Olympiad problems

2003 Oral Moscow Geometry Olympiad, 3

Inside the segment $AC$, an arbitrary point $B$ is selected and circles with diameters $AB$ and $BC$ are constructed. Points $M$ and $L$ are chosen on the circles (in one half-plane with respect to $AC$), respectively, so that $\angle MBA = \angle LBC$. Points $K$ and $F$ are marked, respectively, on rays $BM$ and $BL$ so that $BK = BC$ and $BF = AB$. Prove that points $M, K, F$ and $L$ lie on the same circle.

2020 Greece National Olympiad, 2

Tags: parallelogram , geometry , concurrency , equal angles

Given a line segment $AB$ and a point $C$ lies inside it such that $AB=3 \cdot AC$ . Construct a parallelogram $ACDE$ such that $AC=DE=CE>AR$. Let $Z$ be a point on $AC$ such that $\angle AEZ=\angle ACE =\omega$. Prove that the line passing through point $B$ and perpendicular on side $EC$, and the line passing through point $D$ and perpendicular on side $AB$, intersect on point , let it be $K$, lying on line $EZ$.

1975 Spain Mathematical Olympiad, 5

Tags: geometry , construction , equal angles

In the plane we have a line $r$ and two points $A$ and $B$ outside the line and in the same half plane. Determine a point $M$ on the line such that the angle of $r$ with $AM$ is double that of $r$ with $BM$. (Consider the smaller angle of two lines of the angles they form).

2021 Sharygin Geometry Olympiad, 8.7

Tags: geometry , bisects segment , equal angles

Let $ABCDE$ be a convex pentagon such that angles $CAB$, $BCA$, $ECD$, $DEC$ and $AEC$ are equal. Prove that $CE$ bisects $BD$.

Geometry Mathley 2011-12, 13.1

Tags: fixed , circles , geometry , equal angles

Let $ABC$ be a triangle with no right angle, $E$ on the line $BC$ such that $\angle AEB = \angle BAC$ and $\Delta_A$ the perpendicular to $BC$ at $E$. Let the circle $\gamma$ with diameter $BC$ intersect $BA$ again at $D$. For each point $M$ on $\gamma$ ($M$ is distinct from $B$), the line $BM$ meets $\Delta_A$ at $M'$ and the line $AM$ meets $\gamma$ again at $M''$. (a) Show that $p(A) = AM' \times DM''$ is independent of the chosen $M$. (b) Keeping $B,C$ fixed, and let $A$ vary. Show that $\frac{p(A)}{d(A,\Delta_A)}$ is independent of $A$. Michel Bataille

2005 Estonia National Olympiad, 2

Tags: polygon , convex , equal angles , geometry

Consider a convex $n$-gon in the plane with $n$ being odd. Prove that if one may find a point in the plane from which all the sides of the $n$-gon are viewed at equal angles, then this point is unique. (We say that segment $AB$ is viewed at angle $\gamma$ from point $O$ iff $\angle AOB =\gamma$ .)

2017 Singapore Junior Math Olympiad, 3

Tags: equal segments , angle , equal angles , isosceles

Let $ABC$ be a triangle with $AB=AC$. Let $D$ be a point on $BC$, and $E$ a point on $AD$ such that $\angle BED=\angle BAC=2\angle CED$. Prove that $BD=2CD$.

2002 District Olympiad, 4

Tags: rectangle , square , geometry , equal angles

Given the rectangle $ABCD$. The points $E ,F$ lie on the segments $(BC) , (DC)$ respectively, such that $\angle DAF = \angle FAE$. Proce that if $DF + BE = AE$ then $ABCD$ is square.

1994 Austrian-Polish Competition, 9

Tags: geometry , construction , equal angles

On the plane are given four distinct points $A,B,C,D$ on a line $g$ in this order, at the mutual distances $AB = a, BC = b, CD = c$. (a) Construct (if possible) a point $P$ outside line $g$ such that $\angle APB =\angle BPC =\angle CPD$. (b) Prove that such a point $P$ exists if and only if $ (a+b)(b+c) < 4ac$

2010 Sharygin Geometry Olympiad, 4

Tags: circles , common tangent , equal angles , geometry , tangent

Circles $\omega_1$ and $\omega_2$ inscribed into equal angles $X_1OY$ and $Y OX_2$ touch lines $OX_1$ and $OX_2$ at points $A_1$ and $A_2$ respectively. Also they touch $OY$ at points $B_1$ and $B_2$. Let $C_1$ be the second common point of $A_1B_2$ and $\omega_1, C_2$ be the second common point of $A_2B_1$ and $\omega_2$. Prove that $C_1C_2$ is the common tangent of two circles.

2012 Korea Junior Math Olympiad, 5

Tags: equal segments , tangent , equal angles , geometry , cyclic quadrilateral

Let $ABCD$ be a cyclic quadrilateral inscirbed in a circle $O$ ($AB> AD$), and let $E$ be a point on segment $AB$ such that $AE = AD$. Let $AC \cap DE = F$, and $DE \cap O = K(\ne D)$. The tangent to the circle passing through $C,F,E$ at $E$ hits $AK$ at $L$. Prove that $AL = AD$ if and only if $\angle KCE = \angle ALE$.

Durer Math Competition CD Finals - geometry, 2021.C3

Tags: isosceles , equal angles , geometry

In the isosceles triangle $ABC$ we have $AC = BC$. Let $X$ be an arbitrary point of the segment $AB$. The line parallel to $BC$ and passing through $X$ intersects the segment $AC$ in $N$, and the line parallel to $AC$ and passing through $BC$ intersects the segment $BC$ in $M$. Let $k_1$ be the circle with center $N$ and radius $NA$. Similarly, let $k_2$ be the circle with center $M$ and radius $MB$. Let $T$ be the intersection of the circles $k_1$ and $k_2$ different from $X$. Show that the angles $\angle NCM$ and $\angle NTM$ are equal.

Kyiv City MO Juniors 2003+ geometry, 2016.8.51

Tags: equal segments , geometry , equal angles

In the quadrilateral $ABCD$, shown in fig. , the equations are true: $\angle ABC = \angle BCD$ and $2AB = CD$. On the side $BC$, a point $X$ is selected such that $\angle BAX = \angle CDA$. Prove that $AX = AD$. [img]https://cdn.artofproblemsolving.com/attachments/2/9/0884eb311d1e40300c1e5980fd53eaadfa7a25.png[/img]

2011 Danube Mathematical Competition, 1

Tags: parallelogram , quadrilateral , equal angles , geometry

Let $ABCM$ be a quadrilateral and $D$ be an interior point such that $ABCD$ is a parallelogram. It is known that $\angle AMB =\angle CMD$. Prove that $\angle MAD =\angle MCD$.

2020 Dutch IMO TST, 1

Tags: equal angles , incenter , geometry

In acute-angled triangle $ABC, I$ is the center of the inscribed circle and holds $| AC | + | AI | = | BC |$. Prove that $\angle BAC = 2 \angle ABC$.

2003 District Olympiad, 4

Tags: geometry , angle , 3d geometry , tetrahedron , equal angles

a) Let $MNP$ be a triangle such that $\angle MNP> 60^o$. Show that the side $MP$ cannot be the smallest side of the triangle $MNP$. b) In a plane the equilateral triangle $ABC$ is considered. The point $V$ that does not belong to the plane $(ABC)$ is chosen so that $\angle VAB = \angle VBC = \angle VCA$. Show that if $VA = AB$, the tetrahedron $VABC$ is regular. Valentin Vornicu

2015 Caucasus Mathematical Olympiad, 5

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

Let $AA_1$ and $CC_1$ be the altitudes of the acute-angled triangle $ABC$. Let $K,L$ and $M$ be the midpoints of the sides $AB,BC$ and $CA$ respectively. Prove that if $\angle C_1MA_1 =\angle ABC$, then $C_1 K = A_1L$.

2017 Ukrainian Geometry Olympiad, 1

Tags: geometry , angle chasing , midpoint , equal angles

In the triangle $ABC$, ${{A}_{1}}$ and ${{C}_{1}} $ are the midpoints of sides $BC $ and $AB$ respectively. Point $P$ lies inside the triangle. Let $\angle BP {{C}_{1}} = \angle PCA$. Prove that $\angle BP {{A}_{1}} = \angle PAC $.

Russian TST 2019, P3

Tags: circles , equal angles , geometry

Let $\Omega$ be the circumcircle of an acute-angled triangle $ABC$. A point $D$ is chosen on the internal bisector of $\angle ACB$ so that the points $D$ and $C$ are separated by $AB$. A circle $\omega$ centered at $D$ is tangent to the segment $AB$ at $E$. The tangents to $\omega$ through $C$ meet the segment $AB$ at $K$ and $L$, where $K$ lies on the segment $AL$. A circle $\Omega_1$ is tangent to the segments $AL, CL$, and also to $\Omega$ at point $M$. Similarly, a circle $\Omega_2$ is tangent to the segments $BK, CK$, and also to $\Omega$ at point $N$. The lines $LM$ and $KN$ meet at $P$. Prove that $\angle KCE = \angle LCP$. Poland

Ukraine Correspondence MO - geometry, 2008.7

Tags: equal segments , equal angles , 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 $ABC$.

VMEO II 2005, 6

Tags: geometry , equal angles

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 $

2009 IMAR Test, 3

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

2013 BMT Spring, 15

Tags: geometry , equal angles

Let $ABCD$ be a convex quadrilateral with $\angle ABD = \angle BCD$, $AD = 1000$, $BD = 2000$, $BC = 2001$, and $DC = 1999$. Point $E$ is chosen on segment $DB$ such that $\angle ABD = \angle ECD$. Find $AE$.

1998 All-Russian Olympiad Regional Round, 9.2

Tags: geometry , circles , equal angles

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]

2014 Iranian Geometry Olympiad (junior), P4

Tags: geometry , equal angles

In a triangle ABC we have $\angle C = \angle A + 90^o$. The point $D$ on the continuation of $BC$ is given such that $AC = AD$. A point $E$ in the side of $BC$ in which $A$ doesn’t lie is chosen such that $\angle EBC = \angle A, \angle EDC = \frac{1}{2} \angle A$ . Prove that $\angle CED = \angle ABC$. by Morteza Saghafian