Olympiad problems

Durer Math Competition CD Finals - geometry, 2019.D3

a) Does there exist a quadrilateral with (both of) the following properties: three of its edges are of the same length, but the fourth one is different, and three of its angles are equal, but the fourth one is different? b) Does there exist a pentagon with (both of) the following properties: four of its edges are of the same length, but the fifth one is different, and four of its angles are equal, but the fifth one is different?

Geometry Mathley 2011-12, 11.2

Tags: equal segments , geometry

Let $ABC$ be a triangle inscribed in the circle $(O)$. Tangents at $B,C$ of the circles $(O)$ meet at $T$ . Let $M,N$ be the points on the rays $BT,CT$ respectively such that $BM = BC = CN$. The line through $M$ and $N$ intersects $CA,AB$ at $E, F$ respectively; $BE$ meets $CT$ at $P, CF$ intersects $BT$ at $Q$. Prove that $AP = AQ$. Trần Quang Hùng

Kyiv City MO Juniors 2003+ geometry, 2020.9.41

Tags: equal segments , angle , geometry

The points $A, B, C, D$ are selected on the circle as followed so that $AB = BC = CD$. Bisectors of $\angle ABD$ and $\angle ACD$ intersect at point $E$. Find $\angle ABC$, if it is known that $AE \parallel CD$.

2001 Mexico National Olympiad, 3

Tags: geometry , cyclic quadrilateral , equal segments

$ABCD$ is a cyclic quadrilateral. $M$ is the midpoint of $CD$. The diagonals meet at $P$. The circle through $P$ which touches $CD$ at $M$ meets $AC$ again at $R$ and $BD$ again at $Q$. The point $S$ on $BD$ is such that $BS = DQ$. The line through $S$ parallel to $AB$ meets $AC$ at $T$. Show that $AT = RC$.

Denmark (Mohr) - geometry, 2016.3

Tags: right angle , area , geometry , equal segments

Prove that all quadrilaterals $ABCD$ where $\angle B = \angle D = 90^o$, $|AB| = |BC|$ and $|AD| + |DC| = 1$, have the same area. [img]https://1.bp.blogspot.com/-55lHuAKYEtI/XzRzDdRGDPI/AAAAAAAAMUk/n8lYt3fzFaAB410PQI4nMEz7cSSrfHEgQCLcBGAsYHQ/s0/2016%2Bmohr%2Bp3.png[/img]

1990 Swedish Mathematical Competition, 4

Tags: angle bisector , geometry , equal segments

$ABCD$ is a quadrilateral. The bisectors of $\angle A$ and $\angle B$ meet at $E$. The line through $E$ parallel to $CD$ meets $AD$ at $L$ and $BC$ at $M$. Show that $LM = AL + BM$.

2020 BMT Fall, 20

Tags: equal segments , equal angles , geometry

Non-degenerate quadrilateral $ABCD$ with $AB = AD$ and $BC = CD$ has integer side lengths, and $\angle ABC = \angle BCD = \angle CDA$. If $AB = 3$ and $B \ne D$, how many possible lengths are there for $BC$?

Novosibirsk Oral Geo Oly VIII, 2017.5

Tags: equal segments , rectangle , geometry

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 $AK + BM = CM$.

2009 Dutch IMO TST, 5

Tags: rational , polygon , geometry , analytic geometry , equal segments

Suppose that we are given an $n$-gon of which all sides have the same length, and of which all the vertices have rational coordinates. Prove that $n$ is even.

2009 All-Russian Olympiad Regional Round, 11.6

Tags: equal segments , geometry

Point $D$ on side $BC$ of acute triangle ABC is such that $AB=AD$. The circumcircle of triangle $ABD$ intersects side $AC$ at points $A$ and $K$. Line $DK$ intersects the perpendicular drawn from $B$ on $AC$, at the point $L$. Prove that $CL= BC$

Kyiv City MO Juniors 2003+ geometry, 2020.7.41

Tags: perpendicular bisector , geometry , concurrency , equal segments

In the quadrilateral $ABCD$, $AB = BC$ . The point $E$ lies on the line $AB$ is such that $BD= BE$ and $AD \perp DE$. Prove that the perpendicular bisectors to segments $AD, CD$ and $CE$ intersect at one point.

2009 Thailand Mathematical Olympiad, 3

Tags: parallel , geometry , equal segments

Let $ABCD$ be a convex quadrilateral with the property that $MA \cdot MC + MA \cdot CD = MB \cdot MD$, where $M$ is the intersection of the diagonals $AC$ and $BD$. The angle bisector of $\angle ACD$ is drawn intersecting ray $\overrightarrow{BA}$ at $K$. Prove that $BC = DK$ if and only if $AB \parallel CD$.

2016 Czech-Polish-Slovak Junior Match, 4

Tags: equal segments , geometry , perpendicular bisector

We are given an acute-angled triangle $ABC$ with $AB < AC < BC$. Points $K$ and $L$ are chosen on segments $AC$ and $BC$, respectively, so that $AB = CK = CL$. Perpendicular bisectors of segments $AK$ and $BL$ intersect the line $AB$ at points $P$ and $Q$, respectively. Segments $KP$ and $LQ$ intersect at point $M$. Prove that $AK + KM = BL + LM$. Poland

2004 District Olympiad, 2

Tags: equal segments , geometry , parallel , angle bisector

Let $ABC$ be a triangle and $D$ a point on the side $BC$. The angle bisectors of $\angle ADB ,\angle ADC$ intersect $AB ,AC$ at points $M ,N$ respectively. The angle bisectors of $\angle ABD , \angle ACD$ intersects $DM , DN$ at points $K , L$ respectively. Prove that $AM = AN$ if and only if $MN$ and $KL$ are parallel.

2020 Ukrainian Geometry Olympiad - April, 4

Tags: angle , equal segments , equal angles , geometry , square

On the sides $AB$ and $AD$ of the square $ABCD$, the points $N$ and $P$ are selected respectively such that $NC=NP$. The point $Q$ is chosen on the segment $AN$ so that $\angle QPN = \angle NCB$. Prove that $2\angle BCQ = \angle AQP$.

2018 Saudi Arabia IMO TST, 3

Tags: parallel , cyclic quadrilateral , equal segments , angle bisector , perpendicular , geometry

Let $ABCD$ be a convex quadrilateral inscibed in circle $(O)$ such that $DB = DA + DC$. The point $P$ lies on the ray $AC$ such that $AP = BC$. The point $E$ is on $(O)$ such that $BE \perp AD$. Prove that $DP$ is parallel to the angle bisector of $\angle BEC$.

2015 Dutch IMO TST, 1

Tags: right angle , equal segments , equal angles , geometry

In a quadrilateral $ABCD$ we have $\angle A = \angle C = 90^o$. Let $E$ be a point in the interior of $ABCD$. Let $M$ be the midpoint of $BE$. Prove that $\angle ADB = \angle EDC$ if and only if $|MA| = |MC|$.

2011 Sharygin Geometry Olympiad, 7

Tags: circles , equal segments , common tangent , angle , geometry

Circles $\omega$ and $\Omega$ are inscribed into the same angle. Line $\ell$ meets the sides of angles, $\omega$ and $\Omega$ in points $A$ and $F, B$ and $C, D$ and $E$ respectively (the order of points on the line is $A,B,C,D,E, F$). It is known that$ BC = DE$. Prove that $AB = EF$.

2016 Hanoi Open Mathematics Competitions, 11

Tags: equal segments , geometry

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 $IE = I F$.

2013 Saudi Arabia BMO TST, 6

Tags: midpoint , equal segments , incenter , geometry

Let $ABC$ be a triangle with incenter $I,$ and let $D,E,F$ be the midpoints of sides $BC, CA, AB$, respectively. Lines $BI$ and $DE$ meet at $P $ and lines $CI$ and $DF$ meet at $Q$. Line $PQ$ meets sides $AB$ and $AC$ at $T$ and $S$, respectively. Prove that $AS = AT$

Kyiv City MO Juniors 2003+ geometry, 2018.7.4

Tags: right angle , angle , equal segments , geometry

Inside the triangle $ABC $, the point $P $ is selected so that $BC = AP $ and $\angle APC = 180 {} ^ \circ - \angle ABC $. On the side $AB $ there is a point $K $, for which $AK = KB + PC $. Prove that $\angle AKC = 90 {} ^ \circ $. (Danilo Hilko)

Denmark (Mohr) - geometry, 2012.5

Tags: equilateral , equal segments , geometry , equal angles , hexagon

In the hexagon $ABCDEF$, all angles are equally large. The side lengths satisfy $AB = CD = EF = 3$ and $BC = DE = F A = 2$. The diagonals $AD$ and $CF$ intersect each other in the point $G$. The point $H$ lies on the side $CD$ so that $DH = 1$. Prove that triangle $EGH$ is equilateral.

2012 Belarus Team Selection Test, 2

Tags: geometry , isosceles , inscirbed , equal segments

$A, B, C, D, E$ are five points on the same circle, so that $ABCDE$ is convex and we have $AB = BC$ and $CD = DE$. Suppose that the lines $(AD)$ and $(BE)$ intersect at $P$, and that the line $(BD)$ meets line $(CA)$ at $Q$ and line $(CE)$ at $T$. Prove that the triangle $PQT$ is isosceles. (I. Voronovich)

2016 Ecuador NMO (OMEC), 3

Tags: angle , equilateral , geometry , equal segments

Let $A, B, C, D$ be four different points on a line $\ell$, such that $AB = BC = CD$. In one of the semiplanes determined by the line $\ell$, the points $P$ and $Q$ are chosen in such a way that the triangle $CPQ$ is equilateral with its vertices named clockwise. Let $M$ and $N$ be two points on the plane such that the triangles $MAP$ and $NQD$ are equilateral (the vertices are also named clockwise). Find the measure of the angle $\angle MBN$.

1991 Austrian-Polish Competition, 3

Tags: midpoint , equal segments , combinatorial geometry , geometry , combinatorics

Given two distinct points $A_1,A_2$ in the plane, determine all possible positions of a point $A_3$ with the following property: There exists an array of (not necessarily distinct) points $P_1,P_2,...,P_n$ for some $n \ge 3$ such that the segments $P_1P_2,P_2P_3,...,P_nP_1$ have equal lengths and their midpoints are $A_1, A_2, A_3, A_1, A_2, A_3, ...$ in this order.