Olympiad problems

1979 Chisinau City MO, 173

The inner angles of the pentagon inscribed in the circle are equal to each other. Prove that this pentagon is regular.

1967 Spain Mathematical Olympiad, 5

Tags: geometry , equal angles

Let $\gamma$ be a semicircle with diameter $AB$ . A creek is built with origin in $A$ , which has its vertices alternately in the diameter $AB$ and in the semicircle $\gamma$ , so that its sides make equal angles $\alpha$ with the diameter (but alternately in either direction). It is requested: a) Values of the angle $\alpha$ for the ravine to pass through the other end $B$ of the diameter. b) The total length of the ravine, in the case that it ends in $B$ , as a function of the length $d$ of the diameter and of the angle $\alpha$ . [img]https://cdn.artofproblemsolving.com/attachments/3/c/54c71cdf1bf8fbb3bdfa38f4b14a1dc961c5fe.png[/img]

2015 Dutch BxMO/EGMO TST, 4

Tags: geometry , angle bisector , angle chasing , circumcircle , equal angles

In a triangle $ABC$ the point $D$ is the intersection of the interior angle bisector of $\angle BAC$ and side $BC$. Let $P$ be the second intersection point of the exterior angle bisector of $\angle BAC$ with the circumcircle of $\angle ABC$. A circle through $A$ and $P$ intersects line segment $BP$ internally in $E$ and line segment $CP$ internally in $F$. Prove that $\angle DEP = \angle DFP$.

Ukrainian TYM Qualifying - geometry, 2014.8

Tags: equal angles , geometry

In the triangle $ABC$ on the ray $BA$ mark the point $K$ so that $\angle BCA= \angle KCA$ , and on the median $BM$ mark the point $T$ so that $\angle CTK=90^o$ . Prove that $\angle MTC=\angle MCB$ .

Novosibirsk Oral Geo Oly VII, 2019.5

Tags: geometry , equal angles

Given a triangle $ABC$, in which the angle $B$ is three times the angle $C$. On the side $AC$, point $D$ is chosen such that the angle $BDC$ is twice the angle $C$. Prove that $BD + BA = AC$.

2013 Regional Competition For Advanced Students, 4

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

We call a pentagon [i]distinguished [/i] if either all side lengths or all angles are equal. We call it [i]very distinguished[/i] if in addition two of the other parts are equal. i.e. $5$ sides and $2$ angles or $2$ sides and $5$ angles.Show that every very distinguished pentagon has an axis of symmetry.

Estonia Open Junior - geometry, 2012.2.3

Tags: geometry , circles , equal angles

Two circles $c$ and $c'$ with centers $O$ and $O'$ lie completely outside each other. Points $A, B$, and $C$ lie on the circle $c$ and points $A', B'$, and $C$ lie on the circle $c'$ so that segment $AB\parallel A'B'$, $BC \parallel B'C'$, and $\angle ABC = \angle A'B'C'$. The lines $AA', BB$', and $CC'$ are all different and intersect in one point $P$, which does not coincide with any of the vertices of the triangles $ABC$ or $A'B'C'$. Prove that $\angle AOB = \angle A'O'B'$.

2003 IMO, 3

Tags: geometry , hexagon , equal angles

Each pair of opposite sides of a convex hexagon has the following property: the distance between their midpoints is equal to $\dfrac{\sqrt{3}}{2}$ times the sum of their lengths. Prove that all the angles of the hexagon are equal.

2011 Ukraine Team Selection Test, 10

Tags: orthocenter , geometry , concurrency , equal angles

Let $ H $ be the point of intersection of the altitudes $ AP $ and $ CQ $ of the acute-angled triangle $ABC$. The points $ E $ and $ F $ are marked on the median $ BM $ such that $ \angle APE = \angle BAC $, $ \angle CQF = \angle BCA $, with point $ E $ lying inside the triangle $APB$ and point $ F $ is inside the triangle $CQB$. Prove that the lines $AE, CF$, and $BH$ intersect at one point.

Croatia MO (HMO) - geometry, 2010.3

Tags: parallel , equal angles , geometry

Let $D$ be a point on the side $AC$ of triangle $ABC$. Let $E$ and $F$ be points on the segments $BD$ and $BC$ respectively, such that $\angle BAE = \angle CAF$. Let $P$ and $Q$ be points on the segments $BC$ and $BD$ respectively, such that $EP \parallel CD$ and $FQ \parallel CD$. Prove that $\angle BAP = \angle CAQ$.

2022 Regional Olympiad of Mexico West, 4

Tags: geometry , fixed , equal angles , ratio

Prove that in all triangles $\vartriangle ABC$ with $\angle A = 2 \angle B$ it holds that, if $D$ is the foot of the perpendicular from $C$ to the perpendicular bisector of $AB$, $\frac{AC}{DC}$ is constant for any value of $\angle B$.

2019 IFYM, Sozopol, 4

Tags: geometry , equal angles , power of a point

The diagonals $AC$ and $BD$ of a convex quadrilateral $ABCD$ intersect in point $M$. The angle bisector of $\angle ACD$ intersects the ray $\overrightarrow{BA}$ in point $K$. If $MA.MC+MA.CD=MB.MD$, prove that $\angle BKC=\angle CDB$.

2012 Dutch IMO TST, 4

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

Let $\vartriangle ABC$ be a triangle. The angle bisector of $\angle CAB$ intersects$ BC$ at $L$. On the interior of line segments $AC$ and $AB$, two points, $M$ and $N$, respectively, are chosen in such a way that the lines $AL, BM$ and $CN$ are concurrent, and such that $\angle AMN = \angle ALB$. Prove that $\angle NML = 90^o$.

2008 Hanoi Open Mathematics Competitions, 9

Tags: parallelogram , construction , equal angles , geometry

Consider a triangle $ABC$. For every point M $\in BC$ ,we define $N \in CA$ and $P \in AB$ such that $APMN$ is a parallelogram. Let $O$ be the intersection of $BN$ and $CP$. Find $M \in BC$ such that $\angle PMO=\angle OMN$

2022 Regional Olympiad of Mexico West, 3

Tags: isosceles , geometry , equal angles

In my isosceles triangle $\vartriangle ABC$ with $AB = CA$, we draw $D$ the midpoint of $BC$. Let $E$ be a point on $AC$ such that $\angle CDE = 60^o$ and $M$ the midpoint of $DE$. Prove that $\angle AME = \angle BMD$.

Kyiv City MO Juniors 2003+ geometry, 2011.89.4

Tags: geometry , midpoint , cyclic , equal angles

Let $ABCD$ be an inscribed quadrilateral. Denote the midpoints of the sides $AB, BC, CD$ and $DA$ through $M, L, N$ and $K$, respectively. It turned out that $\angle BM N = \angle MNC$. Prove that: i) $\angle DKL = \angle CLK$. ii) in the quadrilateral $ABCD$ there is a pair of parallel sides.

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?

2007 Bosnia and Herzegovina Junior BMO TST, 4

Tags: geometry , circumcircle , incenter , arc midpoint , equal angles

Let $I$ be the incenter of the triangle $ABC$ ($AB < BC$). Let $M$ be the midpoint of $AC$, and let $N$ be the midpoint of the arc $AC$ of the circumcircle of $ABC$ which contains $B$. Prove that $\angle IMA = \angle INB$.

2011 Saudi Arabia BMO TST, 3

Tags: geometry , pentagon , bisects segment , equal angles

Let $ABCDE$ be a convex pentagon such that $\angle BAC = \angle CAD = \angle DAE$ and $\angle ABC = \angle ACD = \angle ADE$. Diagonals $BD$ and $CE$ meet at $P$. Prove that $AP$ bisects side $CD$.

2020 SAFEST Olympiad, 4

Tags: geometry , orthocenter , equal angles , circumcenter

Let $O$ be the circumcenter and $H$ the orthocenter of an acute-triangle $ABC$. The perpendicular bisector of $AO$ intersects the line $BC$ at point $S$. Let $L$ be the midpoint of $OH$. Prove that $\angle OAH = \angle LSA$.

2016 Auckland Mathematical Olympiad, 2

Tags: geometry , equal angles , square

In square $ABCD$, $\overline{AC}$ and $\overline{BD}$ meet at point $E$. Point $F$ is on $\overline{CD}$ and $\angle CAF = \angle FAD$. If $\overline{AF}$ meets $\overline{ED}$ at point $G$, and if $\overline{EG} = 24$ cm, then find the length of $\overline{CF}$.

2011 Balkan MO Shortlist, G2

Tags: geometry , circumcircle , angle bisector , circles , symmedian , equal angles

Let $ABC$ be a triangle and let $O$ be its circumcentre. The internal and external bisectrices of the angle $BAC$ meet the line $BC$ at points $D$ and $E$, respectively. Let further $M$ and $L$ respectively denote the midpoints of the segments $BC$ and $DE$. The circles $ABC$ and $ALO$ meet again at point $N$. Show that the angles $BAN$ and $CAM$ are equal.

2015 Caucasus Mathematical Olympiad, 2

Tags: equal segments , convex quadrilateral , equal angles , geometry

In the convex quadrilateral $ABCD$, point $K$ is the midpoint of $AB$, point $L$ is the midpoint of $BC$, point $M$ is the midpoint of CD, and point $N$ is the midpoint of $DA$. Let $S$ be a point lying inside the quadrilateral $ABCD$ such that $KS = LS$ and $NS = MS$ .Prove that $\angle KSN = \angle MSL$.

2016 Irish Math Olympiad, 4

Tags: circles , equal angles , geometry

Let $ABC$ be a triangle with $|AC| \ne |BC|$. Let $P$ and $Q$ be the intersection points of the line $AB$ with the internal and external angle bisectors at $C$, so that $P$ is between $A$ and $B$. Prove that if $M$ is any point on the circle with diameter $PQ$, then $\angle AMP = \angle BMP$.

2013 Argentina National Olympiad, 2

Tags: angle bisector , geometry , ratio , convex quadrilateral , equal angles

In a convex quadrilateral $ABCD$ the angles $\angle A$ and $\angle C$ are equal and the bisector of $\angle B$ passes through the midpoint of the side $CD$. If it is known that $CD = 3AD$, calculate $\frac{AB}{BC}$.