Olympiad problems

2012 Danube Mathematical Competition, 3

Let $ABC$ be a triangle with $\angle BAC = 90^o$. Angle bisector of the $\angle CBA$ intersects the segment $(AB)$ at point $E$. If there exists $D \in (CE)$ so that $\angle DAC = \angle BDE =x^o$ , calculate $x$.

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 $MNK$.

VMEO II 2005, 6

Tags: equal angles , geometry

For a given cyclic quadrilateral $ABCD$, let $I$ be a variable point on the diagonal $AC$ such that $I$ and $A$ are on the same side of the diagonal $BD$. Assume $E,F$ lie on the diagonal $BD$ such that $IE\parallel AB$ and $IF\parallel AD$. Show that $\angle BIE =\angle DCF $

2015 Sharygin Geometry Olympiad, P17

Tags: simson line , geometry , equal angles

Let $O$ be the circumcenter of a triangle $ABC$. The projections of points $D$ and $X$ to the sidelines of the triangle lie on lines $\ell $ and $L $ such that $\ell // XO$. Prove that the angles formed by $L$ and by the diagonals of quadrilateral $ABCD$ are equal.

Champions Tournament Seniors - geometry, 2005.2

Tags: circumcircle , symmedian , equal angles , angle , geometry

Given a triangle $ABC$, the line passing through the vertex $A$ symmetric to the median $AM$ wrt the line containing the bisector of the angle $\angle BAC$ intersects the circle circumscribed around the triangle $ABC$ at points $A$ and $K$. Let $L$ be the midpoint of the segment $AK$. Prove that $\angle BLC=2\angle BAC$.

2019 Estonia Team Selection Test, 7

Tags: altitude , geometry , equal angles

An acute-angled triangle $ABC$ has two altitudes $BE$ and $CF$. The circle with diameter $AC$ intersects the segment $BE$ at point $P$. A circle with diameter $AB$ intersects the segment $CF$ at point $Q$ and the extension of this altitude at point $Q'$. Prove that $\angle PQ'Q = \angle PQB$.

2008 Postal Coaching, 5

Tags: geometry , incircle , cyclic quadrilateral , equal angles , angle

A convex quadrilateral $ABCD$ is given. There rays $BA$ and $CD$ meet in $P$, and the rays $BC$ and $AD$ meet in $Q$. Let $H$ be the projection of $D$ on $PQ$. Prove that $ABCD$ is cyclic if and only if the angle between the rays beginning at $H$ and tangent to the incircle of triangle $ADP$ is equal to the angle between the rays beginning at $H$ and tangent to the incircle of triangle $CDQ$. Also find out whether $ABCD$ is inscribable or circumscribable and justify.

2015 Oral Moscow Geometry Olympiad, 4

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

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 Silk Road, 1

Tags: altitude , geometry , angle chasing , 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 $.

1996 North Macedonia National Olympiad, 1

Tags: geometry , parallelogram , equal angles , perpendicular

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 $\angle DEA = \angle CEB$.

Kyiv City MO Juniors 2003+ geometry, 2021.8.4

Tags: equal angles , angle , geometry

Let $BM$ be the median of the triangle $ABC$, in which $AB> BC$. Point $P$ is chosen so that $AB \parallel PC$ and $PM \perp BM$. Prove that $\angle ABM = \angle MBP$. (Mikhail Standenko)

2016 Irish Math Olympiad, 4

Tags: geometry , equal angles , circles

Let $ABC$ be a triangle with $|AC| \ne |BC|$. Let $P$ and $Q$ be the intersection points of the line $AB$ with the internal and external angle bisectors at $C$, so that $P$ is between $A$ and $B$. Prove that if $M$ is any point on the circle with diameter $PQ$, then $\angle AMP = \angle BMP$.

2007 Sharygin Geometry Olympiad, 14

Tags: diagonal , geometry , equal angles , trapezoid

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 $\angle PBA = \angle DBC$.

2013 Czech-Polish-Slovak Junior Match, 4

Tags: geometry , circumcircle , equal angles , angle chasing

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 $\angle ADB =\angle PDC$.

Champions Tournament Seniors - geometry, 2018.3

Tags: geometry , parallelogram , equal angles , equal segments

The vertex $F$ of the parallelogram $ACEF$ lies on the side $BC$ of parallelogram $ABCD$. It is known that $AC = AD$ and $AE = 2CD$. Prove that $\angle CDE = \angle BEF$.

2019 Romania National Olympiad, 3

Tags: geometry , equal segments , equal angles , angle

Let $ABC$ be a triangle in which $\angle ABC = 45^o$ and $\angle BAC > 90^o$. Let $O$ be the midpoint of the side $[BC]$. Consider the point $M \in (AC)$ such that $\angle COM =\angle CAB$. Perpendicular from $M$ on $AC$ intersects line $AB$ at point $P$. a) Find the measure of the angle $\angle BCP$. b) Show that if $\angle BAC = 105^o$, then $PB = 2MO$.

Novosibirsk Oral Geo Oly IX, 2020.6

Tags: tangent , geometry , equal angles

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

2015 BMT Spring, 7

Tags: geometry , equal angles , equal segments

$X_1, X_2, . . . , X_{2015}$ 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 $X_3$.

2021 Yasinsky Geometry Olympiad, 3

Tags: equal angles , geometry , equal segments , angle

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)

2015 Ukraine Team Selection Test, 1

Tags: angle bisector , equal angles , circumcircle , geometry

Let $O$ be the circumcenter of the triangle $ABC, A'$ be a point symmetric of $A$ wrt line $BC, X$ is an arbitrary point on the ray $AA'$ ($X \ne A$). Angle bisector of angle $BAC$ intersects the circumcircle of triangle $ABC$ at point $D$ ($D \ne A$). Let $M$ be the midpoint of the segment $DX$. A line passing through point $O$ parallel to $AD$, intersects $DX$ at point $N$. Prove that angles $BAM$ and $CAN$ angles are equal.

2021 Durer Math Competition Finals, 13

Tags: geometry , trapezoid , equal angles

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 $CD$?

May Olympiad L2 - geometry, 2019.3

Tags: geometry , equal angles

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 $\angle APR =\angle RPQ$.

2013 Costa Rica - Final Round, G2

Tags: parallelogram , equal angles , geometry

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

2015 Dutch IMO TST, 1

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

In a quadrilateral $ABCD$ we have $\angle A = \angle C = 90^o$. Let $E$ be a point in the interior of $ABCD$. Let $M$ be the midpoint of $BE$. Prove that $\angle ADB = \angle EDC$ if and only if $|MA| = |MC|$.

Denmark (Mohr) - geometry, 2012.5

Tags: geometry , hexagon , equal angles , equilateral , equal segments

In the hexagon $ABCDEF$, all angles are equally large. The side lengths satisfy $AB = CD = EF = 3$ and $BC = DE = F A = 2$. The diagonals $AD$ and $CF$ intersect each other in the point $G$. The point $H$ lies on the side $CD$ so that $DH = 1$. Prove that triangle $EGH$ is equilateral.