Olympiad problems

Ukraine Correspondence MO - geometry, 2014.12

Let $\omega$ be the circumscribed circle of triangle $ABC$, and let $\omega'$ 'be the circle tangent to the side $BC$ and the extensions of the sides $AB$ and $AC$. The common tangents to the circles $\omega$ and $\omega'$ intersect the line $BC$ at points $D$ and $E$. Prove that $\angle BAD = \angle CAE$.

2019 Saudi Arabia JBMO TST, 2

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

In triangle $ABC$ point $M$ is the midpoint of side $AB$, and point $D$ is the foot of altitude $CD$. Prove that $\angle A = 2\angle B$ if and only if $AC = 2MD$

Estonia Open Junior - geometry, 2013.2.3

Tags: right triangle , geometry , equal angles , isosceles

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 $\angle KPL = \angle MPN = 45^o$.

2017 Yasinsky Geometry Olympiad, 2

Tags: 3-dimensional geometry , 3d geometry , geometry , equal angles , tetrahedron , perpendicular

In the tetrahedron $DABC, AB=BC, \angle DBC =\angle DBA$. Prove that $AC \perp DB$.

Kyiv City MO Juniors 2003+ geometry, 2011.8.41

Tags: median , geometry , equal angles

The medians $AL, BM$, and $CN$ are drawn in the triangle $ABC$. Prove that $\angle ANC = \angle ALB$ if and only if $\angle ABM =\angle LAC$. (Veklich Bogdan)

2012 Switzerland - Final Round, 3

Tags: circles , geometry , equal angles , tangent

The circles $k_1$ and $k_2$ intersect at points $D$ and $P$. The common tangent of the two circles on the side of $D$ touches $k_1$ at $A$ and $k_2$ at $B$. The straight line $AD$ intersects $k_2$ for a second time at $C$. Let $M$ be the center of the segment $BC$. Show that $ \angle DPM = \angle BDC$ .

2015 BMT Spring, 7

Tags: equal segments , geometry , equal angles

$X_1, X_2, . . . , X_{2015}$ are $2015$ points in the plane such that for all $1 \le i, j \le 2015$, the line segment $X_iX_{i+1} = X_jX_{j+1}$ and angle $\angle X_iX_{i+1}X_{i+2} = \angle X_jX_{j+1}X_{j+2}$ (with cyclic indices such that $X_{2016} = X_1$ and $X_{2017} = X_2$). Given fixed $X_1$ and $X_2$, determine the number of possible locations for $X_3$.

2003 Oral Moscow Geometry Olympiad, 2

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

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 $CD$?

2021 Durer Math Competition Finals, 13

Tags: trapezoid , equal angles , geometry

The trapezoid $ABCD$ satisfies $AB \parallel CD$, $AB = 70$, $AD = 32$ and $BC = 49$. We also know that $\angle ABC = 3 \angle ADC$. How long is the base $CD$?

2019 Silk Road, 1

Tags: angle chasing , altitude , geometry , equal angles

The altitudes of the acute-angled non-isosceles triangle $ ABC $ intersect at the point $ H $. On the segment $ C_1H $, where $ CC_1 $ is the altitude of the triangle, the point $ K $ is marked. Points $ L $ and $ M $ are the feet of perpendiculars from point $ K $ on straight lines $ AC $ and $ BC $, respectively. The lines $ AM $ and $ BL $ intersect at $ N $. Prove that $ \angle ANK = \angle HNL $.

2018 Switzerland - Final Round, 4

Tags: geometry , concurrency , equal angles

Let $D$ be a point inside an acute triangle $ABC$, such that $\angle BAD = \angle DBC$ and $\angle DAC = \angle BCD$. Let $P$ be a point on the circumcircle of the triangle $ADB$. Suppose $P$ are itself outside the triangle $ABC$. A line through $P$ intersects the ray $BA$ in $X$ and ray $CA$ in $Y$, so that $\angle XPB = \angle PDB$. Show that $BY$ and $CX$ intersect on $AD$.

2015 BMT Spring, 16

Tags: 3d geometry , equal angles , geometry , equal segments

Five points $A, B, C, D$, and $E$ in three-dimensional Euclidean space have the property that $AB = BC = CD = DE = EA = 1$ and $\angle ABC = \angle BCD =\angle CDE = \angle DEA = 90^o$ . Find all possible $\cos(\angle EAB)$.

2010 Junior Balkan Team Selection Tests - Romania, 2

Tags: equal angles , convex quadrilateral , angle , geometry

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 $AB = AD$.

May Olympiad L2 - geometry, 2019.3

Tags: equal angles , geometry

On the sides $AB, BC$ and $CA$ of a triangle $ABC$ are located the points $P, Q$ and $R$ respectively, such that $BQ = 2QC, CR = 2RA$ and $\angle PRQ = 90^o$. Show that $\angle APR =\angle RPQ$.

1986 All Soviet Union Mathematical Olympiad, 428

Tags: line , geometry , equal angles

A line is drawn through the $A$ vertex of triangle $ABC$ with $|AB|\ne|AC|$. Prove that the line can not contain more than one point $M$ such, that $M$ is not a triangle vertex, and $\angle ABM = \angle ACM$. What lines do not contain such a point $M$ at all?

2011 Belarus Team Selection Test, 2

Tags: equal angles , geometry

Two different points $X,Y$ are marked on the side $AB$ of a triangle $ABC$ so that $\frac{AX \cdot BX}{CX^2}=\frac{AY \cdot BY}{CY^2}$ . Prove that $\angle ACX=\angle BCY$. I.Zhuk

2013 Switzerland - Final Round, 3

Tags: projection , equal angles , geometry

Let $ABCD$ be a cyclic quadrilateral with $\angle ADC = \angle DBA$. Furthermore, let $E$ be the projection of $A$ on $BD$. Show that $BC = DE - BE$ .

2022 OMpD, 2

Tags: angle , equal angles , geometry , rectangle

Let $ABCD$ be a rectangle. The point $E$ lies on side $ \overline{AB}$ and the point $F$ is lies side $ \overline{AD}$, such that $\angle FEC=\angle CEB$ and $\angle DFC=\angle CFE$. Determine the measure of the angle $\angle FCE$ and the ratio $AD/AB$.

2013 Czech-Polish-Slovak Junior Match, 4

Tags: angle chasing , geometry , equal angles , circumcircle

Let $ABCD$ be a convex quadrilateral with $\angle DAB =\angle ABC =\angle BCD > 90^o$. The circle circumscribed around the triangle $ABC$ intersects the sides $AD$ and $CD$ at points $K$ and $L$, respectively, different from any vertex of the quadrilateral $ABCD$ . Segments $AL$ and $CK$ intersect at point $P$. Prove that $\angle ADB =\angle PDC$.

2018 Regional Olympiad of Mexico West, 3

Tags: similarity , similar triangles , equal angles , geometry

A scalene acute triangle $ABC$ is drawn on the plane, in which $BC$ is the longest side. Points $P$ and $D$ are constructed, the first inside $ABC$ and the second outside, so that $\angle ABC = \angle CBD$, $\angle ACP = \angle BCD$ and that the area of triangle $ABC$ is equal to the area of quadrilateral $BPCD$. Prove that triangles $BCD$ and $ACP$ are similar.

2005 JBMO Shortlist, 4

Tags: equal angles , isosceles , geometry

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$.

2021 Poland - Second Round, 2

Tags: equal segments , equal angles , parallelogram , 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 $AO = OC$.

2019 Switzerland - Final Round, 7

Tags: angle , equal segments , equal angles , geometry

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 $\angle ACB = 3 \angle DCB$.

2010 Junior Balkan Team Selection Tests - Romania, 4

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

Let $ABC$ be an isosceles triangle with $AB = AC$ and let $n$ be a natural number, $n>1$. On the side $AB$ we consider the point $M$ such that $n \cdot AM = AB$. On the side $BC$ we consider the points $P_1, P_2, ....., P_ {n-1}$ such that $BP_1 = P_1P_2 = .... = P_ {n-1} C = \frac{1}{n} BC$. Show that: $\angle {MP_1A} + \angle {MP_2A} + .... + \angle {MP_ {n-1} A} = \frac{1} {2} \angle {BAC}$.

2018 Yasinsky Geometry Olympiad, 1

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

In the triangle $ABC$, $AD$ is altitude, $M$ is the midpoint of $BC$. It is known that $\angle BAD = \angle DAM = \angle MAC$. Find the values of the angles of the triangle $ABC$