Olympiad problems

2014 Junior Balkan Team Selection Tests - Moldova, 7

Let the isosceles right triangle $ABC$ with $\angle A= 90^o$ . The points $E$ and $F$ are taken on the ray $AC$ so that $\angle ABE = 15^o$ and $CE = CF$. Determine the measure of the angle $CBF$.

1951 Moscow Mathematical Olympiad, 189

Tags: angle , equal segments , convex , inequalities

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 $\angle B < \angle B', \angle C > \angle C', \angle D < \angle D'$.

2011 Saudi Arabia BMO TST, 4

Tags: equal segments , geometry

Let $ABC$ be a triangle with circumcenter $O$. Points $P$ and $Q$ are interior to sides $CA$ and $AB$, respectively. Circle $\omega$ passes through the midpoints of segments $BP$, $CQ$, $PQ$. Prove that if line $PQ$ is tangent to circle $\omega$, then $OP = OQ$.

2007 Thailand Mathematical Olympiad, 5

Tags: equal segments , geometry

A triangle $\vartriangle ABC$ has $\angle A = 90^o$, and a point $D$ is chosen on $AC$. Point $F$ is the foot of altitude from $A$ to $BC$. Suppose that $BD = DC = CF = 2$. Compute $AC$.

2022 Federal Competition For Advanced Students, P2, 2

Tags: equal segments , orthocenter , geometry

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]

2020 Yasinsky Geometry Olympiad, 4

Tags: equal segments , median , geometry , equal angles

The median $AM$ is drawn in the triangle $ABC$ ($AB \ne AC$). The point $P$ is the foot of the perpendicular drawn on the segment $AM$ from the point $B$. On the segment $AM$ we chose such a point $Q$ that $AQ = 2PM$. Prove that $\angle CQM = \angle BAM$.

2011 Saudi Arabia Pre-TST, 1.4

Tags: equal segments , equal angles , geometry

Let $ABC$ be a triangle with $AB=AC$ and $\angle BAC = 40^o$. Points $S$ and $T$ lie on the sides $AB$ and $BC$, such that $\angle BAT = \angle BCS = 10^o$. Lines $AT$ and $CS$ meet at $P$. Prove that $BT = 2PT$.

2011 Oral Moscow Geometry Olympiad, 1

Tags: rectangle , equal segments , perpendicular , geometry

The bisector of angle $B$ and the bisector of external angle $D$ of rectangle $ABCD$ intersect side $AD$ and line $AB$ at points $M$ and $K$, respectively. Prove that the segment $MK$ is equal and perpendicular to the diagonal of the rectangle.

2000 Tournament Of Towns, 2

Tags: equal segments , circumcenter , chord , circumcircle , geometry

The chords $AC$ and $BD$ of a, circle with centre $O$ intersect at the point $K$. The circumcentres of triangles $AKB$ and $CKD$ are $M$ and $N$ respectively. Prove that $OM = KN$. (A Zaslavsky )

Kyiv City MO Juniors Round2 2010+ geometry, 2019.8.41

Tags: geometry , equal segments , parallelogram

Through the vertices $A, B$ of the parallelogram $ABCD$ passes a circle that intersects for the second time diagonals $BD$ and $AC$ at points $X$ and $Y$, respectively. The circumsccribed circle of $\vartriangle ADX$ intersects diagonal $AC$ for the second time at the point $Z$. Prove that $AY = CZ$.

Kharkiv City MO Seniors - geometry, 2015.10.3

Tags: geometry , circumcircle , equal segments , circles

On side $AB$ of triangle $ABC$, point $M$ is selected. A straight line passing through $M$ intersects the segment $AC$ at point $N$ and the ray $CB$ at point $K$. The circumscribed circle of the triangle $AMN$ intersects $\omega$, the circumscribed circle of the triangle $ABC$, at points $A$ and $S$. Straight lines $SM$ and $SK$ intersect with $\omega$ for the second time at points $P$ and $Q$, respectively. Prove that $AC = PQ$.

1998 Singapore MO Open, 1

Tags: equal segments , tangent , geometry , circles

In Fig. , $PA$ and $QB$ are tangents to the circle at $A$ and $B$ respectively. The line $AB$ is extended to meet $PQ$ at $S$. Suppose that $PA = QB$. Prove that $QS = SP$. [img]https://cdn.artofproblemsolving.com/attachments/6/f/f21c0c70b37768f3e80e9ee909ef34c57635d5.png[/img]

Denmark (Mohr) - geometry, 2019.5

Tags: equilateral , geometry , right triangle , equal segments

In the figure below the triangles $BCD, CAE$ and $ABF$ are equilateral, and the triangle $ABC$ is right-angled with $\angle A = 90^o$. Prove that $|AD| = |EF|$. [img]https://1.bp.blogspot.com/-QMMhRdej1x8/XzP18QbsXOI/AAAAAAAAMUI/n53OsE8rwZcjB_zpKUXWXq6bg3o8GUfSwCLcBGAsYHQ/s0/2019%2Bmohr%2Bp5.png[/img]

Kyiv City MO Juniors Round2 2010+ geometry, 2019.7.3

Tags: equal segments , geometry , equal angles

In the quadrilateral $ABCD$ it is known that $\angle ABD= \angle DBC$ and $AD= CD$. Let $DH$ be the altitude of $\vartriangle ABD$. Prove that $| BC - BH | = HA$. (Hilko Danilo)

2010 Greece Junior Math Olympiad, 2

Tags: equilateral , equal segments , perpendicular , geometry , rectangle

Let $ABCD$ be a rectangle with sides $AB=a$ and $BC=b$. Let $O$ be the intersection point of it's diagonals. Extent side $BA$ towards $A$ at a segment $AE=AO$, and diagonal $DB$ towards $B$ at a segment $BZ=BO$. If the triangle $EZC$ is an equilateral, then prove that: i) $b=a\sqrt3$ ii) $AZ=EO$ iii) $EO \perp ZD$

2015 Denmark MO - Mohr Contest, 3

Tags: equal segments , equilateral , geometry

Triangle $ABC$ is equilateral. The point $D$ lies on the extension of $AB$ beyond $B$, the point $E$ lies on the extension of $CB$ beyond $B$, and $|CD| = |DE|$. Prove that $|AD| = |BE|$. [img]https://1.bp.blogspot.com/-QnAXFw3ijn0/XzR0YjqBQ3I/AAAAAAAAMU0/0TvhMQtBNjolYHtgXsQo2OPGJzEYSfCwACLcBGAsYHQ/s0/2015%2BMohr%2Bp3.png[/img]

2010 IFYM, Sozopol, 8

Tags: trapezoid , equal angles , geometry , equal segments

In the trapezoid $ABCD, AB // CD$ and the diagonals intersect at $O$. The points $P, Q$ are on $AD, BC$ respectively such that $\angle AP B = \angle CP D$ and $\angle AQB = \angle CQD$. Show that $OP = OQ$.

2017 Novosibirsk Oral Olympiad in Geometry, 5

Tags: geometry , rectangle , equal segments

Point $K$ is marked on the diagonal $AC$ in rectangle $ABCD$ so that $CK = BC$. On the side $BC$, point $M$ is marked so that $KM = CM$. Prove that $AK + BM = CM$.

2016 Dutch IMO TST, 3

Tags: equal segments , geometry , isosceles , angle bisector

Let $\vartriangle ABC$ be an isosceles triangle with $|AB| = |AC|$. Let $D, E$ and $F$ be points on line segments $BC, CA$ and $AB$, respectively, such that $|BF| = |BE|$ and such that $ED$ is the internal angle bisector of $\angle BEC$. Prove that $|BD|= |EF|$ if and only if $|AF| = |EC|$.

2018 Thailand Mathematical Olympiad, 9

Tags: equal segments , circumcircle , geometry , incenter

In $\vartriangle ABC$ the incircle is tangent to $AB$ at $D$. Let $P$ be a point on $BC$ different from $B$ and $C$, and let $K$ and $L$ be incenters of $\vartriangle ABP$ and $\vartriangle ACP$ respectively. Suppose that the circumcircle of $\vartriangle KP L$ cuts $AP$ again at $Q$. Prove that $AD = AQ$.

Kyiv City MO 1984-93 - geometry, 1993.8.3

Tags: geometry , equal segments , angle bisector

In the triangle $ABC$, $\angle .ACB = 60^o$, and the bisectors $AA_1$ and $BB_1$ intersect at the point $M$. Prove that $MB_1 = MA_1$.

2016 Romania National Olympiad, 2

Tags: geometry , angle , equal segments , angle bisector

Consider the triangle $ABC$, where $\angle B= 30^o, \angle C = 15^o$, and $M$ is the midpoint of the side $[BC]$. Let point $N \in (BC)$ be such that $[NC] = [AB]$. Show that $[AN$ is the angle bisector of $MAC$

2019 Denmark MO - Mohr Contest, 5

Tags: equal segments , equilateral , right triangle , geometry

In the figure below the triangles $BCD, CAE$ and $ABF$ are equilateral, and the triangle $ABC$ is right-angled with $\angle A = 90^o$. Prove that $|AD| = |EF|$. [img]https://1.bp.blogspot.com/-QMMhRdej1x8/XzP18QbsXOI/AAAAAAAAMUI/n53OsE8rwZcjB_zpKUXWXq6bg3o8GUfSwCLcBGAsYHQ/s0/2019%2Bmohr%2Bp5.png[/img]

Kyiv City MO Juniors Round2 2010+ geometry, 2018.8.3

Tags: equal segments , angle , geometry

In the triangle $ABC$ it is known that $\angle ACB> 90 {} ^ \circ$, $\angle CBA> 45 {} ^ \circ$. On the sides $AC$ and $AB$, respectively, there are points $P$ and $T$ such that $ABC$ and $PT = BC$. The points ${{P} _ {1}}$ and ${{T} _ {1}}$ on the sides $AC$ and $AB$ are such that $AP = C {{P} _ {1}}$ and $AT = B {{T} _ {1}}$. Prove that $\angle CBA- \angle {{P} _ {1}} {{T} _ {1}} A = 45 {} ^ \circ$. (Anton Trygub)

2011 Tournament of Towns, 1

Tags: circumcenter , incenter , geometry , equal segments

$P$ and $Q$ are points on the longest side $AB$ of triangle $ABC$ such that $AQ = AC$ and $BP = BC$. Prove that the circumcentre of triangle $CPQ$ coincides with the incentre of triangle $ABC$.