Olympiad problems

1999 USAMO, 6

Tags: geometry , trapezoid , geometric transformation , reflection , incenter , symmetry , isosceles triangle

Let $ABCD$ be an isosceles trapezoid with $AB \parallel CD$. The inscribed circle $\omega$ of triangle $BCD$ meets $CD$ at $E$. Let $F$ be a point on the (internal) angle bisector of $\angle DAC$ such that $EF \perp CD$. Let the circumscribed circle of triangle $ACF$ meet line $CD$ at $C$ and $G$. Prove that the triangle $AFG$ is isosceles.

2014 Taiwan TST Round 3, 4

Tags: circumcircle , isosceles triangle , angle chasing , inversion , geometry

Let $ABC$ be a triangle with $\angle B > \angle C$. Let $P$ and $Q$ be two different points on line $AC$ such that $\angle PBA = \angle QBA = \angle ACB $ and $A$ is located between $P$ and $C$. Suppose that there exists an interior point $D$ of segment $BQ$ for which $PD=PB$. Let the ray $AD$ intersect the circle $ABC$ at $R \neq A$. Prove that $QB = QR$.

2017 Hanoi Open Mathematics Competitions, 15

Tags: isosceles triangle , isosceles , quadrilateral , geometry

Show that an arbitrary quadrilateral can be divided into nine isosceles triangles.

2021 Thailand TSTST, 1

Tags: right triangle , isosceles triangle , geometry

Let $ABC$ be an isosceles triangle with $BC=CA$, and let $D$ be a point inside side $AB$ such that $AD< DB$. Let $P$ and $Q$ be two points inside sides $BC$ and $CA$, respectively, such that $\angle DPB = \angle DQA = 90^{\circ}$. Let the perpendicular bisector of $PQ$ meet line segment $CQ$ at $E$, and let the circumcircles of triangles $ABC$ and $CPQ$ meet again at point $F$, different from $C$. Suppose that $P$, $E$, $F$ are collinear. Prove that $\angle ACB = 90^{\circ}$.

India EGMO 2021 TST, 5

Tags: right triangle , isosceles triangle , geometry

Let $ABC$ be an isosceles triangle with $BC=CA$, and let $D$ be a point inside side $AB$ such that $AD< DB$. Let $P$ and $Q$ be two points inside sides $BC$ and $CA$, respectively, such that $\angle DPB = \angle DQA = 90^{\circ}$. Let the perpendicular bisector of $PQ$ meet line segment $CQ$ at $E$, and let the circumcircles of triangles $ABC$ and $CPQ$ meet again at point $F$, different from $C$. Suppose that $P$, $E$, $F$ are collinear. Prove that $\angle ACB = 90^{\circ}$.

2014 May Olympiad, 4

Tags: geometry , right triangle , isosceles triangle , equilateral triangle , angle

Let $ABC$ be a right triangle and isosceles, with $\angle C = 90^o$. Let $M$ be the midpoint of $AB$ and $N$ the midpoint of $AC$. Let $ P$ be such that $MNP$ is an equilateral triangle with $ P$ inside the quadrilateral $MBCN$. Calculate the measure of $\angle CAP$

2020 Candian MO, 2#

Tags: trigonometry , law of cosines , law of sines , reflection , geometry , isosceles triangle

Let $ABC$ be a triangle, and $M$ an interior point such that $\angle MAB=10^\circ$, $\angle MBA=20^\circ$, $\angle MAC=40^\circ$ and $\angle MCA=30^\circ$. Prove that the triangle is isosceles.

2021 Estonia Team Selection Test, 2

Tags: right triangle , isosceles triangle , geometry

Let $ABC$ be an isosceles triangle with $BC=CA$, and let $D$ be a point inside side $AB$ such that $AD< DB$. Let $P$ and $Q$ be two points inside sides $BC$ and $CA$, respectively, such that $\angle DPB = \angle DQA = 90^{\circ}$. Let the perpendicular bisector of $PQ$ meet line segment $CQ$ at $E$, and let the circumcircles of triangles $ABC$ and $CPQ$ meet again at point $F$, different from $C$. Suppose that $P$, $E$, $F$ are collinear. Prove that $\angle ACB = 90^{\circ}$.

2021 Taiwan TST Round 2, 4

Tags: geometry , right triangle , isosceles triangle

Let $ABC$ be an isosceles triangle with $BC=CA$, and let $D$ be a point inside side $AB$ such that $AD< DB$. Let $P$ and $Q$ be two points inside sides $BC$ and $CA$, respectively, such that $\angle DPB = \angle DQA = 90^{\circ}$. Let the perpendicular bisector of $PQ$ meet line segment $CQ$ at $E$, and let the circumcircles of triangles $ABC$ and $CPQ$ meet again at point $F$, different from $C$. Suppose that $P$, $E$, $F$ are collinear. Prove that $\angle ACB = 90^{\circ}$.

2024 Czech-Polish-Slovak Junior Match, 3

Tags: isosceles , isosceles triangle , geometry

Determine the possible interior angles of isosceles triangles that can be subdivided in two isosceles triangles with disjoint interior.

2014 India IMO Training Camp, 3

Tags: circumcircle , isosceles triangle , angle chasing , inversion , geometry

Let $ABC$ be a triangle with $\angle B > \angle C$. Let $P$ and $Q$ be two different points on line $AC$ such that $\angle PBA = \angle QBA = \angle ACB $ and $A$ is located between $P$ and $C$. Suppose that there exists an interior point $D$ of segment $BQ$ for which $PD=PB$. Let the ray $AD$ intersect the circle $ABC$ at $R \neq A$. Prove that $QB = QR$.

2013 Singapore Senior Math Olympiad, 1

Tags: geometry , circumcircle , altitude , isosceles triangle , right triangle , length , triangle

In the Triangle ABC AB>AC, the extension of the altitude AD with D lying inside BC intersects the circum-circle of the Triangle ABC at P. The circle through P and tangent to BC at D intersects the circum-circle of Triangle ABC at Q distinct from P with PQ=DQ. Prove that AD=BD-DC

2021 Saudi Arabia IMO TST, 7

Tags: geometry , right triangle , isosceles triangle

Let $ABC$ be an isosceles triangle with $BC=CA$, and let $D$ be a point inside side $AB$ such that $AD< DB$. Let $P$ and $Q$ be two points inside sides $BC$ and $CA$, respectively, such that $\angle DPB = \angle DQA = 90^{\circ}$. Let the perpendicular bisector of $PQ$ meet line segment $CQ$ at $E$, and let the circumcircles of triangles $ABC$ and $CPQ$ meet again at point $F$, different from $C$. Suppose that $P$, $E$, $F$ are collinear. Prove that $\angle ACB = 90^{\circ}$.

1979 IMO Shortlist, 24

Tags: tangency , isosceles triangle , circles , geometry

A circle $C$ with center $O$ on base $BC$ of an isosceles triangle $ABC$ is tangent to the equal sides $AB,AC$. If point $P$ on $AB$ and point $Q$ on $AC$ are selected such that $PB \times CQ = (\frac{BC}{2})^2$, prove that line segment $PQ$ is tangent to circle $C$, and prove the converse.

2012 Bundeswettbewerb Mathematik, 4

Tags: combinatorial geometry , regular polygon , isosceles , trapezoid , isosceles triangle , combinatorics , geometry

From the vertices of a regular 27-gon, seven are chosen arbitrarily. Prove that among these seven points there are three points that form an isosceles triangle or four points that form an isosceles trapezoid.

2021 Germany Team Selection Test, 2

Tags: geometry , right triangle , isosceles triangle

Let $ABC$ be an isosceles triangle with $BC=CA$, and let $D$ be a point inside side $AB$ such that $AD< DB$. Let $P$ and $Q$ be two points inside sides $BC$ and $CA$, respectively, such that $\angle DPB = \angle DQA = 90^{\circ}$. Let the perpendicular bisector of $PQ$ meet line segment $CQ$ at $E$, and let the circumcircles of triangles $ABC$ and $CPQ$ meet again at point $F$, different from $C$. Suppose that $P$, $E$, $F$ are collinear. Prove that $\angle ACB = 90^{\circ}$.

2021 Brazil Team Selection Test, 2

Tags: geometry , right triangle , isosceles triangle

Let $ABC$ be an isosceles triangle with $BC=CA$, and let $D$ be a point inside side $AB$ such that $AD< DB$. Let $P$ and $Q$ be two points inside sides $BC$ and $CA$, respectively, such that $\angle DPB = \angle DQA = 90^{\circ}$. Let the perpendicular bisector of $PQ$ meet line segment $CQ$ at $E$, and let the circumcircles of triangles $ABC$ and $CPQ$ meet again at point $F$, different from $C$. Suppose that $P$, $E$, $F$ are collinear. Prove that $\angle ACB = 90^{\circ}$.

2012 Kyiv Mathematical Festival, 3

Tags: geometry , isosceles triangle

Let $O$ be the circumcenter of triangle $ABC$: Points $D$ and $E$ are chosen at sides $AB$ and $AC$ respectively such that $\angle ADO = \angle AEO = 60^o$ and $BDEC$ is inscribed quadrangle. Prove or disprove that $ABC$ is isosceles triangle.

2014 Brazil Team Selection Test, 3

Tags: circumcircle , isosceles triangle , angle chasing , inversion , geometry

Let $ABC$ be a triangle with $\angle B > \angle C$. Let $P$ and $Q$ be two different points on line $AC$ such that $\angle PBA = \angle QBA = \angle ACB $ and $A$ is located between $P$ and $C$. Suppose that there exists an interior point $D$ of segment $BQ$ for which $PD=PB$. Let the ray $AD$ intersect the circle $ABC$ at $R \neq A$. Prove that $QB = QR$.

2012 Chile National Olympiad, 4

Tags: ratio , isosceles triangle , isosceles , geometry , angle

Consider an isosceles triangle $ABC$, where $AB = AC$. $D$ is a point on the $AC$ side and $P$ a point on the segment $BD$ so that the angle $\angle APC = 90^o$ and $ \angle ABP = \angle BCP $. Determine the ratio $AD: DC$.

2011 Sharygin Geometry Olympiad, 3

Tags: isosceles , construction , isosceles triangle , geometry

Restore the isosceles triangle $ABC$ ($AB = AC$) if the common points $I, M, H$ of bisectors, medians and altitudes respectively are given.

2021 Azerbaijan IMO TST, 1

Tags: right triangle , isosceles triangle , geometry

Let $ABC$ be an isosceles triangle with $BC=CA$, and let $D$ be a point inside side $AB$ such that $AD< DB$. Let $P$ and $Q$ be two points inside sides $BC$ and $CA$, respectively, such that $\angle DPB = \angle DQA = 90^{\circ}$. Let the perpendicular bisector of $PQ$ meet line segment $CQ$ at $E$, and let the circumcircles of triangles $ABC$ and $CPQ$ meet again at point $F$, different from $C$. Suppose that $P$, $E$, $F$ are collinear. Prove that $\angle ACB = 90^{\circ}$.

May Olympiad L1 - geometry, 2014.4

Tags: geometry , right triangle , isosceles triangle , equilateral triangle , angle

Let $ABC$ be a right triangle and isosceles, with $\angle C = 90^o$. Let $M$ be the midpoint of $AB$ and $N$ the midpoint of $AC$. Let $ P$ be such that $MNP$ is an equilateral triangle with $ P$ inside the quadrilateral $MBCN$. Calculate the measure of $\angle CAP$

2018 Brazil Team Selection Test, 4

Tags: geometry , isosceles , tangent , isosceles triangle , conditional geometry , circumcenter

Consider an isosceles triangle $ABC$ with $AB = AC$. Let $\omega(XYZ)$ be the circumcircle of the triangle $XY Z$. The tangents to $\omega(ABC)$ through $B$ and $C$ meet at the point $D$. The point $F$ is marked on the arc $AB$ of $\omega(ABC)$ that does not contain $C$. Let $K$ be the point of intersection of lines $AF$ and $BD$ and $L$ the point of intersection of the lines $AB$ and $CF$. Let $T$ and $S$ be the centers of the circles $\omega(BLC)$ and $\omega(BLK)$, respectively. Suppose that the circles $\omega(BTS)$ and $\omega(CFK)$ are tangent to each other at the point $P$. Prove that $P$ belongs to the line $AB$.

2021 Estonia Team Selection Test, 2

Tags: geometry , right triangle , isosceles triangle

Let $ABC$ be an isosceles triangle with $BC=CA$, and let $D$ be a point inside side $AB$ such that $AD< DB$. Let $P$ and $Q$ be two points inside sides $BC$ and $CA$, respectively, such that $\angle DPB = \angle DQA = 90^{\circ}$. Let the perpendicular bisector of $PQ$ meet line segment $CQ$ at $E$, and let the circumcircles of triangles $ABC$ and $CPQ$ meet again at point $F$, different from $C$. Suppose that $P$, $E$, $F$ are collinear. Prove that $\angle ACB = 90^{\circ}$.