Olympiad problems

2022 Dutch BxMO TST, 2

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 $\angle BXD = \angle CYD$.

2008 Singapore Senior Math Olympiad, 1

Tags: trapezium , trapezoid , equal angles , geometry

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 $\angle BLA = \angle CKD$.

2000 Mexico National Olympiad, 6

Tags: geometry , equal angles

Let $ABC$ be a triangle with $\angle B > 90^o$ such that there is a point $H$ on side $AC$ with $AH = BH$ and BH perpendicular to $BC$. Let $D$ and $E$ be the midpoints of $AB$ and $BC$ respectively. A line through $H$ parallel to $AB$ cuts $DE$ at $F$. Prove that $\angle BCF = \angle ACD$.

2004 Oral Moscow Geometry Olympiad, 3

Tags: equal angles , locus , geometry

Given a square $ABCD$. Find the locus of points $M$ such that $\angle AMB = \angle CMD$.

2023 China Western Mathematical Olympiad, 3

Tags: geometry , equal segments , equal angles

In $\triangle ABC$, points $P,Q$ satisfy $\angle PBC = \angle QBA$ and $\angle PCB = \angle QCA$, $D$ is a point on $BC$ such that $\angle PDB=\angle QDC$. Let $X,Y$ be the reflections of $A$ with respect to lines $BP$ and $CP$, respectively. Prove that $DX=DY$. [img]https://cdn.artofproblemsolving.com/attachments/a/7/f208f1651afc0fef9eef4c68ba36bf77556058.jpg[/img]

2010 Grand Duchy of Lithuania, 4

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

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 $BK = 2DC$.

2010 Junior Balkan Team Selection Tests - Romania, 4

Tags: equal segments , equal angles , isosceles , angle , 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: $\angle {MP_1A} + \angle {MP_2A} + .... + \angle {MP_ {n-1} A} = \frac{1} {2} \angle {BAC}$.

2022 International Zhautykov Olympiad, 3

Tags: geometry , parallelogram , equal angles

In parallelogram $ABCD$ with acute angle $A$ a point $N$ is chosen on the segment $AD$, and a point $M$ on the segment $CN$ so that $AB = BM = CM$. Point $K$ is the reflection of $N$ in line $MD$. The line $MK$ meets the segment $AD$ at point $L$. Let $P$ be the common point of the circumcircles of $AMD$ and $CNK$ such that $A$ and $P$ share the same side of the line $MK$. Prove that $\angle CPM = \angle DPL$.

1998 All-Russian Olympiad Regional Round, 9.2

Tags: geometry , equal angles , circles

Two circles intersect at points $P$ and $Q$. The straight line intersects these circles at points $A$, $B$, $C$, $D$, as shown in fig. . Prove that $\angle APB = \angle CQD$. [img]https://cdn.artofproblemsolving.com/attachments/1/a/a581e11be68bbb628db5b5b8e75c7ff6e196c5.png[/img]

Ukraine Correspondence MO - geometry, 2008.7

Tags: equal angles , equal segments , geometry

On the sides $AC$ and $AB$ of the triangle $ABC$, the points $D$ and $E$ were chosen such that $\angle ABD =\angle CBD$ and $3 \angle ACE = 2\angle BCE$. Let $H$ be the point of intersection of $BD$ and $CE$, and $CD = DE = CH$. Find the angles of triangle $ABC$.

2015 BMT Spring, 16

Tags: geometry , 3d geometry , equal angles , 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 $\cos(\angle EAB)$.

2022 Regional Olympiad of Mexico West, 4

Tags: ratio , geometry , fixed , equal angles

Prove that in all triangles $\vartriangle ABC$ with $\angle A = 2 \angle B$ it holds that, if $D$ is the foot of the perpendicular from $C$ to the perpendicular bisector of $AB$, $\frac{AC}{DC}$ is constant for any value of $\angle B$.

2015 Romania National Olympiad, 3

Tags: angle bisector , midpoint , equal angles , geometry

In the convex quadrilateral $ABCD$ we have that $\angle BCD = \angle ADC \ge 90 ^o$. The bisectors of $\angle BAD$ and $\angle ABC$ intersect in $M$. Prove that if $M \in CD$, then $M$ is the middle of $CD$.

Swiss NMO - geometry, 2013.3

Tags: geometry , projection , equal angles

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

Swiss NMO - geometry, 2016.1

Tags: geometry , equal angles , circumcircle , angle

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 $\angle BAP$.

Swiss NMO - geometry, 2016.5

Tags: equal angles , geometry , concurrency

Let $ABC$ be a right triangle with $\angle ACB = 90^o$ and M the center of $AB$. Let $G$ br any point on the line $MC$ and $P$ a point on the line $AG$, such that $\angle CPA = \angle BAC$ . Further let $Q$ be a point on the straight line $BG$, such that $\angle BQC = \angle CBA$ . Show that the circles of the triangles $AQG$ and $BPG$ intersect on the segment $AB$.

2004 District Olympiad, 4

Tags: geometry , equal angles , isosceles , right triangle , angle

Consider the isosceles right triangle $ABC$ ($AB = AC$) and the points $M, P \in [AB]$ so that $AM = BP$. Let $D$ be the midpoint of the side $BC$ and $R, Q$ the intersections of the perpendicular from $A$ on$ CM$ with $CM$ and $BC$ respectively. Prove that a) $\angle AQC = \angle PQB$ b) $\angle DRQ = 45^o$

Novosibirsk Oral Geo Oly IX, 2020.4

Tags: equal angles , square

Points $E$ and $F$ are the midpoints of sides $BC$ and $CD$ of square $ABCD$, respectively. Lines $AE$ and $BF$ meet at point $P$. Prove that $\angle PDA = \angle AED$.

2010 Junior Balkan Team Selection Tests - Romania, 2

Tags: midpoint , equal angles , angle , geometry

Let $ABC$ be a triangle and $D, E, F$ the midpoints of the sides $BC, CA, AB$ respectively. Show that $\angle DAC = \angle ABE$ if and only if $\angle AFC = \angle BDA$

2009 IMAR Test, 3

Tags: angle chasing , convex quadrilateral , equal angles , angle , geometry

Consider a convex quadrilateral $ABCD$ with $AB=CB$ and $\angle ABC +2 \angle CDA = \pi$ and let $E$ be the midpoint of $AC$. Show that $\angle CDE =\angle BDA$. Paolo Leonetti

Oliforum Contest III 2012, 3

Tags: hexagon , equal angles , geometry

Show that if equiangular hexagon has sides $a, b, c, d, e, f$ in order then $a - d = e - b = c - f$.

2020 Czech-Austrian-Polish-Slovak Match, 6

Tags: equal angles , fixed , fixed point , geometry

Let $ABC$ be an acute triangle. Let $P$ be a point such that $PB$ and $PC$ are tangent to circumcircle of $ABC$. Let $X$ and $Y$ be variable points on $AB$ and $AC$, respectively, such that $\angle XPY = 2\angle BAC$ and $P$ lies in the interior of triangle $AXY$. Let $Z$ be the reflection of $A$ across $XY$. Prove that the circumcircle of $XYZ$ passes through a fixed point. (Dominik Burek, Poland)

2022 China Northern MO, 1

Tags: isosceles right triangle , equal angles , geometry

As shown in the figure, given $\vartriangle ABC$ with $AB \perp AC$, $AB=BC$, $D$ is the midpoint of the side $AB$, $DF\perp DE$, $DE=DF$ and $BE \perp EC$. Prove that $\angle AFD= \angle CEF$. [img]https://cdn.artofproblemsolving.com/attachments/9/2/f16a8c8c463874f3ccb333d91cdef913c34189.png[/img]

Estonia Open Junior - geometry, 2008.2.2

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

In a right triangle $ABC$, $K$ is the midpoint of the hypotenuse $AB$ and $M$ such a point on the $BC$ that $| B M | = 2 | MC |$. Prove that $\angle MAB = \angle MKC$.

2012 Romania National Olympiad, 2

Tags: geometry , equal angles , right triangle

Let $ABC$ be a triangle with right $\angle A$. Consider points $D \in (AC)$ and $E \in (BD)$ such that $\angle ABC = \angle ECD = \angle CED$. Prove that $BE = 2 \cdot AD$