Olympiad problems

2005 Oral Moscow Geometry Olympiad, 4

Tags: geometry , perpendicular , hexagon , equal segments , right angle

Given a hexagon $ABCDEF$, in which $AB = BC, CD = DE, EF = FA$, and angles $A$ and $C$ are right. Prove that lines $FD$ and $BE$ are perpendicular. (B. Kukushkin)

Novosibirsk Oral Geo Oly VIII, 2020.4

Tags: geometry , perpendicular , equal segments

Point $P$ is chosen inside triangle $ABC$ so that $\angle APC+\angle ABC=180^o$ and $BC=AP.$ On the side $AB$, a point $K$ is chosen such that $AK = KB + PC$. Prove that $CK \perp AB$.

2022 Austrian Junior Regional Competition, 3

Tags: geometry , equal segments , semicircle

A semicircle is erected over the segment $AB$ with center $M$. Let $P$ be one point different from $A$ and $B$ on the semicircle and $Q$ the midpoint of the arc of the circle $AP$. The point of intersection of the straight line $BP$ with the parallel to $P Q$ through $M$ is $S$. Prove that $PM = PS$ holds. [i](Karl Czakler)[/i]

2014 Contests, 2

Tags: geometry , parallelogram , equal segments

Let $ABCD$ be a parallelogram. On side $AB$, point $M$ is taken so that $AD = DM$. On side $AD$ point $N$ is taken so that $AB = BN$. Prove that $CM = CN$.

OIFMAT II 2012, 4

Tags: geometry , circumcenter , orthocenter , equal segments

Given a $ \vartriangle ABC $ with $ AB> AC $ and $ \angle BAC = 60^o$. Denote the circumcenter and orthocenter as $ O $ and $ H $ respectively. We also have that $ OH $ intersects $ AB $ in $ P $ and $ AC $ in $ Q $. Prove that $ PO = HQ $.

2022 Federal Competition For Advanced Students, P2, 2

Tags: geometry , equal segments , orthocenter

Let $ ABC$ be an acute-angled, non-isosceles triangle with orthocenter $H$, $M$ midpoint of side $AB$ and $w$ bisector of angle $\angle ACB$. Let $S$ be the point of intersection of the perpendicular bisector of side $AB$ with $w$ and $F$ the foot of the perpendicular from $H$ on $w$. Prove that the segments $MS$ and $MF$ are equal. [i](Karl Czakler)[/i]

2016 Hanoi Open Mathematics Competitions, 11

Tags: geometry , equal segments

Let be given a triangle $ABC$, and let $I$ be the midpoint of $BC$. The straight line $d$ passing $I$ intersects $AB,AC$ at $M,N$ , respectively. The straight line $d'$ ($\ne d$) passing $I$ intersects $AB, AC$ at $Q, P$ , respectively. Suppose $M, P$ are on the same side of $BC$ and $MP , NQ$ intersect $BC$ at $E$ and $F$, respectively. Prove that $IE = I F$.

2019 Regional Olympiad of Mexico Center Zone, 3

Tags: geometry , equal segments , equal angles

Let $ABC$ be an acute triangle and $D$ a point on the side $BC$ such that $\angle BAD = \angle DAC$. The circumcircles of the triangles $ABD$ and $ACD$ intersect the segments $AC$ and $AB$ at $E$ and $F$, respectively. The internal bisectors of $\angle BDF$ and $\angle CDE$ intersect the sides $AB$ and $AC$ at $P$ and $Q$, respectively. Points $X$ and $Y$ are chosen on the side $BC$ such that $PX$ is parallel to $AC$ and $QY$ is parallel to $AB$. Finally, let $Z$ be the point of intersection of $BE$ and $CF$. Prove that $ZX = ZY$.

2000 Kazakhstan National Olympiad, 2

Tags: geometry , equal segments , circles

Given a circle centered at $ O $ and two points $ A $ and $ B $ lying on it. $ A $ and $ B $ do not form a diameter. The point $ C $ is chosen on the circle so that the line $ AC $ divides the segment $ OB $ in half. Let lines $ AB $ and $ OC $ intersect at $ D $, and let lines $ BC $ and $ AO $ intersect at $ F $. Prove that $ AF = CD $.

2020 Ukrainian Geometry Olympiad - April, 5

Tags: geometry , parallelogram , pentagon , equal angles , equal segments

Given a convex pentagon $ABCDE$, with $\angle BAC = \angle ABE = \angle DEA - 90^o$, $\angle BCA = \angle ADE$ and also $BC = ED$. Prove that $BCDE$ is parallelogram.

2011 Grand Duchy of Lithuania, 4

Tags: geometry , orthocenter , equal segments , cyclic quadrilateral

In the cyclic quadrilateral $ABCD$ with $AB = AD$, points $M$ and $N$ lie on the sides $CD$ and $BC$ respectively so that $MN = BN + DM$. Lines $AM$ and $AN$ meet the circumcircle of $ABCD$ again at points $P$ and $Q$ respectively. Prove that the orthocenter of the triangle $APQ$ lies on the segment $MN$.

2008 Federal Competition For Advanced Students, P1, 4

Tags: geometry , isosceles , isosceles triangle , equal segments , midpoint

In a triangle $ABC$ let $E$ be the midpoint of the side $AC$ and $F$ the midpoint of the side $BC$. Let $G$ be the foot of the perpendicular from $C$ to $ AB$. Show that $\vartriangle EFG$ is isosceles if and only if $\vartriangle ABC$ is isosceles.

2021 Saudi Arabia JBMO TST, 2

Tags: geometry , equal segments , equal angles

In a triangle $ABC$, let $K$ be a point on the median $BM$ such that $CM = CK$. It turned out that $\angle CBM = 2\angle ABM$. Show that $BC = KM$.

Swiss NMO - geometry, 2019.7

Tags: geometry , equal angles , equal segments , angle

Let $ABC$ be a triangle with $\angle CAB = 2 \angle ABC$. Assume that a point $D$ is inside the triangle $ABC$ exists such that $AD = BD$ and $CD = AC$. Show that $\angle ACB = 3 \angle DCB$.

2015 Caucasus Mathematical Olympiad, 5

Tags: midpoint , altitude , geometry , equal angles , equal segments

Let $AA_1$ and $CC_1$ be the altitudes of the acute-angled triangle $ABC$. Let $K,L$ and $M$ be the midpoints of the sides $AB,BC$ and $CA$ respectively. Prove that if $\angle C_1MA_1 =\angle ABC$, then $C_1 K = A_1L$.

Indonesia MO Shortlist - geometry, g9

Tags: geometry , angle bisector , cyclic quadrilateral , parallelogram , equal segments

It is known that $ABCD$ is a parallelogram. The point $E$ is taken so that $BCED$ is a cyclic quadrilateral. Let $\ell$ be a line that passes through $A$, intersects the segment $DC$ at point $F$ and intersects the extension of the line $BC$ at $G$. Given $EF = EG = EC$. Prove that $\ell$ is the bisector of the angle $\angle BAD$.

Kyiv City MO Juniors 2003+ geometry, 2020.8.51

Tags: geometry , cyclic , hexagon , equal segments , concurrency

Let $ABCDEF$ be a hexagon inscribed in a circle in which $AB = BC, CD = DE$ and $EF = FA$. Prove that the lines $AD, BE$ and $CF$ intersect at one point.

2007 Sharygin Geometry Olympiad, 13

Tags: geometry , fixed point , circumcircle , equal segments

On the side $AB$ of a triangle $ABC$, two points $X, Y$ are chosen so that $AX = BY$. Lines $CX$ and $CY$ meet the circumcircle of the triangle, for the second time, at points $U$ and $V$. Prove that all lines $UV$ (for all $X, Y$, given $A, B, C$) have a common point.

2015 Oral Moscow Geometry Olympiad, 3

Tags: geometry , construction , equal segments

In triangle $ABC$, points $D, E$, and $F$ are marked on sides $AC, BC$, and $AB$ respectively, so that $AD = AB$, $EC = DC$, $BF = BE$. After that, they erased everything except points $E, F$ and $D$. Reconstruct the triangle $ABC$ (no study required).

2014 Thailand Mathematical Olympiad, 1

Tags: equal segments , geometry

Let $\vartriangle ABC$ be an isosceles triangle with $\angle BAC = 100^o$. Let $D, E$ be points on ray $\overrightarrow{AB}$ so that $BC = AD = BE$. Show that $BC \cdot DE = BD \cdot CE$

2014 Gulf Math Olympiad, 3

Tags: geometry , equal segments , right triangle , semicircle , angle

(i) $ABC$ is a triangle with a right angle at $A$, and $P$ is a point on the hypotenuse $BC$. The line $AP$ produced beyond $P$ meets the line through $B$ which is perpendicular to $BC$ at $U$. Prove that $BU = BA$ if, and only if, $CP = CA$. (ii) $A$ is a point on the semicircle $CB$, and points $X$ and $Y$ are on the line segment $BC$. The line $AX$, produced beyond $X$, meets the line through $B$ which is perpendicular to $BC$ at $U$. Also the line $AY$, produced beyond $Y$, meets the line through $C$ which is perpendicular to $BC$ at $V$. Given that $BY = BA$ and $CX = CA$, determine the angle $\angle VAU$.

1998 Tournament Of Towns, 2

Tags: geometry , parallelogram , equal segments , equal angles

$ABCD$ is a parallelogram. A point $M$ is found on the side $AB$ or its extension such that $\angle MAD = \angle AMO$ where $O$ is the intersection point of the diagonals of the parallelogram. Prove that $MD = MG$. (M Smurov)

2021 Novosibirsk Oral Olympiad in Geometry, 6

Tags: equal segments , equilateral , geometry

Inside the equilateral triangle $ABC$, points $P$ and $Q$ are chosen so that the quadrilateral $APQC$ is convex, $AP = PQ = QC$ and $\angle PBQ = 30^o$. Prove that $AQ = BP$.

1951 Moscow Mathematical Olympiad, 189

Tags: equal segments , convex , inequalities , angle

Let $ABCD$ and $A'B'C'D'$ be two convex quadrilaterals whose corresponding sides are equal, i.e., $AB = A'B', BC = B'C'$, etc. Prove that if $\angle A > \angle A'$, then $\angle B < \angle B', \angle C > \angle C', \angle D < \angle D'$.

Ukrainian From Tasks to Tasks - geometry, 2015.14

Tags: geometry , concyclic , equal segments

On the side $AB$ of the triangle $ABC$ mark the points $M$ and $N$, such that $BM = BC$ and $AN = AC$. Then on the sides $BC$ and $AC$ mark the points$ P$ and $Q$, respectively, such that $BP = BN$ and $AQ = AM$. Prove that the points $C, Q, M, N$ and $P$ lie on the same circle.