Found problems: 58
2014 Contests, 2 juniors
Let $ABCD$ be a parallelogram with an acute angle at $A$. Let $G$ be a point on the line $AB$, distinct from $B$, such that $|CG| = |CB|$. Let $H$ be a point on the line $BC$, distinct from $B$, such that $|AB| =|AH|$. Prove that triangle $DGH$ is isosceles.
[asy]
unitsize(1.5 cm);
pair A, B, C, D, G, H;
A = (0,0);
B = (2,0);
D = (0.5,1.5);
C = B + D - A;
G = reflect(A,B)*(C) + C - B;
H = reflect(B,C)*(H) + A - B;
draw(H--A--D--C--G);
draw(interp(A,G,-0.1)--interp(A,G,1.1));
draw(interp(C,H,-0.1)--interp(C,H,1.1));
draw(D--G--H--cycle, dashed);
dot("$A$", A, SW);
dot("$B$", B, SE);
dot("$C$", C, E);
dot("$D$", D, NW);
dot("$G$", G, NE);
dot("$H$", H, SE);
[/asy]
2014 India IMO Training Camp, 3
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 $QB = QR$.
2021 Germany Team Selection Test, 2
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 Brazil Team Selection Test, 3
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 Kyiv Mathematical Festival, 3
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.
2018 Yasinsky Geometry Olympiad, 6
Given a triangle $ABC$, in which $AB = BC$. Point $O$ is the center of the circumcircle, point $I$ is the center of the incircle. Point $D$ lies on the side $BC$, such that the lines $DI$ and $AB$ parallel. Prove that the lines $DO$ and $CI$ are perpendicular.
(Vyacheslav Yasinsky)
India EGMO 2023 TST, 6
Let $ABC$ be an isosceles triangle with $AB = AC$. Suppose $P,Q,R$ are points on segments $AC, AB, BC$ respectively such that $AP = QB$, $\angle PBC = 90^\circ - \angle BAC$ and $RP = RQ$. Let $O_1, O_2$ be the circumcenters of $\triangle APQ$ and $\triangle CRP$. Prove that $BR = O_1O_2$.
[i]Proposed by Atul Shatavart Nadig[/i]
2023 Sharygin Geometry Olympiad, 9.5
A point $D$ lie on the lateral side $BC$ of an isosceles triangle $ABC$. The ray $AD$ meets the line passing through $B$ and parallel to the base $AC$ at point $E$. Prove that the tangent to the circumcircle of triangle $ABD$ at $B$ bisects $EC$.
2021 Thailand TSTST, 1
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 Longlists, 67
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.
2001 Tuymaada Olympiad, 6
On the side $AB$ of an isosceles triangle $AB$ ($AC=BC$) lie points $P$ and $Q$ such that $\angle PCQ \le \frac{1}{2} \angle ACB$. Prove that $PQ \le \frac{1}{2} AB$.
2024 Czech-Polish-Slovak Junior Match, 3
Determine the possible interior angles of isosceles triangles that can be subdivided in two isosceles triangles with disjoint interior.
2010 Sharygin Geometry Olympiad, 8
Bisectrices $AA_1$ and $BB_1$ of triangle $ABC$ meet in $I$. Segments $A_1I$ and $B_1I$ are the bases of isosceles triangles with opposite vertices $A_2$ and $B_2$ lying on line $AB$. It is known that line $CI$ bisects segment $A_2B_2$. Is it true that triangle $ABC$ is isosceles?
2021 Taiwan TST Round 2, 4
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 $\angle ACB = 90^{\circ}$.
2007 Sharygin Geometry Olympiad, 2
Each diagonal of a quadrangle divides it into two isosceles triangles. Is it true that the quadrangle is a diamond?
2014 Taiwan TST Round 3, 4
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 $QB = QR$.
2019 Singapore Junior Math Olympiad, 1
In the triangle $ABC, AC=BC, \angle C=90^o, D$ is the midpoint of $BC, E$ is the point on $AB$ such that $AD$ is perpendicular to $CE$. Prove that $AE=2EB$.
2012 Chile National Olympiad, 4
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 $AD: DC$.
2014 May Olympiad, 4
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 $\angle CAP$
1990 Chile National Olympiad, 1
Show that any triangle can be subdivided into isosceles triangles.
2021 Azerbaijan IMO TST, 1
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 $\angle ACB = 90^{\circ}$.
2005 Sharygin Geometry Olympiad, 9.2
Find all isosceles triangles that cannot be cut into three isosceles triangles with the same sides.
2021 Germany Team Selection Test, 2
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
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 $\angle ACB = 90^{\circ}$.
2017 Hanoi Open Mathematics Competitions, 15
Show that an arbitrary quadrilateral can be divided into nine isosceles triangles.