Olympiad problems

Indonesia MO Shortlist - geometry, g7

Given an isosceles trapezoid $ABCD$ with base $AB$. The diagonals $AC$ and $BD$ intersect at point $S$. Let $M$ the midpoint of $BC$ and the bisector of the angle $BSC$ intersect $BC$ at $N$. Prove that $\angle AMD = \angle AND$.

2007 Sharygin Geometry Olympiad, 14

Tags: diagonal , trapezoid , equal angles , geometry

In a trapezium with bases $AD$ and $BC$, let $P$ and $Q$ be the middles of diagonals $AC$ and $BD$ respectively. Prove that if $\angle DAQ = \angle CAB$ then $\angle PBA = \angle DBC$.

2011 Sharygin Geometry Olympiad, 12

Tags: altitude , geometry , construction , equal angles

Let $AP$ and $BQ$ be the altitudes of acute-angled triangle $ABC$. Using a compass and a ruler, construct a point $M$ on side $AB$ such that $\angle AQM = \angle BPM$.

1998 Tournament Of Towns, 2

Tags: parallelogram , geometry , equal segments , equal angles

$ABCD$ 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)

2007 Regional Olympiad of Mexico Center Zone, 5

Tags: equal angles , geometry

Consider a triangle $ABC$ with $\angle ACB = 2 \angle CAB $ and $\angle ABC> 90 ^ \circ$. Consider the perpendicular on $AC$ that passes through $A$ and intersects $BC$ at $D$, prove that $$\frac {1} {BC} - \frac {2} {DC} = \frac {1} {CA} $$

Russian TST 2018, P1

Tags: geometry , equal angles

Let $ABC$ be an isosceles triangle with $AB = AC$. Let P be a point in the interior of $ABC$ such that $PB > PC$ and $\angle PBA = \angle PCB$. Let $M$ be the midpoint of the side $BC$. Let $O$ be the circumcenter of the triangle $APM$. Prove that $\angle OAC=2 \angle BPM$ .

2021 Final Mathematical Cup, 2

Tags: geometry , circumcenter , equal angles

Let $ABC$ be an acute triangle, where $AB$ is the smallest side and let $D$ be the midpoint of $AB$. Let $P$ be a point in the interior of the triangle $ABC$ such that $\angle CAP = \angle CBP = \angle ACB$. From the point $P$, we draw perpendicular lines on $BC$ and $AC$ where the intersection point with $BC$ is $M$, and with $AC$ is $N$ . Through the point $M$ we draw a line parallel to $AC$, and through $N$ parallel to $BC$. These lines intercept at the point $K$. Prove that $D$ is the center of the circumscribed circle for the triangle $MNK$.

2016 Saudi Arabia IMO TST, 2

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

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

2001 Argentina National Olympiad, 2

Tags: geometry , angle , angle bisector , equal angles

Let $\vartriangle ABC$ be a triangle such that angle $\angle ABC$ is less than angle $\angle ACB$. The bisector of angle $\angle BAC$ cuts side $BC$ at $D$. Let $E$ be on side $AB$ such that $\angle EDB = 90^o$ and $F$ on side $AC$ such that $\angle BED = \angle DEF$. Prove that $\angle BAD = \angle FDC$.

Kyiv City MO Juniors Round2 2010+ geometry, 2013.8.3

Tags: angle , parallel , geometry , equal angles

Inside $\angle BAC = 45 {} ^ \circ$ the point $P$ is selected that the conditions $\angle APB = \angle APC = 45 {} ^ \circ $ are fulfilled. Let the points $M$ and $N$ be the projections of the point $P$ on the lines $AB$ and $AC$, respectively. Prove that $BC\parallel MN $. (Serdyuk Nazar)

1996 North Macedonia National Olympiad, 1

Tags: geometry , equal angles , perpendicular , parallelogram

Let $ABCD$ be a parallelogram which is not a rectangle and $E$ be the point in its plane such that $AE \perp AB$ and $CE \perp CB$. Prove that $\angle DEA = \angle CEB$.

2018 Yasinsky Geometry Olympiad, 6

Tags: angle chasing , altitude , projection , equal angles , geometry

$AH$ is the altitude of the acute triangle $ABC$, $K$ and $L$ are the feet of the perpendiculars, from point $H$ on sides $AB$ and $AC$ respectively. Prove that the angles $BKC$ and $BLC$ are equal.

2010 Grand Duchy of Lithuania, 4

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

In the triangle $ABC$ angle $C$ is a right angle. On the side $AC$ point $D$ has been found, and on the segment $BD$ point K has been found such that $\angle ABC = \angle KAD = \angle AKD$. Prove that $BK = 2DC$.

2020 Yasinsky Geometry Olympiad, 2

Tags: trapezoid , geometry , equal angles

On the midline $MN$ of the trapezoid $ABCD$ ($AD\parallel BC$) the points $F$ and $G$ are chosen so that $\angle ABF =\angle CBG$. Prove that then $\angle BAF = \angle DAG$. (Dmitry Prokopenko)

2013 Costa Rica - Final Round, G2

Tags: parallelogram , geometry , equal angles

Consider the triangle $ABC$. Let $P, Q$ inside the angle $A$ such that $\angle BAP=\angle CAQ$ and $PBQC$ is a parallelogram. Show that $\angle ABP=\angle ACP.$

2019 Romanian Master of Mathematics Shortlist, G4 ver.I

Tags: geometry , circles , equal angles

Let $\Omega$ be the circumcircle of an acute-angled triangle $ABC$. Let $D$ be the midpoint of the minor arc $AB$ of $\Omega$. A circle $\omega$ centered at $D$ is tangent to $AB$ at $E$. The tangents to $\omega$ through $C$ meet the segment $AB$ at $K$ and $L$, where $K$ lies on the segment $AL$. A circle $\Omega_1$ is tangent to the segments $AL, CL$, and also to $ \Omega$ at point $M$. Similarly, a circle $\Omega_2$ is tangent to the segments $BK, CK$, and also to $\Omega$ at point $N$. The lines $LM$ and $KN$ meet at $P$. Prove that $\angle KCE = \angle LCP$. Poland

Indonesia MO Shortlist - geometry, g5

Tags: midpoint , geometry , equal angles

Given a cyclic quadrilateral $ABCD$. Suppose $E, F, G, H$ are respectively the midpoint of the sides $AB, BC, CD, DA$. The line passing through $G$ and perpendicular on $AB$ intersects the line passing through $H$ and perpendicular on $BC$ at point $K$. Prove that $\angle EKF = \angle ABC$.

VMEO IV 2015, 10.2

Tags: geometry , equal angles , equal segments

Given a triangle $ABC$ with obtuse $\angle A$ and attitude $AH$ with $H \in BC$. Let $E,F$ on $CA$, $AB$ satisfying $\angle BEH = \angle C$ and $\angle CFH = \angle B$. Let $BE$ cut $CF$ at $D$. Prove that $DE = DF$.

2018 Irish Math Olympiad, 8

Tags: equilateral triangle , geometry , equal angles

Let $M$ be the midpoint of side $BC$ of an equilateral triangle $ABC$. The point $D$ is on $CA$ extended such that $A$ is between $D$ and $C$. The point $E$ is on $AB$ extended such that $B$ is between $A$ and $E$, and $|MD| = |ME|$. The point $F$ is the intersection of $MD$ and $AB$. Prove that $\angle BFM = \angle BME$.

Kyiv City MO Juniors Round2 2010+ geometry, 2013.7.3

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

In the square $ABCD$ on the sides $AD$ and $DC$, the points $M$ and $N$ are selected so that $\angle BMA = \angle NMD = 60 { } ^ \circ $. Find the value of the angle $MBN$.

2021 Yasinsky Geometry Olympiad, 3

Tags: equal segments , angle , equal angles , geometry

The segments $AC$ and $BD$ are perpendicular, and $AC$ is twice as large as $BD$ and intersects $BD$ in it in the midpoint. Find the value of the angle $BAD$, if we know that $\angle CAD = \angle CDB$. (Gregory Filippovsky)

2020 Durer Math Competition Finals, 5

Tags: hexagon , equal angles , geometry

The hexagon $ABCDEF$ has all angles equal . We know that four consecutive sides of the hexagon have lengths $7, 6, 3$ and $5$ in this order. What is the sum of the lengths of the two remaining sides?

2010 Dutch IMO TST, 4

Tags: cyclic quadrilateral , midpoint , equal angles , geometry

Let $ABCD$ be a cyclic quadrilateral satisfying $\angle ABD = \angle DBC$. Let $E$ be the intersection of the diagonals $AC$ and $BD$. Let $M$ be the midpoint of $AE$, and $N$ be the midpoint of $DC$. Show that $MBCN$ is a cyclic quadrilateral.

2013 Saudi Arabia GMO TST, 2

Tags: polygon , geometry , combinatorial geometry , angle , equal angles , combinatorics

Find all values of $n$ for which there exists a convex cyclic non-regular polygon with $n$ vertices such that the measures of all its internal angles are equal.

2015 Romania Team Selection Tests, 1

Tags: miquel point , spiral similarity , equal angles , geometry

Let $ABC$ be a triangle. Let $P_1$ and $P_2$ be points on the side $AB$ such that $P_2$ lies on the segment $BP_1$ and $AP_1 = BP_2$; similarly, let $Q_1$ and $Q_2$ be points on the side $BC$ such that $Q_2$ lies on the segment $BQ_1$ and $BQ_1 = CQ_2$. The segments $P_1Q_2$ and $P_2Q_1$ meet at $R$, and the circles $P_1P_2R$ and $Q_1Q_2R$ meet again at $S$, situated inside triangle $P_1Q_1R$. Finally, let $M$ be the midpoint of the side $AC$. Prove that the angles $P_1RS$ and $Q_1RM$ are equal.