Olympiad problems

2010 IFYM, Sozopol, 8

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 $OP = OQ$.

2023-24 IOQM India, 24

Tags: geometry , trapezoid

A trapezium in the plane is a quadrilateral in which a pair of opposite sides are parallel. A trapezium is said to be non-degenerate if it has positive area. Find the number of mutually non-congruent, non-degenerate trapeziums whose sides are four distinct integers from the set $\{5,6,7,8,9,10\}$

2022-2023 OMMC, 16

Tags: geometry , trapezoid

Let $ABCD$ be an isosceles trapezoid with $AB=5$, $CD = 8$, and $BC = DA = 6$. There exists an angle $\theta$ such that there is only one point $X$ satisfying $\angle AXD = 180^{\circ} - \angle BXC = \theta$. Find $\sin(\theta)^2$.

2001 Estonia National Olympiad, 5

Tags: combinatorics , combinatorial geometry , lattice , trapezoid

Consider all trapezoids in a coordinate plane with interior angles of $90^o, 90^o, 45^o$ and $135^o$ whose bases are parallel to a coordinate axis and whose vertices have integer coordinates. Define the [i]size [/i] of such a trapezoid as the total number of points with integer coordinates inside and on the boundary of the trapezoid. (a) How many pairwise non-congruent such trapezoids of size $2001$ are there? (b) Find all positive integers not greater than $50$ that do not appear as sizes of any such trapezoid.

2005 China Team Selection Test, 3

Tags: rhombus , trapezoid , geometry , circumcircle , perpendicular bisector

Find the least positive integer $n$ ($n\geq 3$), such that among any $n$ points (no three are collinear) in the plane, there exist three points which are the vertices of a non-isoscele triangle.

2012 CentroAmerican, 2

Tags: perpendicular bisector , geometry , trapezoid , circumcircle

Let $\gamma$ be the circumcircle of the acute triangle $ABC$. Let $P$ be the midpoint of the minor arc $BC$. The parallel to $AB$ through $P$ cuts $BC, AC$ and $\gamma$ at points $R,S$ and $T$, respectively. Let $K \equiv AP \cap BT$ and $L \equiv BS \cap AR$. Show that $KL$ passes through the midpoint of $AB$ if and only if $CS = PR$.

2017 Ukrainian Geometry Olympiad, 2

Tags: angle chasing , trapezoid , equal angles , geometry

Point $M$ is the midpoint of the base $BC$ of trapezoid $ABCD$. On base $AD$, point $P$ is selected. Line $PM$ intersects line $DC$ at point $Q$, and the perpendicular from $P$ on the bases intersects line $BQ$ at point $K$. Prove that $\angle QBC = \angle KDA$.

2016 Sharygin Geometry Olympiad, 6

Tags: trapezoid , geometry

The sidelines $AB$ and $CD$ of a trapezoid meet at point $P$, and the diagonals of this trapezoid meet at point $Q$. Point $M$ on the smallest base $BC$ is such that $AM=MD$. Prove that $\angle PMB=\angle QMB$.

1981 AMC 12/AHSME, 5

Tags: geometry , trapezoid

In trapezoid $ABCD$, sides $AB$ and $CD$ are parallel, and diagonal $BD$ and side $AD$ have equal length. If $m\angle DBC=110^\circ$ and $m\angle CBD =30^\circ$, then $m \angle ADB=$ $\text{(A)}\ 80^\circ \qquad \text{(B)}\ 90^\circ \qquad \text{(C)}\ 100^\circ \qquad \text{(D)}\ 110^\circ \qquad \text{(E)}\ 120^\circ$

1965 AMC 12/AHSME, 8

Tags: trapezoid , ratio , geometry

One side of a given triangle is $ 18$ inches. Inside the triangle a line segment is drawn parallel to this side forming a trapezoid whose area is one-third of that of the triangle. The length of this segment, in inches, is: $ \textbf{(A)}\ 6\sqrt {6} \qquad \textbf{(B)}\ 9\sqrt {2} \qquad \textbf{(C)}\ 12 \qquad \textbf{(D)}\ 6\sqrt {3} \qquad \textbf{(E)}\ 9$

1975 AMC 12/AHSME, 26

Tags: angle bisector , trapezoid , geometry

In acute triangle $ABC$ the bisector of $\measuredangle A$ meets side $BC$ at $D$. The circle with center $B$ and radius $BD$ intersects side $AB$ at $M$; and the circle with center $C$ and radius $CD$ intersects side $AC$ at $N$. Then it is always true that $ \textbf{(A)}\ \measuredangle CND+\measuredangle BMD-\measuredangle DAC =120^{\circ} \qquad\textbf{(B)}\ AMDN\ \text{is a trapezoid} \qquad\textbf{(C)}\ BC\ \text{is parallel to}\ MN \\ \qquad\textbf{(D)}\ AM-AN=\frac{3(DB-DC)}{2} \qquad\textbf{(E)}\ AB-AC=\frac{3(DB-DC)}{2}$

2024 China Team Selection Test, 19

Tags: combinatorics , trapezoid

$n$ is a positive integer. An equilateral triangle of side length $3n$ is split into $9n^2$ unit equilateral triangles, each colored one of red, yellow, blue, such that each color appears $3n^2$ times. We call a trapezoid formed by three unit equilateral triangles as a "standard trapezoid". If a "standard trapezoid" contains all three colors, we call it a "colorful trapezoid". Find the maximum possible number of "colorful trapezoids".

1995 IberoAmerican, 2

Tags: geometry , similar triangles , trapezoid , trigonometry

The incircle of a triangle $ABC$ touches the sides $BC$, $CA$, $AB$ at the points $D$, $E$, $F$ respectively. Let the line $AD$ intersect this incircle of triangle $ABC$ at a point $X$ (apart from $D$). Assume that this point $X$ is the midpoint of the segment $AD$, this means, $AX = XD$. Let the line $BX$ meet the incircle of triangle $ABC$ at a point $Y$ (apart from $X$), and let the line $CX$ meet the incircle of triangle $ABC$ at a point $Z$ (apart from $X$). Show that $EY = FZ$.

2004 Junior Balkan Team Selection Tests - Romania, 2

Tags: perpendicular bisector , trigonometry , rectangle , geometry , angle bisector , trapezoid , incenter

Let $ABC$ be an isosceles triangle with $AB=AC$. Consider a variable point $P$ on the extension of the segment $BC$ beyound $B$ (in other words, $P$ lies on the line $BC$ such that the point $B$ lies inside the segment $PC$). Let $r_{1}$ be the radius of the incircle of the triangle $APB$, and let $r_{2}$ be the radius of the $P$-excircle of the triangle $APC$. Prove that the sum $r_{1}+r_{2}$ of these two radii remains constant when the point $P$ varies. [i]Remark.[/i] The $P$-excircle of the triangle $APC$ is defined as the circle which touches the side $AC$ and the [i]extensions[/i] of the sides $AP$ and $CP$.

1955 AMC 12/AHSME, 31

Tags: trapezoid , geometry

An equilateral triangle whose side is $ 2$ is divided into a triangle and a trapezoid by a line drawn parallel to one of its sides. If the area of the trapezoid equals one-half of the area of the original triangle, the length of the median of the trapezoid is: $ \textbf{(A)}\ \frac{\sqrt{6}}{2} \qquad \textbf{(B)}\ \sqrt{2} \qquad \textbf{(C)}\ 2\plus{}\sqrt{2} \qquad \textbf{(D)}\ \frac{2\plus{}\sqrt{2}}{2} \qquad \textbf{(E)}\ \frac{2\sqrt{3}\minus{}\sqrt{6}}{2}$

2012 Tuymaada Olympiad, 3

Tags: incenter , geometric transformation , symmetry , reflection , trapezoid , circumcircle , geometry

Point $P$ is taken in the interior of the triangle $ABC$, so that \[\angle PAB = \angle PCB = \dfrac {1} {4} (\angle A + \angle C).\] Let $L$ be the foot of the angle bisector of $\angle B$. The line $PL$ meets the circumcircle of $\triangle APC$ at point $Q$. Prove that $QB$ is the angle bisector of $\angle AQC$. [i]Proposed by S. Berlov[/i]

2009 Tuymaada Olympiad, 2

Tags: geometry , homothety , angle bisector , trapezoid , geometric transformation

$ M$ is the midpoint of base $ BC$ in a trapezoid $ ABCD$. A point $ P$ is chosen on the base $ AD$. The line $ PM$ meets the line $ CD$ at a point $ Q$ such that $ C$ lies between $ Q$ and $ D$. The perpendicular to the bases drawn through $ P$ meets the line $ BQ$ at $ K$. Prove that $ \angle QBC \equal{} \angle KDA$. [i]Proposed by S. Berlov[/i]

2005 Purple Comet Problems, 18

Tags: trapezoid , number theory , ratio , geometry , relatively prime

The side lengths of a trapezoid are $\sqrt[4]{3}, \sqrt[4]{3}, \sqrt[4]{3}$, and $2 \cdot \sqrt[4]{3}$. Its area is the ratio of two relatively prime positive integers, $m$ and $n$. Find $m + n$.

2021 Yasinsky Geometry Olympiad, 5

Tags: isosceles , construction , trapezoid , geometry

Construct an equilateral trapezoid given the height and the midline, if it is known that the midline is divided by diagonals into three equal parts. (Grigory Filippovsky)

2018 Serbia Team Selection Test, 4

Tags: trapezoid , combinatorics

An isosceles trapezium is called [i]right[/i] if only one pair of its sides are parallel (i.e parallelograms are not right). A dissection of a rectangle into $n$ (can be different shapes) right isosceles trapeziums is called [i]strict[/i] if the union of any $i,(2\leq i \leq n)$ trapeziums in the dissection do not form a right isosceles trapezium. Prove that for any $n, n\geq 9$ there is a strict dissection of a $2017 \times 2018$ rectangle into $n$ right isosceles trapeziums. [i]Proposed by Bojan Basic[/i]

1992 IMTS, 1

Tags: trapezoid , geometry

In trapezoid $ABCD$, the diagonals intersect at $E$, the area of $\triangle ABE$ is 72 and the area of $\triangle CDE$ is 50. What is the area of trapezoid $ABCD$?

2007 IMO Shortlist, 3

Tags: trapezoid , geometry , homothety , ddit

The diagonals of a trapezoid $ ABCD$ intersect at point $ P$. Point $ Q$ lies between the parallel lines $ BC$ and $ AD$ such that $ \angle AQD \equal{} \angle CQB$, and line $ CD$ separates points $ P$ and $ Q$. Prove that $ \angle BQP \equal{} \angle DAQ$. [i]Author: Vyacheslav Yasinskiy, Ukraine[/i]

1997 Austrian-Polish Competition, 4

Tags: trapezoid , midpoint , orthocenter , geometry

In a trapezoid $ABCD$ with $AB // CD$, the diagonals $AC$ and $BD$ intersect at point $E$. Let $F$ and $G$ be the orthocenters of the triangles $EBC$ and $EAD$. Prove that the midpoint of $GF$ lies on the perpendicular from $E$ to $AB$.

1988 AMC 12/AHSME, 6

Tags: rectangle , rhombus , trapezoid , geometry , parallelogram

A figure is an equiangular parallelogram if and only if it is a $ \textbf{(A)}\ \text{rectangle}\qquad\textbf{(B)}\ \text{regular polygon}\qquad\textbf{(C)}\ \text{rhombus}\qquad\textbf{(D)}\ \text{square}\qquad\textbf{(E)}\ \text{trapezoid} $

2013 India Regional Mathematical Olympiad, 1

Tags: geometry , trapezoid , circumcircle

Let $ABC$ be an isosceles triangle with $AB=AC$ and let $\Gamma$ denote its circumcircle. A point $D$ is on arc $AB$ of $\Gamma$ not containing $C$. A point $E$ is on arc $AC$ of $\Gamma$ not containing $B$. If $AD=CE$ prove that $BE$ is parallel to $AD$.