Olympiad problems

2001 BAMO, 2

Let $JHIZ$ be a rectangle, and let $A$ and $C$ be points on sides $ZI$ and $ZJ,$ respectively. The perpendicular from $A$ to $CH$ intersects line $HI$ in $X$ and the perpendicular from $C$ to $AH$ intersects line $HJ$ in $Y.$ Prove that $X,$ $Y,$ and $Z$ are collinear (lie on the same line).

2020 Yasinsky Geometry Olympiad, 2

Tags: perpendicular , square , geometry

Let $ABCD$ be a square, point $E$ be the midpoint of the side $BC$. On the side $AB$ mark a point $F$ such that $FE \perp DE$. Prove that $AF + BE = DF$. (Ercole Suppa, Italy)

2014 Romania National Olympiad, 2

Tags: geometry , rhombus , square , perpendicular , isosceles , right triangle

Outside the square $ABCD$, the rhombus $BCMN$ is constructed with angle $BCM$ obtuse . Let $P$ be the intersection point of the lines $BM$ and $AN$ . Prove that $DM \perp CP$ and the triangle $DPM$ is right isosceles .

2020-21 IOQM India, 22

Tags: geometry , area , angle bisector , perpendicular

In triangle $ABC$, let $P$ and $R$ be the feet of the perpendiculars from $A$ onto the external and internal bisectors of $\angle ABC$, respectively; and let $Q$ and $S$ be the feet of the perpendiculars from $A$ onto the internal and external bisectors of $\angle ACB$, respectively. If $PQ = 7, QR = 6$ and $RS = 8$, what is the area of triangle $ABC$?

2017 Hanoi Open Mathematics Competitions, 14

Tags: geometry , trapezoid , collinear , collinearity , perpendicular

Given trapezoid $ABCD$ with bases $AB \parallel CD$ ($AB < CD$). Let $O$ be the intersection of $AC$ and $BD$. Two straight lines from $D$ and $C$ are perpendicular to $AC$ and $BD$ intersect at $E$ , i.e. $CE \perp BD$ and $DE \perp AC$ . By analogy, $AF \perp BD$ and $BF \perp AC$ . Are three points $E , O, F$ located on the same line?

2003 District Olympiad, 2

Tags: geometry , equal segments , parallel , perpendicular , right triangle

In the right triangle $ABC$ ( $\angle A = 90^o$), $D$ is the intersection of the bisector of the angle $A$ with the side $(BC)$, and $P$ and $Q$ are the projections of the point $D$ on the sides $(AB),(AC)$ respectively . If $BQ \cap DP=\{M\}$, $CP \cap DQ=\{N\}$, $BQ\cap CP=\{H\}$, show that: a) $PM = DN$ b) $MN \parallel BC$ c) $AH \perp BC$.

2016 Novosibirsk Oral Olympiad in Geometry, 5

Tags: geometry , parallelogram , perpendicular

In the parallelogram $CMNP$ extend the bisectors of angles $MCN$ and $PCN$ and intersect with extensions of sides PN and $MN$ at points $A$ and $B$, respectively. Prove that the bisector of the original angle $C$ of the the parallelogram is perpendicular to $AB$. [img]https://cdn.artofproblemsolving.com/attachments/f/3/fde8ef133758e06b1faf8bdd815056173f9233.png[/img]

2011 Belarus Team Selection Test, 1

Tags: geometry , incenter , orthocenter , perpendicular

In an acute-angled triangle $ABC$, the orthocenter is $H$. $I_H$ is the incenter of $\vartriangle BHC$. The bisector of $\angle BAC$ intersects the perpendicular from $I_H$ to the side $BC$ at point $K$. Let $F$ be the foot of the perpendicular from $K$ to $AB$. Prove that $2KF+BC=BH +HC$ A. Voidelevich

2015 Indonesia MO Shortlist, G2

Tags: geometry , circles , perpendicular , collinear , tangent circle

Two circles that are not equal are tangent externally at point $R$. Suppose point $P$ is the intersection of the external common tangents of the two circles. Let $A$ and $B$ are two points on different circles so that $RA$ is perpendicular to $RB$. Show that the line $AB$ passes through $P$.

2014 Contests, 1

Tags: geometry , perpendicular , isosceles

As shown in the figure, given $\vartriangle ABC$ with $\angle B$, $\angle C$ acute angles, $AD \perp BC$, $DE \perp AC$, $M$ midpoint of $DE$, $AM \perp BE$. Prove that $\vartriangle ABC$ is isosceles. [img]https://cdn.artofproblemsolving.com/attachments/a/8/f553c33557979f6f7b799935c3bde743edcc3c.png[/img]

2014 Contests, 2

Tags: geometry , circumcircle , perpendicular

The center of the circumcircle of the acute triangle $ABC$ is $M$, and the circumcircle of $ABM$ meets $BC$ and $AC$ at $P$ and $Q$ ($P\ne B$). Show that the extension of the line segment $CM$ is perpendicular to $PQ$.

1977 IMO Longlists, 22

Tags: geometry , convex quadrilateral , perpendicular

Let $S$ be a convex quadrilateral $ABCD$ and $O$ a point inside it. The feet of the perpendiculars from $O$ to $AB, BC, CD, DA$ are $A_1, B_1, C_1, D_1$ respectively. The feet of the perpendiculars from $O$ to the sides of $S_i$, the quadrilateral $A_iB_iC_iD_i$, are $A_{i+1}B_{i+1}C_{i+1}D_{i+1}$, where $i = 1, 2, 3.$ Prove that $S_4$ is similar to S.

2022 Poland - Second Round, 2

Tags: geometry , circumcircle , perpendicular

Given a cyclic quadriteral $ABCD$. The circumcenter lies in the quadriteral $ABCD$. Diagonals $AC$ and $BD$ intersects at $S$. Points $P$ and $Q$ are the midpoints of $AD$ and $BC$. Let $p$ be a line perpendicular to $AC$ through $P$, $q$ perpendicular line to $BD$ through $Q$ and $s$ perpendicular to $CD$ through $S$. Prove that $p,q,s$ intersects at one point.

1987 Tournament Of Towns, (148) 5

Tags: geometry , locus , perpendicular , equilateral

Perpendiculars are drawn from an interior point $M$ of the equilateral triangle $ABC$ to its sides , intersecting them at points $D, E$ and $F$ . Find the locus of all points $M$ such that $DEF$ is a right triangle . (J . Tabov , Sofia)

Kyiv City MO Juniors 2003+ geometry, 2018.7.41

Tags: geometry , perpendicular , angle , midpoint

In the quadrilateral $ABCD$ point $E$ - the midpoint of the side $AB$, point $F$ - the midpoint of the side $BC$, point $G$ - the midpoint $AD$ . It turned out that the segment $GE$ is perpendicular to $AB$, and the segment $GF$ is perpendicular to the segment $BC$. Find the value of the angle $GCD$, if it is known that $\angle ADC = 70 {} ^ \circ$.

1994 Mexico National Olympiad, 3

Tags: parallelogram , equal angles , perpendicular , geometry

$ABCD$ is a parallelogram. Take $E$ on the line $AB$ so that $BE = BC$ and $B$ lies between $A$ and $E$. Let the line through $C$ perpendicular to $BD$ and the line through $E$ perpendicular to $AB$ meet at $F$. Show that $\angle DAF = \angle BAF$.

Ukrainian From Tasks to Tasks - geometry, 2016.13

Tags: geometry , perpendicular , isosceles

Let $ABC$ be an isosceles acute triangle ($AB = BC$). On the side $BC$ we mark a point $P$, such that $\angle PAC = 45^o$, and $Q$ is the point of intersection of the perpendicular bisector of the segment $AP$ with the side $AB$. Prove that $PQ \perp BC$.

2013 Oral Moscow Geometry Olympiad, 1

Tags: geometry , angle bisector , perpendicular , median

In triangle $ABC$ the angle bisector $AK$ is perpendicular on the median is $CL$. Prove that in the triangle $BKL$ also one of angle bisectors are perpendicular to one of the medians.

2018 Regional Olympiad of Mexico Northeast, 2

Tags: geometry , perpendicular , incenter

Let $ABC$ be a triangle such that $M$ and $N$ are the midpoints of $AC$ and $BC$, respectively. Let $I$ be the incenter of $ABC$ and $E$ be the intersection of $MN$ with $Bl$. Let $P$ be a point such that $EP$ is perpendicular to $MN$ and $NP$ parallel to $IA$. Prove that $IP$ is perpendicular to $BC$.

2016 Saudi Arabia BMO TST, 2

Tags: geometry , circumcircle , incircle , perpendicular

Let $ABC$ be a triangle with $AB \ne AC$. The incirle of triangle $ABC$ is tangent to $BC, CA, AB$ at $D, E, F$, respectively. The perpendicular line from $D$ to $EF$ intersects $AB$ at $X$. The second intersection point of circumcircles of triangles $AEF$ and $ABC$ is $T$. Prove that $TX \perp T F$

2014 Oral Moscow Geometry Olympiad, 4

Tags: geometry , euler line , collinear , centroid , orthocenter , perpendicular

The medians $AA_0, BB_0$, and $CC_0$ of the acute-angled triangle $ABC$ intersect at the point $M$, and heights $AA_1, BB_1$ and $CC_1$ at point $H$. Tangent to the circumscribed circle of triangle $A_1B_1C_1$ at $C_1$ intersects the line $A_0B_0$ at the point $C'$. Points $A'$ and $B'$ are defined similarly. Prove that $A', B'$ and $C'$ lie on one line perpendicular to the line $MH$.

1996 North Macedonia National Olympiad, 1

Tags: geometry , parallelogram , equal angles , perpendicular

Let $ABCD$ be a parallelogram which is not a rectangle and $E$ be the point in its plane such that $AE \perp AB$ and $CE \perp CB$. Prove that $\angle DEA = \angle CEB$.

1995 Abels Math Contest (Norwegian MO), 2a

Tags: geometry , circles , perpendicular , tangent circle

Two circles $k_1,k_2$ touch each other at $P$, and touch a line $\ell$ at $A$ and $B$ respectively. Line $AP$ meets $k_2$ at $C$. Prove that $BC$ is perpendicular to $\ell$.

2021 Regional Olympiad of Mexico West, 5

Tags: perpendicular , geometry

Let $ABC$ be a triangle such that $AC$ is its shortest side. A point $P$ is inside it and satisfies that $BP = AC$. Let $R$ be the midpoint of $BC$ and let $M$ be the midpoint of $AP$. Let $E$ be the intersection of $BP$ and $AC$. Prove that the bisector of angle $\angle BE A$ is perpendicular to segment $MR$.

1998 Estonia National Olympiad, 3

Tags: ratio , perpendicular , triangle area , geometry , irrational , area

In a triangle $ABC$, the bisector of the largest angle $\angle A$ meets $BC$ at point $D$. Let $E$ and $F$ be the feet of perpendiculars from $D$ to $AC$ and $AB$, respectively. Let $R$ denote the ratio between the areas of triangles $DEB$ and $DFC$. (a) Prove that, for every real number $r > 0$, one can construct a triangle ABC for which $R$ is equal to $r$. (b) Prove that if $R$ is irrational, then at least one side length of $\vartriangle ABC$ is irrational. (c) Give an example of a triangle $ABC$ with exactly two sides of irrational length, but with rational $R$.