Olympiad problems

Kyiv City MO Juniors 2003+ geometry, 2016.9.5

On the sides $BC$ and $AB$ of the triangle $ABC$ the points ${{A} _ {1}}$ and ${{C} _ {1}} $ are selected accordingly so that the segments $A {{A} _ {1}}$ and $C {{C} _ {1}}$ are equal and perpendicular. Prove that if $\angle ABC = 45 {} ^ \circ$, then $AC = A {{A} _ {1}} $. (Gogolev Andrew)

2023 Romania National Olympiad, 3

Tags: metric relation , perpendicularity , geometry

We consider triangle $ABC$ with $\angle BAC = 90^{\circ}$ and $\angle ABC = 60^{\circ}.$ Let $ D \in (AC) , E \in (AB),$ such that $CD = 2 \cdot DA$ and $DE $ is bisector of $\angle ADB.$ Denote by $M$ the intersection of $CE$ and $BD$, and by $P$ the intersection of $DE$ and $AM$. a) Show that $AM \perp BD$. b) Show that $3 \cdot PB = 2 \cdot CM$.

the 4th XMO, 1

Tags: perpendicularity , perpendicular , geometry

As shown in the figure, it is known that $BC= AC$ in $\vartriangle ABC$, $M$ is the midpoint of $AB$, points $D$, $E$ lie on $AB$ such that $\angle DCE= \angle MCB$, the circumscribed circle of $\vartriangle BDC$ and the circumscribed circle of $\vartriangle AEC$ intersect at point $F $(different from point $C$), point $H$ lies on $AB$ such that the straight line $CM$ bisects the line segment $HF$. Let the circumcenters of $\vartriangle HFE$ and $\vartriangle BFM$ be $O_1$, $O_2$ respectively. Prove that $O_1O_2 \perp CF$. [img]https://cdn.artofproblemsolving.com/attachments/8/c/62d4ecbc18458fb4f2bf88258d5024cddbc3b0.jpg[/img]

2019 Oral Moscow Geometry Olympiad, 6

Tags: excenter , excircle , perpendicularity , perpendicular , geometry

In the acute triangle $ABC$, the point $I_c$ is the center of excircle on the side $AB$, $A_1$ and $B_1$ are the tangency points of the other two excircles with sides $BC$ and $CA$, respectively, $C'$ is the point on the circumcircle diametrically opposite to point $C$. Prove that the lines $I_cC'$ and $A_1B_1$ are perpendicular.

2019 Azerbaijan BMO TST, 2

Tags: perpendicular , tangent , perpendicularity , orthocenter , perpendicular bisector , geometry , circumcircle

Let $ABC$ be a triangle inscribed in circle $\Gamma$ with center $O$. Let $H$ be the orthocenter of triangle $ABC$ and let $K$ be the midpoint of $OH$. Tangent of $\Gamma$ at $B$ intersects the perpendicular bisector of $AC$ at $L$. Tangent of $\Gamma$ at $C$ intersects the perpendicular bisector of $AB$ at $M$. Prove that $AK$ and $LM$ are perpendicular. by Michael Sarantis, Greece

2018 Yasinsky Geometry Olympiad, 6

Tags: perpendicular , isosceles triangle , geometry , perpendicularity , circumcircle , incenter , isosceles

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)

2017 Ukrainian Geometry Olympiad, 2

Tags: angle chasing , perpendicularity , geometry

On the side $AC$ of a triangle $ABC$, let a $K$ be a point such that $AK = 2KC$ and $\angle ABK = 2 \angle KBC$. Let $F$ be the midpoint of $AC$, $L$ be the projection of $A$ on $BK$. Prove that $FL \bot BC$.

2022 Oral Moscow Geometry Olympiad, 4

Tags: perpendicularity , perpendicular , geometry

In triangle $ABC$, angle $C$ is equal to $60^o$. Bisectors $AA'$ and $BB'$ intersect at point $I$. Point $K$ is symmetric to $I$ with respect to line $AB$. Prove that lines $CK$ and $A'B'$ are perpendicular. (D. Shvetsov, A. Zaslavsky)

2018 Danube Mathematical Competition, 2

Tags: angle chasing , perpendicularity , geometry , perpendicular

Let $ABC$ be a triangle such that in its interior there exists a point $D$ with $\angle DAC = \angle DCA = 30^o$ and $ \angle DBA = 60^o$. Denote $E$ the midpoint of the segment $BC$, and take $F$ on the segment $AC$ so that $AF = 2FC$. Prove that $DE \perp EF$.

2018 Belarusian National Olympiad, 9.3

Tags: perpendicularity , geometry

The bisector of angle $CAB$ of triangle $ABC$ intersects the side $CB$ at $L$. The point $D$ is the foot of the perpendicular from $C$ to $AL$ and the point $E$ is the foot of perpendicular from $L$ to $AB$. The lines $CB$ and $DE$ meet at $F$. Prove that $AF$ is an altitude of triangle $ABC$.

2023 Puerto Rico Team Selection Test, 2

Tags: geometry , perpendicularity , perpendicular

Let $I$ be the incenter of a triangle $ABC$ and let $D$ and $E$ be the touchpoints of the incircle with sides $BC$ and $AC$, respectively. The lines $DE$ and $BI$ intersect at point $P$. Prove that $AP$ is perpendicular to $BP$.

1996 Mexico National Olympiad, 6

Tags: perpendicularity , collinearity , geometry

In a triangle $ABC$ with $AB < BC < AC$, points $A' ,B' ,C'$ are such that $AA' \perp BC$ and $AA' = BC, BB' \perp CA$ and $BB'=CA$, and $CC' \perp AB$ and $CC'= AB$, as shown on the picture. Suppose that $\angle AC'B$ is a right angle. Prove that the points $A',B' ,C' $ are collinear.

2018 Balkan MO Shortlist, G2

Tags: tangent , orthocenter , perpendicularity , perpendicular , perpendicular bisector , circumcircle , geometry

Let $ABC$ be a triangle inscribed in circle $\Gamma$ with center $O$. Let $H$ be the orthocenter of triangle $ABC$ and let $K$ be the midpoint of $OH$. Tangent of $\Gamma$ at $B$ intersects the perpendicular bisector of $AC$ at $L$. Tangent of $\Gamma$ at $C$ intersects the perpendicular bisector of $AB$ at $M$. Prove that $AK$ and $LM$ are perpendicular. by Michael Sarantis, Greece

XMO (China) 2-15 - geometry, 4.1

Tags: geometry , perpendicular , perpendicularity

As shown in the figure, it is known that $BC= AC$ in $\vartriangle ABC$, $M$ is the midpoint of $AB$, points $D$, $E$ lie on $AB$ such that $\angle DCE= \angle MCB$, the circumscribed circle of $\vartriangle BDC$ and the circumscribed circle of $\vartriangle AEC$ intersect at point $F $(different from point $C$), point $H$ lies on $AB$ such that the straight line $CM$ bisects the line segment $HF$. Let the circumcenters of $\vartriangle HFE$ and $\vartriangle BFM$ be $O_1$, $O_2$ respectively. Prove that $O_1O_2 \perp CF$. [img]https://cdn.artofproblemsolving.com/attachments/8/c/62d4ecbc18458fb4f2bf88258d5024cddbc3b0.jpg[/img]

2013 Romania National Olympiad, 3

Tags: rectangle , perpendicularity , geometry

Let $ABCD$ be a rectangle with $5AD <2 AB$ . On the side $AB$ consider the points $S$ and $T$ such that $AS = ST = TB$. Let $M, N$ and $P$ be the projections of points $A, S$ and $T$ on lines $DS, DT$ and $DB$ respectively .Prove that the points $M, N$, and $P$ are collinear if and only if $15 AD^2 = 2 AB^2$.

2022 Costa Rica - Final Round, 1

Tags: geometry , rhombus , perpendicular , perpendicularity

Let $\Gamma$ be a circle with center $O$. Consider the points $A$, $B$, $C$, $D$, $E$ and $F$ in $\Gamma$, with $D$ and $E$ in the (minor) arc $BC$ and $C$ in the (minor) arc $EF$, such that $DEFO$ is a rhombus and $\vartriangle ABC$ It is equilateral. Show that $\overleftrightarrow{BD}$ and $\overleftrightarrow{CE}$ are perpendicular.