Olympiad problems

2015 Caucasus Mathematical Olympiad, 2

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 $\angle KSN = \angle MSL$.

2024 Moldova EGMO TST, 9

Tags: equal angles , geometry

Given a convex quadrilateral $ KLMN $, in which $ \angle NKL = {{90} ^ {\circ}} $. Let $ P $ be the midpoint of the segment $ LM $. It turns out that $ \angle KNL = \angle MKP $. Prove that $ \angle KNM = \angle LKP $.

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.

2020 Ukrainian Geometry Olympiad - April, 2

Tags: geometry , equal angles , circles , tangent , symmetry

Let $\Gamma$ be a circle and $P$ be a point outside, $PA$ and $PB$ be tangents to $\Gamma$ , $A, B \in \Gamma$ . Point $K$ is an arbitrary point on the segment $AB$. The circumscirbed circle of $\vartriangle PKB$ intersects $\Gamma$ for the second time at point $T$, point $P'$ is symmetric to point $P$ wrt point $A$. Prove that $\angle PBT = \angle P'KA$.

Durer Math Competition CD Finals - geometry, 2021.C3

Tags: equal angles , geometry , isosceles

In the isosceles triangle $ABC$ we have $AC = BC$. Let $X$ be an arbitrary point of the segment $AB$. The line parallel to $BC$ and passing through $X$ intersects the segment $AC$ in $N$, and the line parallel to $AC$ and passing through $BC$ intersects the segment $BC$ in $M$. Let $k_1$ be the circle with center $N$ and radius $NA$. Similarly, let $k_2$ be the circle with center $M$ and radius $MB$. Let $T$ be the intersection of the circles $k_1$ and $k_2$ different from $X$. Show that the angles $\angle NCM$ and $\angle NTM$ are equal.

2005 JBMO Shortlist, 4

Tags: geometry , isosceles , equal angles

Let $ABC$ be an isosceles triangle $(AB=AC)$ so that $\angle A< 2 \angle B$ . Let $D,Z $ points on the extension of height $AM$ so that $\angle CBD = \angle A$ and $\angle ZBA = 90^\circ$. Let $E$ the orthogonal projection of $M$ on height $BF$, and let $K$ the orthogonal projection of $Z$ on $AE$. Prove that $ \angle KDZ = \angle KDB = \angle KZB$.

1999 Ukraine Team Selection Test, 5

Tags: pentagon , equal angles , geometry

A convex pentagon $ABCDE$ with $DC = DE$ and $\angle DCB = \angle DEA = 90^o$ is given. Let $F$ be a point on the segment $AB$ such that $AF : BF = AE : BC$. Prove that $\angle FCE = \angle ADE$ and $\angle FEC = \angle BDC$.

Kyiv City MO Juniors Round2 2010+ geometry, 2018.8.31

Tags: geometry , isosceles , equal angles , equal segments

On the sides $AB$, $BC$ and $CA$ of the isosceles triangle $ABC$ with the vertex at the point $B$ marked the points $M$, $D$ and $K$ respectively so that $AM = 2DC$ and $\angle AMD = \angle KDC$. Prove that $MD = KD$.

Kyiv City MO Seniors 2003+ geometry, 2018.11.4.1

Tags: orthocenter , geometry , equal angles

In the quadrilateral $ABCD$, the diagonal $AC$ is the bisector $\angle BAD$ and $\angle ADC = \angle ACB$. The points $X, \, \, Y$ are the feet of the perpendiculars drawn from the point $A$ on the lines $BC, \, \, CD$, respectively. Prove that the orthocenter $\Delta AXY$ lies on the line $BD$.

2021 Sharygin Geometry Olympiad, 9.3

Tags: geometry , equal angles , angle

Let $ABC$ be an acute-angled scalene triangle and $T$ be a point inside it such that $\angle ATB = \angle BTC = 120^o$. A circle centered at point $E$ passes through the midpoints of the sides of $ABC$. For $B, T, E$ collinear, find angle $ABC$.

2012 Korea Junior Math Olympiad, 5

Tags: geometry , equal angles , equal segments , tangent , cyclic quadrilateral

Let $ABCD$ be a cyclic quadrilateral inscirbed in a circle $O$ ($AB> AD$), and let $E$ be a point on segment $AB$ such that $AE = AD$. Let $AC \cap DE = F$, and $DE \cap O = K(\ne D)$. The tangent to the circle passing through $C,F,E$ at $E$ hits $AK$ at $L$. Prove that $AL = AD$ if and only if $\angle KCE = \angle ALE$.

1988 Tournament Of Towns, (189) 2

Tags: geometry , equal angles , angle , square

A point $M$ is chosen inside the square $ABCD$ in such a way that $\angle MAC = \angle MCD = x$ . Find $\angle ABM$.

2012 Balkan MO Shortlist, G6

Tags: perpendicular , equal angles , geometry

Let $P$ and $Q$ be points inside a triangle $ABC$ such that $\angle PAC = \angle QAB$ and $\angle PBC = \angle QBA$. Let $D$ and $E$ be the feet of the perpendiculars from $P$ to the lines $BC$ and $AC$, and $F$ be the foot of perpendicular from $Q$ to the line $AB$. Let $M$ be intersection of the lines $DE$ and $AB$. Prove that $MP \perp CF$

2020 Ukrainian Geometry Olympiad - April, 3

Tags: tangent , geometry , equal angles , circles

The angle $POQ$ is given ($OP$ and $OQ$ are rays). Let $M$ and $N$ be points inside the angle $POQ$ such that $\angle POM = \angle QON$ and $\angle POM < \angle PON$. Consider two circles: one touches the rays $OP$ and $ON$, the other touches the rays $OM$ and $OQ$. Denote by $B$ and $C$ the points of their intersection. Prove that $\angle POC = \angle QOB$.

2021 Poland - Second Round, 2

Tags: equal segments , parallelogram , equal angles , geometry

The point P lies on the side $CD$ of the parallelogram $ABCD$ with $\angle DBA = \angle CBP$. Point $O$ is the center of the circle passing through the points $D$ and $P$ and tangent to the straight line $AD$ at point $D$. Prove that $AO = OC$.

2018 Puerto Rico Team Selection Test, 3

Tags: equal angles , geometry

Let $M$ be the point of intersection of diagonals $AC$ and $BD$ of the convex quadrilateral $ABCD$. Let $K$ be the point of intersection of the extension of side $AB$ (beyond$A$) with the bisector of the angle $ACD$. Let $L$ be the intersection of $KC$ and $BD$. If $MA \cdot CD = MB \cdot LD$, prove that the angle $BKC$ is equal to the angle $CDB$.

2019 Switzerland - Final Round, 7

Tags: geometry , equal angles , equal segments , angle

Let $ABC$ be a triangle with $\angle CAB = 2 \angle ABC$. Assume that a point $D$ is inside the triangle $ABC$ exists such that $AD = BD$ and $CD = AC$. Show that $\angle ACB = 3 \angle DCB$.

1998 Czech and Slovak Match, 1

Tags: parallelogram , equal angles , geometry

Let $P$ be an interior point of the parallelogram $ABCD$. Prove that $\angle APB+ \angle CPD = 180^\circ$ if and only if $\angle PDC = \angle PBC$.

2003 All-Russian Olympiad Regional Round, 11.2

Tags: geometry , equal angles , angle

On the diagonal $AC$ of a convex quadrilateral $ABCD$ is chosen such a point $K$ such that $KD = DC$, $\angle BAC = \frac12 \angle KDC$, $\angle DAC = \frac12 \angle KBC$. Prove that $\angle KDA = \angle BCA$ or $\angle KDA = \angle KBA$.

2020 BMT Fall, 20

Tags: equal segments , equal angles , geometry

Non-degenerate quadrilateral $ABCD$ with $AB = AD$ and $BC = CD$ has integer side lengths, and $\angle ABC = \angle BCD = \angle CDA$. If $AB = 3$ and $B \ne D$, how many possible lengths are there for $BC$?

2020 Dutch IMO TST, 1

Tags: geometry , incenter , equal angles

In acute-angled triangle $ABC, I$ is the center of the inscribed circle and holds $| AC | + | AI | = | BC |$. Prove that $\angle BAC = 2 \angle ABC$.

2003 Oral Moscow Geometry Olympiad, 2

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

In a convex quadrilateral $ABCD$, $\angle ABC = 90^o$ , $\angle BAC = \angle CAD$, $AC = AD, DH$ is the alltitude of the triangle $ACD$. In what ratio does the line $BH$ divide the segment $CD$?

2011 Ukraine Team Selection Test, 10

Tags: concurrency , equal angles , geometry , orthocenter

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.

2017 JBMO Shortlist, G3

Tags: geometry , circumcircle , equal angles , angle

Consider triangle $ABC$ such that $AB \le AC$. Point $D$ on the arc $BC$ of thecircumcirle of $ABC$ not containing point $A$ and point $E$ on side $BC$ are such that $\angle BAD = \angle CAE < \frac12 \angle BAC$ . Let $S$ be the midpoint of segment $AD$. If $\angle ADE = \angle ABC - \angle ACB$ prove that $\angle BSC = 2 \angle BAC$ .

1999 Romania National Olympiad, 1

Tags: equal angles , geometry

Let $AD$ be the bisector of angle $A$ of the triangle $ABC$. One considers the points M, N on the half-lines $(AB$ and $(AC$, respectively, such that $\angle MDA = \angle B$ and $\angle NDA = \angle C$. Let $AD \cap MN=\{P\}$. Prove that: $$AD^3 = AB \cdot AC\cdot AP$$