Olympiad problems

Kyiv City MO Juniors Round2 2010+ geometry, 2020.8.2

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

Estonia Open Junior - geometry, 2013.2.3

Tags: geometry , right triangle , isosceles , equal angles

In an isosceles right triangle $ABC$ the right angle is at vertex $C$. On the side $AC$ points $K, L$ and on the side $BC$ points $M, N$ are chosen so that they divide the corresponding side into three equal segments. Prove that there is exactly one point $P$ inside the triangle $ABC$ such that $\angle KPL = \angle MPN = 45^o$.

2012 Dutch IMO TST, 4

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

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 $\angle NML = 90^o$.

Cono Sur Shortlist - geometry, 2018.G3

Tags: equal angles , geometry , pentagon , equal segments

Consider the pentagon $ABCDE$ such that $AB = AE = x$, $AC = AD = y$, $\angle BAE = 90^o$ and $\angle ACB = \angle ADE = 135^o$. It is known that $C$ and $D$ are inside the triangle $BAE$. Determine the length of $CD$ in terms of $x$ and $y$.

2019 Istmo Centroamericano MO, 3

Tags: equal angles , tangent circle , geometry

Let $ABC$ be an acute triangle, with $AB <AC$. Let $M$ be the midpoint of $AB$, $H$ the foot of the altitude from $A$, and $Q$ be point on side $AC$ such that $\angle ABQ = \angle BCA$. Show that the circumcircles of the triangles $ABQ$ and $BHM$ are tangent.

2004 Junior Balkan Team Selection Tests - Romania, 1

Tags: geometry , congruent , circumcircle , equal angles , incircle

Let $ABC$ be a triangle inscribed in the circle $K$ and consider a point $M$ on the arc $BC$ that do not contain $A$. The tangents from $M$ to the incircle of $ABC$ intersect the circle $K$ at the points $N$ and $P$. Prove that if $\angle BAC = \angle NMP$, then triangles $ABC$ and $MNP$ are congruent. Valentin Vornicu [hide= about Romania JBMO TST 2004 in aops]I found the Romania JBMO TST 2004 links [url=https://artofproblemsolving.com/community/c6h5462p17656]here [/url] but they were inactive. So I am asking for solution for the only geo I couldn't find using search. The problems were found [url=https://artofproblemsolving.com/community/c6h5135p16284]here[/url].[/hide]

1974 Chisinau City MO, 82

Tags: equal angles , angle , geometry

Is there a moment in a day when three hands - hour, minute and second - of a clock running correctly form angles of $120^o$ in pairs?

2011 Argentina National Olympiad, 3

Tags: equal angles , right triangle , geometry

Let $ABC$ be a triangle with $\angle A = 90^o, \angle B = 75^o$ and $AB = 2$. The points $P$ and $Q$ on the sides $AC$ and $BC$ respectively are such that $\angle APB = \angle CPQ$ and $\angle BQA = \angle CQP$ . Calculate the measurement of the segment $QA $.

Kyiv City MO Juniors 2003+ geometry, 2020.9.4

Tags: geometry , circumcircle , fixed , fixed point , equal angles

Let the point $D$ lie on the arc $AC$ of the circumcircle of the triangle $ABC$ ($AB < BC$), which does not contain the point $B$. On the side $AC$ are selected an arbitrary point $X$ and a point $X'$ for which $\angle ABX= \angle CBX'$. Prove that regardless of the choice of the point $X$, the circle circumscribed around $\vartriangle DXX'$, passes through a fixed point, which is different from point $D$. (Nikolaev Arseniy)

2015 Dutch BxMO/EGMO TST, 4

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

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 $\angle DEP = \angle DFP$.

1976 Kurschak Competition, 1

Tags: geometry , equal angles , parallelogram

$ABCD$ is a parallelogram. $P$ is a point outside the parallelogram such that angles $\angle PAB$ and $\angle PCB$ have the same value but opposite orientation. Show that $\angle APB = \angle DPC$.

2019 Philippine MO, 4

Tags: concyclic , equal angles , circumcircle , geometry

In acute triangle $ABC $with $\angle BAC > \angle BCA$, let $P$ be the point on side $BC$ such that $\angle PAB = \angle BCA$. The circumcircle of triangle $AP B$ meets side $AC$ again at $Q$. Point $D$ lies on segment $AP$ such that $\angle QDC = \angle CAP$. Point $E$ lies on line $BD$ such that $CE = CD$. The circumcircle of triangle $CQE$ meets segment $CD$ again at $F$, and line $QF$ meets side $BC$ at $G$. Show that $B, D, F$, and $G$ are concyclic

2016 Switzerland - Final Round, 1

Tags: equal angles , circumcircle , angle , geometry

Let $ABC$ be a triangle with $\angle BAC = 60^o$. Let $E$ be the point on the side $BC$ , such that $2 \angle BAE = \angle ACB$ . Let $D$ be the second intersection of $AB$ and the circumcircle of the triangle $AEC$ and $P$ be the second intersection of $CD$ and the circumcircle of the triangle $DBE$. Calculate the angle $\angle BAP$.

2002 Switzerland Team Selection Test, 2

Tags: parallelogram , equal angles , geometry

A point$ O$ inside a parallelogram $ABCD$ satisfies $\angle AOB + \angle COD = \pi$. Prove that $\angle CBO = \angle CDO$.

2010 Junior Balkan Team Selection Tests - Romania, 2

Tags: equal angles , geometry , convex quadrilateral , angle

Let $ABCD$ be a convex quadrilateral with $\angle BCD= 120^o, \angle {CBA} = 45^o, \angle {CBD} = 15^o$ and $\angle {CAB} = 90^o$. Show that $AB = AD$.

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 $\angle BCQ=\frac{1}{2}\angle AQP$.

2010 District Olympiad, 4

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

We consider the quadrilateral $ABCD$, with $AD = CD = CB$ and $AB \parallel CD$. Points $E$ and $F$ belong to the segments $CD$ and $CB$ so that angles $\angle ADE = \angle AEF$. Prove that: a) $4CF \le CB$ , b) if $4CF = CB$, then $AE$ is the bisector of the angle $\angle DAF$.

2010 Saudi Arabia IMO TST, 2

Tags: ratio , geometry , equal angles

Points $M$ and $N$ are considered in the interior of triangle $ABC$ such that $\angle MAB = \angle NAC$ and $\angle MBA = \angle NBC$. Prove that $$\frac{AM \cdot AN}{AB \cdot AC}+ \frac{BM\cdot BN}{BA \cdot BC}+ \frac{CM \cdot CN }{CA \cdot CB}=1$$

2018 Junior Balkan Team Selection Tests - Romania, 3

Tags: geometry , geometric transformation , reflection , circumcircle , equal angles

Let $ABC$ be a triangle with $AB > AC$. Point $P \in (AB)$ is such that $\angle ACP = \angle ABC$. Let $D$ be the reflection of $P$ into the line $AC$ and let $E$ be the point in which the circumcircle of $BCD$ meets again the line $AC$. Prove that $AE = AC$.

2015 Dutch BxMO/EGMO TST, 4

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

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 $\angle DEP = \angle DFP$.

OMMC POTM, 2022 3

Tags: equal angles , geometry

Define acute triangle $ABC$ with circumcircle $\omega.$ Let $Q$ be the midpoint of minor arc $BC$ in $\omega$ and let $Q'$ be the reflection of $Q$ over $BC.$ If the circle with diameter $BC$ is tangent to the external angle bisector of $\angle BAC$ at $P,$ show $\angle BPQ' = \angle CPA.$ [i]Proposed by Evan Chang (squareman), USA[/i] [img]https://cdn.artofproblemsolving.com/attachments/8/1/6333de3458f913477c75882896a40a48cd7ef7.png[/img]

2001 239 Open Mathematical Olympiad, 5

Tags: geometry , circles , equal angles , tangent circle

The circles $ S_1 $ and $ S_2 $ intersect at points $ A $ and $ B $. Circle $ S_3 $ externally touches $ S_1 $ and $ S_2 $ at points $ C $ and $ D $ respectively. Let $ K $ be the midpoint of the chord cut by the line $ AB $ on circles $ S_3 $. Prove that $ \angle CKA = \angle DKA $.

Kyiv City MO Seniors 2003+ geometry, 2005.10.4

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

In a right triangle $ABC $ with a right angle $\angle C $, n the sides $AC$ and $AB$, the points $M$ and $N$ are selected, respectively, that $CM = MN$ and $\angle MNB = \angle CBM$. Let the point $K$ be the projection of the point $C $ on the segment $MB $. Prove that the line $NK$ passes through the midpoint of the segment $BC$. (Alex Klurman)

1990 All Soviet Union Mathematical Olympiad, 512

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

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.

2018 Ukraine Team Selection Test, 10

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

Let $ABC$ be a triangle with $AH$ altitude. The point $K$ is chosen on the segment $AH$ as follows such that $AH =3KH$. Let $O$ be the center of the circle circumscribed around by triangle $ABC, M$ and $N$ be the midpoints of $AC$ and AB respectively. Lines $KO$ and $MN$ intersect at the point $Z$, a perpendicular to $OK$ passing through point $Z$ intersects lines $AC$ and $AB$ at points $X$ and $Y$ respectively. Prove that $\angle XKY =\angle CKB$.