Olympiad problems

2010 Belarus Team Selection Test, 4.2

Let $ABC$ be a triangle. The incircle of $ABC$ touches the sides $AB$ and $AC$ at the points $Z$ and $Y$, respectively. Let $G$ be the point where the lines $BY$ and $CZ$ meet, and let $R$ and $S$ be points such that the two quadrilaterals $BCYR$ and $BCSZ$ are parallelogram. Prove that $GR=GS$. [i]Proposed by Hossein Karke Abadi, Iran[/i]

2007 IMO, 4

Tags: geometry , circumcircle , incenter , Triangle , IMO , IMO 2007

In triangle $ ABC$ the bisector of angle $ BCA$ intersects the circumcircle again at $ R$, the perpendicular bisector of $ BC$ at $ P$, and the perpendicular bisector of $ AC$ at $ Q$. The midpoint of $ BC$ is $ K$ and the midpoint of $ AC$ is $ L$. Prove that the triangles $ RPK$ and $ RQL$ have the same area. [i]Author: Marek Pechal, Czech Republic[/i]

2007 Germany Team Selection Test, 1

Tags: geometry , incenter , Triangle , IMO Shortlist , geometry solved , midpoint , congruent triangles

A point $D$ is chosen on the side $AC$ of a triangle $ABC$ with $\angle C < \angle A < 90^\circ$ in such a way that $BD=BA$. The incircle of $ABC$ is tangent to $AB$ and $AC$ at points $K$ and $L$, respectively. Let $J$ be the incenter of triangle $BCD$. Prove that the line $KL$ intersects the line segment $AJ$ at its midpoint.

2004 India IMO Training Camp, 1

Tags: geometry , circumcircle , Circumcenter , Triangle , IMO Shortlist , excenters , geometry solved

Let $ABC$ be a triangle and let $P$ be a point in its interior. Denote by $D$, $E$, $F$ the feet of the perpendiculars from $P$ to the lines $BC$, $CA$, $AB$, respectively. Suppose that \[AP^2 + PD^2 = BP^2 + PE^2 = CP^2 + PF^2.\] Denote by $I_A$, $I_B$, $I_C$ the excenters of the triangle $ABC$. Prove that $P$ is the circumcenter of the triangle $I_AI_BI_C$. [i]Proposed by C.R. Pranesachar, India [/i]

2009 Jozsef Wildt International Math Competition, w. 24

Tags: Triangle , geometry , inequalities

If $K$, $L$, $M$ denote the midpoints of the sides $AB$, $BC$, $CA$ in triangle $\triangle ABC$, then for all $P$ in the plane of triangle $\triangle ABC$, we have $$\frac{AB}{PK}+\frac{BC}{PL}+\frac{CA}{PM} \geq \frac{AB\cdot BC \cdot CA}{4\cdot PK\cdot PL\cdot PM}$$

2007 Sharygin Geometry Olympiad, 3

Tags: geometry , Triangle

Segments connecting an inner point of a convex non-equilateral n-gon to its vertices divide the n-gon into n equal triangles. What is the least possible n?

1976 Euclid, 1

Tags: Triangle , geometry

Source: 1976 Euclid Part B Problem 1 ----- Triangle $ABC$ has $\angle{B}=30^{\circ}$, $AB=150$, and $AC=50\sqrt{3}$. Determine the length of $BC$.

1997 IMO Shortlist, 5

Tags: geometry , 3D geometry , tetrahedron , Triangle , IMO Shortlist

Let $ ABCD$ be a regular tetrahedron and $ M,N$ distinct points in the planes $ ABC$ and $ ADC$ respectively. Show that the segments $ MN,BN,MD$ are the sides of a triangle.

2020 Germany Team Selection Test, 2

Tags: geometry , IMO Shortlist , IMO Shortlist 2019 , Triangle

Let $P$ be a point inside triangle $ABC$. Let $AP$ meet $BC$ at $A_1$, let $BP$ meet $CA$ at $B_1$, and let $CP$ meet $AB$ at $C_1$. Let $A_2$ be the point such that $A_1$ is the midpoint of $PA_2$, let $B_2$ be the point such that $B_1$ is the midpoint of $PB_2$, and let $C_2$ be the point such that $C_1$ is the midpoint of $PC_2$. Prove that points $A_2, B_2$, and $C_2$ cannot all lie strictly inside the circumcircle of triangle $ABC$. (Australia)

1985 IMO Shortlist, 16

Tags: geometry , construction , Triangle , circles , IMO Shortlist

If possible, construct an equilateral triangle whose three vertices are on three given circles.

1992 IMO Longlists, 50

Tags: geometry , Triangle , circles , IMO Shortlist , IMO Longlist

Let $N$ be a point inside the triangle $ABC$. Through the midpoints of the segments $AN, BN$, and $CN$ the lines parallel to the opposite sides of $\triangle ABC$ are constructed. Let $AN, BN$, and $CN$ be the intersection points of these lines. If $N$ is the orthocenter of the triangle $ABC$, prove that the nine-point circles of $\triangle ABC$ and $\triangle A_NB_NC_N$ coincide. [hide="Remark."]Remark. The statement of the original problem was that the nine-point circles of the triangles $A_NB_NC_N$ and $A_MB_MC_M$ coincide, where $N$ and $M$ are the orthocenter and the centroid of $ABC$. This statement is false.[/hide]

2007 Germany Team Selection Test, 1

Tags: geometry , incenter , Triangle , IMO Shortlist , geometry solved , midpoint , congruent triangles

A point $D$ is chosen on the side $AC$ of a triangle $ABC$ with $\angle C < \angle A < 90^\circ$ in such a way that $BD=BA$. The incircle of $ABC$ is tangent to $AB$ and $AC$ at points $K$ and $L$, respectively. Let $J$ be the incenter of triangle $BCD$. Prove that the line $KL$ intersects the line segment $AJ$ at its midpoint.

1998 Slovenia National Olympiad, Problem 3

Tags: geometry , incenter , circumcircle , Triangle

In a right-angled triangle $ABC$ with the hypotenuse $BC$, $D$ is the foot of the altitude from $A$. The line through the incenters of the triangles $ABD$ and $ADC$ intersects the legs of $\triangle ABC$ at $E$ and $F$. Prove that $A$ is the circumcenter of triangle $DEF$.

1974 IMO Longlists, 15

Tags: trigonometry , circumcircle , geometry , Triangle , Trigonometric inequality , IMO , IMO 1974

Let $ABC$ be a triangle. Prove that there exists a point $D$ on the side $AB$ of the triangle $ABC$, such that $CD$ is the geometric mean of $AD$ and $DB$, iff the triangle $ABC$ satisfies the inequality $\sin A\sin B\le\sin^2\frac{C}{2}$. [hide="Comment"][i]Alternative formulation, from IMO ShortList 1974, Finland 2:[/i] We consider a triangle $ABC$. Prove that: $\sin(A) \sin(B) \leq \sin^2 \left( \frac{C}{2} \right)$ is a necessary and sufficient condition for the existence of a point $D$ on the segment $AB$ so that $CD$ is the geometrical mean of $AD$ and $BD$.[/hide]

2005 IMO Shortlist, 2

Tags: geometry , ratio , IMO , Triangle , IMO 2005 , concurrence

Six points are chosen on the sides of an equilateral triangle $ABC$: $A_1$, $A_2$ on $BC$, $B_1$, $B_2$ on $CA$ and $C_1$, $C_2$ on $AB$, such that they are the vertices of a convex hexagon $A_1A_2B_1B_2C_1C_2$ with equal side lengths. Prove that the lines $A_1B_2$, $B_1C_2$ and $C_1A_2$ are concurrent. [i]Bogdan Enescu, Romania[/i]

1969 IMO Shortlist, 57

Tags: geometry , Centroid , Triangle , IMO Shortlist , IMO Longlist

Given triangle $ ABC $ with points $ M $ and $ N $ are in the sides $ AB $ and $ AC $ respectively. If $ \dfrac{BM}{MA} +\dfrac{CN}{NA} = 1 $ , then prove that the centroid of $ ABC $ lies on $ MN $ .

1972 IMO Shortlist, 2

Tags: geometry , combinatorial geometry , combinatorics , point set , Triangle , IMO Shortlist

We are given $3n$ points $A_1,A_2, \ldots , A_{3n}$ in the plane, no three of them collinear. Prove that one can construct $n$ disjoint triangles with vertices at the points $A_i.$

2016 Peru IMO TST, 5

Tags: geometry , IMO Shortlist , Triangle

Let $ABC$ be an acute triangle with orthocenter $H$. Let $G$ be the point such that the quadrilateral $ABGH$ is a parallelogram. Let $I$ be the point on the line $GH$ such that $AC$ bisects $HI$. Suppose that the line $AC$ intersects the circumcircle of the triangle $GCI$ at $C$ and $J$. Prove that $IJ = AH$.

2021 Belarusian National Olympiad, 10.2

Tags: geometry , Triangle

In a triangle $ABC$ equality $2BC=AB+AC$ holds. The angle bisector of $\angle BAC$ inteesects $BC$ at $L$. A circle, that is tangent to $AL$ at $L$ and passes through $B$ intersects $AB$ for the second time at $X$. A circle, that is tangent to $AL$ at $L$ and passes through $C$ intersects $AC$ for the second time at $Y$ Find all possible values of $XY:BC$

1991 IMO Shortlist, 8

Tags: combinatorics , point set , Triangle , combinatorial geometry , IMO Shortlist

$ S$ be a set of $ n$ points in the plane. No three points of $ S$ are collinear. Prove that there exists a set $ P$ containing $ 2n \minus{} 5$ points satisfying the following condition: In the interior of every triangle whose three vertices are elements of $ S$ lies a point that is an element of $ P.$

2007 France Team Selection Test, 3

Tags: geometry , incenter , Triangle , IMO Shortlist , geometry solved , midpoint , congruent triangles

A point $D$ is chosen on the side $AC$ of a triangle $ABC$ with $\angle C < \angle A < 90^\circ$ in such a way that $BD=BA$. The incircle of $ABC$ is tangent to $AB$ and $AC$ at points $K$ and $L$, respectively. Let $J$ be the incenter of triangle $BCD$. Prove that the line $KL$ intersects the line segment $AJ$ at its midpoint.

1999 IMO Shortlist, 8

Tags: geometry , circumcircle , incenter , angle bisector , Triangle , IMO Shortlist

Given a triangle $ABC$. The points $A$, $B$, $C$ divide the circumcircle $\Omega$ of the triangle $ABC$ into three arcs $BC$, $CA$, $AB$. Let $X$ be a variable point on the arc $AB$, and let $O_{1}$ and $O_{2}$ be the incenters of the triangles $CAX$ and $CBX$. Prove that the circumcircle of the triangle $XO_{1}O_{2}$ intersects the circle $\Omega$ in a fixed point.

2016 IMO, 1

Tags: IMO 2016 , geometry , IMO , Triangle , parallelogram

Triangle $BCF$ has a right angle at $B$. Let $A$ be the point on line $CF$ such that $FA=FB$ and $F$ lies between $A$ and $C$. Point $D$ is chosen so that $DA=DC$ and $AC$ is the bisector of $\angle{DAB}$. Point $E$ is chosen so that $EA=ED$ and $AD$ is the bisector of $\angle{EAC}$. Let $M$ be the midpoint of $CF$. Let $X$ be the point such that $AMXE$ is a parallelogram. Prove that $BD,FX$ and $ME$ are concurrent.

1980 Bundeswettbewerb Mathematik, 2

Tags: geometry , bisectors , distance , Triangle

In a triangle $ABC$, the bisectors of angles $A$ and $B$ meet the opposite sides of the triangle at points $D$ and $E$, respectively. A point $P$ is arbitrarily chosen on the line $DE$. Prove that the distance of $P$ from line $AB$ equals the sum or the difference of the distances of $P$ from lines $AC$ and $BC$.

1980 IMO Longlists, 1

Tags: trigonometry , geometry , Triangle , perpendicular bisector , IMO Shortlist

Let $\alpha, \beta$ and $\gamma$ denote the angles of the triangle $ABC$. The perpendicular bisector of $AB$ intersects $BC$ at the point $X$, the perpendicular bisector of $AC$ intersects it at $Y$. Prove that $\tan(\beta) \cdot \tan(\gamma) = 3$ implies $BC= XY$ (or in other words: Prove that a sufficient condition for $BC = XY$ is $\tan(\beta) \cdot \tan(\gamma) = 3$). Show that this condition is not necessary, and give a necessary and sufficient condition for $BC = XY$.