are equal, but the fifth one is different?

1989 All Soviet Union Mathematical Olympiad, 492

Tags: equal ratio , geometry , similar triangles , equal angles

$ABC$ is a triangle. $A' , B' , C'$ are points on the segments $BC, CA, AB$ respectively. $\angle B' A' C' = \angle A$ , $\frac{AC'}{C'B} = \frac{BA' }{A' C} = \frac{CB'}{B'A}$. Show that $ABC$ and $A'B'C'$ are similar.

Swiss NMO - geometry, 2013.7

Tags: geometry , bisects segment , equal angles

Let $O$ be the center of the circle of the triangle $ABC$ with $AB \ne AC$. Furthermore, let $S$ and $T$ be points on the rays $AB$ and $AC$, such that $\angle ASO = \angle ACO$ and $\angle ATO = \angle ABO$. Show that $ST$ bisects the segment $BC$.

2021 Portugal MO, 2

Tags: geometry , equal segments , equal angles

Let $ABC$ be a triangle such that $AB = AC$. Let $D$ be a point in $[BC]$ and $E$ a point in $[AD]$ such that $\angle BE D = \angle BAC = 2 \angle DEC$. Shows that $DB = 2CD$. [img]https://cdn.artofproblemsolving.com/attachments/d/5/677e19d8e68a89134e17a4ab6051e41f283486.png[/img]

2019 Regional Olympiad of Mexico Center Zone, 3

Tags: geometry , equal segments , equal angles

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 $ZX = ZY$.

2018 NZMOC Camp Selection Problems, 4

Tags: geometry , right angle , equal angles

Let $P$ be a point inside triangle $ABC$ such that $\angle CPA = 90^o$ and $\angle CBP = \angle CAP$. Prove that $\angle P XY = 90^o$, where $X$ and $Y$ are the midpoints of $AB$ and $AC$ respectively.

2016 Saudi Arabia IMO TST, 2

Tags: convex , hexagon , equal segments , equal angles , angle , geometry

Let $ABCDEF$ be a convex hexagon with $AB = CD = EF$, $BC =DE = FA$ and $\angle A+\angle B = \angle C +\angle D = \angle E +\angle F$. Prove that $\angle A=\angle C=\angle E$ and $\angle B=\angle D=\angle F$. Tran Quang Hung

2021 Sharygin Geometry Olympiad, 9.6

Tags: geometry , trapezoid , equal angles

The diagonals of trapezoid $ABCD$ ($BC\parallel AD$) meet at point $O$. Points $M$ and $N$ lie on the segments $BC$ and $AD$ respectively. The tangent to the circle $AMC$ at $C$ meets the ray $NB$ at point $P$; the tangent to the circle $BND$ at $D$ meets the ray $MA$ at point $R$. Prove that $\angle BOP =\angle AOR$.

Kharkiv City MO Seniors - geometry, 2014.10.4

Tags: square , angle chasing , equal angles , geometry , angle

Let $ABCD$ be a square. The points $N$ and $P$ are chosen on the sides $AB$ and $AD$ respectively, such that $NP=NC$. The point $Q$ on the segment $AN$ is such that that $\angle QPN=\angle NCB$. Prove that $\angle BCQ=\frac{1}{2}\angle AQP$.

2022 Sharygin Geometry Olympiad, 10.3

Tags: geometry , equal angles , combinatorial geometry

A line meets a segment $AB$ at point $C$. Which is the maximal number of points $X$ of this line such that one of angles $AXC$ and $BXC$ is equlal to a half of the second one?

2024 Czech and Slovak Olympiad III A, 2

Tags: equal angles , geometry

Let the interior point $P$ of the convex quadrilateral $ABCD$ be such that $$|\angle PAD| = |\angle ADP| = |\angle CBP| = |\angle PCB| = |\angle CPD|.$$ Let $O$ be the center of the circumcircle of the triangle $CPD$. Prove that $|OA| = |OB|$.

Kyiv City MO Juniors Round2 2010+ geometry, 2020.8.2

Tags: geometry , equal segments , equal angles , right angle

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

2021 SAFEST Olympiad, 4

Tags: geometry , equal angles , equal segments

Let $ABC$ be a triangle with $AB > AC$. Let $D$ be a point on the side $AB$ such that $DB = DC$ and let $M$ be the midpoint of $AC$. The line parallel to $BC$ passing through $D$ intersects the line $BM$ in $K$. Show that $\angle KCD = \angle DAC.$

2012 Denmark MO - Mohr Contest, 5

Tags: geometry , hexagon , equal angles , equilateral , equal segments

In the hexagon $ABCDEF$, all angles are equally large. The side lengths satisfy $AB = CD = EF = 3$ and $BC = DE = F A = 2$. The diagonals $AD$ and $CF$ intersect each other in the point $G$. The point $H$ lies on the side $CD$ so that $DH = 1$. Prove that triangle $EGH$ is equilateral.

2019 Saudi Arabia JBMO TST, 3

Tags: geometry , equal angles , angle

Consider a triangle $ABC$ and let $M$ be the midpoint of the side $BC$. Suppose $\angle MAC = \angle ABC$ and $\angle BAM = 105^o$. Find the measure of $\angle ABC$.

2019 Estonia Team Selection Test, 7

Tags: geometry , equal angles , altitude

An acute-angled triangle $ABC$ has two altitudes $BE$ and $CF$. The circle with diameter $AC$ intersects the segment $BE$ at point $P$. A circle with diameter $AB$ intersects the segment $CF$ at point $Q$ and the extension of this altitude at point $Q'$. Prove that $\angle PQ'Q = \angle PQB$.

1998 Tournament Of Towns, 2

Tags: parallelogram , equal segments , equal angles , geometry

$ABCD$ is a parallelogram. A point $M$ is found on the side $AB$ or its extension such that $\angle MAD = \angle AMO$ where $O$ is the intersection point of the diagonals of the parallelogram. Prove that $MD = MG$. (M Smurov)

2009 IMAR Test, 3

Tags: geometry , angle chasing , convex quadrilateral , equal angles , angle

Consider a convex quadrilateral $ABCD$ with $AB=CB$ and $\angle ABC +2 \angle CDA = \pi$ and let $E$ be the midpoint of $AC$. Show that $\angle CDE =\angle BDA$. Paolo Leonetti

2020 Ukrainian Geometry Olympiad - April, 5

Tags: parallelogram , pentagon , equal angles , equal segments , geometry

Given a convex pentagon $ABCDE$, with $\angle BAC = \angle ABE = \angle DEA - 90^o$, $\angle BCA = \angle ADE$ and also $BC = ED$. Prove that $BCDE$ is parallelogram.

2006 Cuba MO, 9

Tags: geometry , perpendicular , equal angles

In the cyclic quadrilateral $ABCD$, the diagonals $AC$ and $BD$ intersect at $P$. Let $O$ be the center of the circumcircle $ABCD$, and $E$ a point of the extension of $OC$ beyond $C$. A parallel line to $CD$ is drawn through $E$ that cuts the extension of $OD$ beyonf $D$ at $F$. Let $Q$ be a point interior to $ABCD$, such that $\angle AFQ = \angle BEQ$ and $\angle FAQ = \angle EBQ$. Prove that $PQ \perp CD$.