Olympiad problems

Novosibirsk Oral Geo Oly VIII, 2016.1

In the quadrilateral $ABCD$, angles $B$ and $C$ are equal to $120^o$, $AB = CD = 1$, $CB = 4$. Find the length $AD$.

2015 Dutch BxMO/EGMO TST, 4

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

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 $\angle DEP = \angle DFP$.

1987 All Soviet Union Mathematical Olympiad, 450

Tags: pentagon , geometry , equal angles

Given a convex pentagon $ABCDE$ with $\angle ABC= \angle ADE$ and $\angle AEC= \angle ADB$ . Prove that $\angle BAC = \angle DAE$ .

2019 Bundeswettbewerb Mathematik, 3

Tags: geometry , equal angles , incircle

Let $ABC$ be atriangle with $\overline{AC}> \overline{BC}$ and incircle $k$. Let $M,W,L$ be the intersections of the median, angle bisector and altitude from point $C$ respectively. The tangent to $k$ passing through $M$, that is different from $AB$, touch $k$ in $T$. Prove that the angles $\angle MTW$ and $\angle TLM$ are equal.

1990 All Soviet Union Mathematical Olympiad, 512

Tags: midpoint , equal segments , equal angles , geometry , diagonal

The line joining the midpoints of two opposite sides of a convex quadrilateral makes equal angles with the diagonals. Show that the diagonals are equal.

2017 Peru IMO TST, 9

Tags: cyclic quadrilateral , geometry , circumcircle , equal angles , tangent circle

Let $ABCD$ be a cyclie quadrilateral, $\omega$ be it's circumcircle and $M$ be the midpoint of the arc $AB$ of $\omega$ which does not contain the vertices $C$ and $D$. The line that passes through $M$ and the intersection point of segments $AC$ and $BD$, intersects again $\omega$ in $N$. Let $P$ and $Q$ be points in the $CD$ segment such that $\angle AQD = \angle DAP$ and $\angle BPC = \angle CBQ$. Prove that the circumcircle of $NPQ$ and $\omega$ are tangent to each other.

2013 Vietnam Team Selection Test, 5

Tags: geometry , equal angles , circumcircle , concyclic

Let $ABC$ be a triangle with $\angle BAC= 45^o$ . Altitudes $AD, BE, CF$ meet at $H$. $EF$ cuts $BC$ at $P$. $I$ is the midpoint of $BC$, $IF$ cuts $PH$ in $Q$. a) Prove that $\angle IQH = \angle AIE$. b) Let $(K)$ be the circumcircle of triangle $ABC$, $(J)$ be the circumcircle of triangle $KPD$. $CK$ cuts circle $(J)$ at $G$, $IG$ cuts $(J)$ at $M$, $JC$ cuts circle of diameter $BC$ at $N$. Prove that $G, N, M, C$ lie on the same circle.

2009 Postal Coaching, 5

Tags: geometry , angle , equal angles , sum

Let $P$ be an interior point of a circle and $A_1,A_2...,A_{10}$ be points on the circle such that $\angle A_1PA_2 = \angle A_2PA_3 = ... = \angle A_{10}PA_1 = 36^o$. Prove that $PA_1 + PA_3 + PA_5 + PA_7 +PA_9 = PA_2 + PA_4 + PA_6 + PA_8 + PA_{10}$.

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 $\angle BAC = 2 \angle ABC$.

2008 Singapore Senior Math Olympiad, 1

Tags: trapezoid , geometry , trapezium , equal angles

Let $ABCD$ be a trapezium with $AD // BC$. Suppose $K$ and $L$ are, respectively, points on the sides $AB$ and $CD$ such that $\angle BAL = \angle CDK$. Prove that $\angle BLA = \angle CKD$.

Ukraine Correspondence MO - geometry, 2019.7

Tags: geometry , construction , equal angles

Given a triangle $ABC$. Construct a point $D$ on the side $AB$ and point $E$ on the side $AC$ so that $BD = CE$ and $\angle ADC = \angle BEC$

2018 Yasinsky Geometry Olympiad, 5

Tags: geometry , incircle , equal angles , isosceles triangle , isosceles

The inscribed circle of the triangle $ABC$ touches its sides $AB, BC, CA$, at points $K,N, M$ respectively. It is known that $\angle ANM = \angle CKM$. Prove that the triangle $ABC$ is isosceles. (Vyacheslav Yasinsky)

2021 Saudi Arabia Training Tests, 13

Tags: geometry , angle , equal angles

Let $ABCD$ be a quadrilateral with $\angle A = \angle B = 90^o$, $AB = AD$. Denote $E$ as the midpoint of $AD$, suppose that $CD = BC + AD$, $AD > BC$. Prove that $\angle ADC = 2\angle ABE$.

2015 Oral Moscow Geometry Olympiad, 4

Tags: circumcircle , equal angles , tangent , midpoint , geometry

In triangle $ABC$, point $M$ is the midpoint of $BC, P$ is the intersection point of the tangents at points $B$ and $C$ of the circumscribed circle, $N$ is the midpoint of the segment $MP$. The segment $AN$ intersects the circumscribed circle at point $Q$. Prove that $\angle PMQ = \angle MAQ$.

2019 District Olympiad, 2

Tags: geometry , equal angles , equal segments , angle

Consider $D$ the midpoint of the base $[BC]$ of the isosceles triangle ABC in which $\angle BAC < 90^o$. On the perpendicular from $B$ on the line $BC$ consider the point $E$ such that $\angle EAB= \angle BAC$, and on the line passing though $C$ parallel to the line $AB$ we consider the point $F$ such that $F$ and $D$ are on different side of the line $AC$ and $\angle FAC = \angle CAD$. Prove that $AE = CF$ and $BF = EF$

2022 Dutch BxMO TST, 2

Tags: geometry , equal angles , angle

Let $ABC$ be an acute triangle, and let $D$ be the foot of the altitude from $A$. The circle with centre $A$ passing through $D$ intersects the circumcircle of triangle $ABC$ in $X$ and $Y$ , in such a way that the order of the points on this circumcircle is: $A, X, B, C, Y$ . Show that $\angle BXD = \angle CYD$.

2018 Puerto Rico Team Selection Test, 3

Tags: geometry , equal angles

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 $CDB$.

Denmark (Mohr) - geometry, 2014.3

Tags: geometry , equal angles , equal segments

The points $C$ and $D$ lie on a halfline from the midpoint $M$ of a segment $AB$, so that $|AC| = |BD|$. Prove that the angles $u = \angle ACM$ and $v = \angle BDM$ are equal. [img]https://1.bp.blogspot.com/-tQEJ1VBCa8U/XzT7IhwlZHI/AAAAAAAAMVI/xpRdlj5Rl64VUt_tCRsQ1UxIsv_SGrMlACLcBGAsYHQ/s0/2014%2BMohr%2Bp3.png[/img]

2004 Olympic Revenge, 1

Tags: geometry , equal angles , equilateral , circumcenter , equal area

$ABC$ is a triangle and $D$ is an internal point such that $\angle DAB=\angle DBC =\angle DCA$. $O_a$ is the circumcenter of $DBC$. $O_b$ is the circumcenter of $DAC$. $O_c$ is the circumcenter of $DAB$. Show that if the area of $ABC$ and $O_aO_bO_c$ are equal then $ABC$ is equilateral.

1998 Czech and Slovak Match, 1

Tags: geometry , parallelogram , equal angles

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

Croatia MO (HMO) - geometry, 2019.3

Tags: geometry , concyclic , equal angles , isosceles

Given an isosceles triangle $ABC$ such that $|AB|=|AC|$ . Let $M$ be the midpoint of the segment $BC$ and let $P$ be a point other than $A$ such that $PA\parallel BC$. The points $X$ and $Y$ are located respectively on rays $PB$ and $PC$, so that the point $B$ is between $P$ and $X$, the point $C$ is between $P$ and $Y$ and $\angle PXM=\angle PYM$. Prove that the points $A,P,X$ and $Y$ are concyclic.

Indonesia Regional MO OSP SMA - geometry, 2015.3

Tags: geometry , equal angles , equal segments , equilateral

Given the isosceles triangle $ABC$, where $AB = AC$. Let $D$ be a point in the segment $BC$ so that $BD = 2DC$. Suppose also that point $P$ lies on the segment $AD$ such that: $\angle BAC = \angle BP D$. Prove that $\angle BAC = 2\angle DP C$.

2009 Puerto Rico Team Selection Test, 4

Tags: parallelogram , geometry , equal angles

The point $ M$ is chosen inside parallelogram $ ABCD$. Show that $ \angle MAB$ is congruent to $ \angle MCB$, if and only if $ \angle MBA$ and $ \angle MDA$ are congruent.

2020 Kazakhstan National Olympiad, 3

Tags: geometry , equal segments , equal angles

A point $ N $ is marked on the median $ CM $ of the triangle $ ABC $ so that $ MN \cdot MC = AB ^ 2/4 $. Lines $ AN $ and $ BN $ intersect the circumcircle $ \triangle ABC $ for the second time at points $ P $ and $ Q $, respectively. $ R $ is the point of segment $ PQ $, nearest to $ Q $, such that $ \angle NRC = \angle BNC $. $ S $ is the point of the segment $ PQ $ closest to $ P $ such that $ \angle NSC = \angle ANC $. Prove that $ RN = SN $.

2003 Junior Balkan Team Selection Tests - Romania, 4

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

Let $E$ be the midpoint of the side $CD$ of a square $ABCD$. Consider the point $M$ inside the square such that $\angle MAB = \angle MBC = \angle BME = x$. Find the angle $x$.