Olympiad problems

2012 Sharygin Geometry Olympiad, 6

Let $ABC$ be an isosceles triangle with $BC = a$ and $AB = AC = b$. Segment $AC$ is the base of an isosceles triangle $ADC$ with $AD = DC = a$ such that points $D$ and $B$ share the opposite sides of AC. Let $CM$ and $CN$ be the bisectors in triangles $ABC$ and $ADC$ respectively. Determine the circumradius of triangle $CMN$. (M.Rozhkova)

2021 Yasinsky Geometry Olympiad, 2

Tags: isosceles , geometry

In the triangle $ABC$, it is known that $AB = BC = 20$ cm, and $AC = 24$ cm. The point $M$ lies on the side $BC$ and is equidistant from sides $AB$ and $AC$. Find this distance. (Alexander Shkolny)

2003 Bosnia and Herzegovina Team Selection Test, 4

Tags: altitude , geometry , isosceles , orthogonal , angle chasing

In triangle $ABC$ $AD$ and $BE$ are altitudes. Let $L$ be a point on $ED$ such that $ED$ is orthogonal to $BL$. If $LB^2=LD\cdot LE$ prove that triangle $ABC$ is isosceles

2008 Greece Junior Math Olympiad, 4

Tags: geometry , trapezoid , isosceles , concurrency

Let $ABCD$ be a trapezoid with $AD=a, AB=2a, BC=3a$ and $\angle A=\angle B =90 ^o$. Let $E,Z$ be the midpoints of the sides $AB ,CD$ respectively and $I$ be the foot of the perpendicular from point $Z$ on $BC$. Prove that : i) triangle $BDZ$ is isosceles ii) midpoint $O$ of $EZ$ is the centroid of triangle $BDZ$ iii) lines $AZ$ and $DI$ intersect at a point lying on line $BO$

2021 Oral Moscow Geometry Olympiad, 5

Tags: geometry , circumcircle , geometric inequality , trapezoid , isosceles

The trapezoid is inscribed in a circle. Prove that the sum of distances from any point of the circle to the midpoints of the lateral sides are not less than the diagonal of the trapezoid.

2010 Contests, 3

Tags: geometry , isosceles , equal segments , angle

Consider triangle $ABC$ with $AB = AC$ and $\angle A = 40 ^o$. The points $S$ and $T$ are on the sides $AB$ and $BC$, respectively, so that $\angle BAT = \angle BCS= 10 ^o$. The lines $AT$ and $CS$ intersect at point $P$. Prove that $BT = 2PT$.

2022 Azerbaijan EGMO/CMO TST, G1

Tags: geometry , tangent , isosceles

Let $ABC$ be an isosceles triangle with $AC = BC$ and circumcircle $k$. The point $D$ lies on the shorter arc of $k$ over the chord $BC$ and is different from $B$ and $C$. Let $E$ denote the intersection of $CD$ and $AB$. Prove that the line through $B$ and $C$ is a tangent of the circumcircle of the triangle $BDE$. (Karl Czakler)

2022 Regional Olympiad of Mexico West, 3

Tags: equal angles , isosceles , geometry

In my isosceles triangle $\vartriangle ABC$ with $AB = CA$, we draw $D$ the midpoint of $BC$. Let $E$ be a point on $AC$ such that $\angle CDE = 60^o$ and $M$ the midpoint of $DE$. Prove that $\angle AME = \angle BMD$.

2010 Thailand Mathematical Olympiad, 2

Tags: geometry , isosceles

Let $\vartriangle ABC$ be an isosceles triangle with $AB = AC$. A circle passing through $B$ and $C$ intersects sides $AB$ and $AC$ at $D$ and $E$ respectively. A point $F$ on this circle is chosen so that $EF\perp BC$. If $BC = x$, $CF = y$, and $BF = z$, find the length of $DF$ in terms of $x, y, z$.

2011 Saudi Arabia IMO TST, 3

Tags: isosceles , angle , geometry

In acute triangle $ABC$, $\angle A = 20^o$. Prove that the triangle is isosceles if and only if $$\sqrt[3]{a^3 + b^3 + c^3 -3abc} = \min\{b, c\}$$, where $a,b, c$ are the side lengths of triangle $ABC$.

2019 Costa Rica - Final Round, 6

Tags: isosceles , right triangle , collinear , geometry

Consider the right isosceles $\vartriangle ABC$ at $ A$. Let $L$ be the intersection of the bisector of $\angle ACB$ with $AB$ and $K$ the intersection point of $CL$ with the bisector of $BC$. Let $X$ be the point on line $AK$ such that $\angle KCX = 90^o$ and let $Y$ be the point of intersection of $CX$ with the circumcircle of $\vartriangle ABC$. Let $Y'$ the reflection of point $Y$ wrt $BC$. Prove that $B - K -Y'$. Notation: $A-B-C$ means than points $A,B,C$ are collinear in that order i.e. $ B$ lies between $ A$ and $C$.

2015 Sharygin Geometry Olympiad, 3

Tags: geometry , isosceles , angle

In triangle $ABC$ we have $AB = BC, \angle B = 20^o$. Point $M$ on $AC$ is such that $AM : MC = 1 : 2$, point $H$ is the projection of $C$ to $BM$. Find angle $AHB$. (M. Yevdokimov)

Swiss NMO - geometry, 2004.1

Tags: geometry , isosceles , tangent , secant

Let $\Gamma$ be a circle and $P$ a point outside of $\Gamma$ . A tangent from $P$ to the circle intersects it in $A$. Another line through $P$ intersects $\Gamma$ at the points $B$ and $C$. The bisector of $\angle APB$ intersects $AB$ at $D$ and $AC$ at $E$. Prove that the triangle $ADE$ is isosceles.

2020 Chile National Olympiad, 3

Tags: ratio , geometry , equal segments , isosceles

Given the isosceles triangle $ABC$ with $| AB | = | AC | = 10$ and $| BC | = 15$. Let points $P$ in $BC$ and $Q$ in $AC$ chosen such that $| AQ | = | QP | = | P C |$. Calculate the ratio of areas of the triangles $(PQA): (ABC)$.

2023 Novosibirsk Oral Olympiad in Geometry, 4

Tags: geometry , trapezoid , isosceles trapezoid , isosceles

In a trapezoid, the length of one of the diagonals is equal to the sum of the lengths of the bases, and the angle between the diagonals is $60$ degrees. Prove that this trapezoid is isosceles.

2012 Ukraine Team Selection Test, 4

Tags: geometry , locus , incircle , isosceles

Given an isosceles triangle $ABC$ ($AB = AC$), the inscribed circle $\omega$ touches its sides $AB$ and $AC$ at points $K$ and $L$, respectively. On the extension of the side of the base $BC$, towards $B$, an arbitrary point $M$. is chosen. Line $M$ intersects $\omega$ at the point $N$ for the second time, line $BN$ intersects the second point $\omega$ at the point $P$. On the line $PK$, there is a point $X$ such that $K$ lies between $P$ and $X$ and $KX = KM$. Determine the locus of the point $X$.

1997 Portugal MO, 2

Tags: geometry , 3d geometry , isosceles , cube

Consider the cube $ABCDEFGH$ and denote by, respectively, $M$ and $N$ the midpoints of $[AB]$ and $[CD]$. Let $P$ be a point on the line defined by $[AE]$ and $Q$ the point of intersection of the lines defined by $[PM]$ and $[BF]$. Prove that the triangle $[PQN]$ is isosceles. [img]https://cdn.artofproblemsolving.com/attachments/0/0/57559efbad87903d087c738df279b055b4aefd.png[/img]

Kyiv City MO Juniors 2003+ geometry, 2012.7.4

Tags: geometry , perimeter , isosceles , geometric inequality

Given an isosceles triangle $ABC$ with a vertex at the point $B$. Based on $AC$, an arbitrary point $D $ is selected, different from the vertices $A$ and $C $. On the line $AC $ select the point $E $ outside the segment $AC$, for which $AE = CD$. Prove that the perimeter $\Delta BDE$ is larger than the perimeter $\Delta ABC$.

2012 Belarus Team Selection Test, 2

Tags: geometry , isosceles , inscirbed , equal segments

$A, B, C, D, E$ are five points on the same circle, so that $ABCDE$ is convex and we have $AB = BC$ and $CD = DE$. Suppose that the lines $(AD)$ and $(BE)$ intersect at $P$, and that the line $(BD)$ meets line $(CA)$ at $Q$ and line $(CE)$ at $T$. Prove that the triangle $PQT$ is isosceles. (I. Voronovich)

Durer Math Competition CD Finals - geometry, 2011.D2

Tags: geometry , right triangle , isosceles

In an right isosceles triangle $ABC$, there are two points on the hypotenuse $AB, K$ and $M$, respectively, such that $KCM$ angle is $45^o$ (point $K$ lies between $A$ and $M$). Prove that $AK^2 + MB^2 = KM^2$ [img]https://cdn.artofproblemsolving.com/attachments/2/c/e7c57e0651e5a4c492cc4ae4b115bf68a7a833.png[/img]

2007 BAMO, 3

Tags: isosceles , geometry

In $\vartriangle ABC, D$ and $E$ are two points on segment $BC$ such that $BD = CE$ and $\angle BAD = \angle CAE$. Prove that $\vartriangle ABC$ is isosceles

2014 Saudi Arabia Pre-TST, 3.3

Tags: geometry , incenter , perpendicular bisector , isosceles

Let $ABC$ be a triangle and $I$ its incenter. The line $AI$ intersects the side $BC$ at $D$ and the perpendicular bisector of $BC$ at $E$. Let $J$ be the incenter of triangle $CDE$. Prove that triangle $CIJ$ is isosceles.

May Olympiad L2 - geometry, 2011.3

Tags: geometry , right triangle , isosceles , angle , perpendicular bisector

In a right triangle rectangle $ABC$ such that $AB = AC$, $M$ is the midpoint of $BC$. Let $P$ be a point on the perpendicular bisector of $AC$, lying in the semi-plane determined by $BC$ that does not contain $A$. Lines $CP$ and $AM$ intersect at $Q$. Calculate the angles that form the lines $AP$ and $BQ$.

Cono Sur Shortlist - geometry, 2009.G2

Tags: trapezoid , isosceles , cyclic , fixed , angle , geometry

The trapezoid $ABCD$, of bases $AB$ and $CD$, is inscribed in a circumference $\Gamma$. Let $X$ a variable point of the arc $AB$ of $\Gamma$ that does not contain $C$ or $D$. We denote $Y$ to the point of intersection of $AB$ and $DX$, and let Z be the point of the segment $CX$ such that $\frac{XZ}{XC}=\frac{AY}{AB}$ . Prove that the measure of $\angle AZX$ does not depend on the choice of $X.$

Kyiv City MO Seniors 2003+ geometry, 2005.11.2

Tags: geometry , isosceles , circles

A circle touches the sides $AC$ and $AB$ of the triangle $ABC $ at the points ${{B}_ {1}} $ and ${{C}_ {1}}$ respectively. The segments $B {{B} _ {1}} $ and $C {{C} _ {1}}$ are equal. Prove that the triangle $ABC $ is isosceles. (Timoshkevich Taras)