Olympiad problems

2013 China Team Selection Test, 2

The circumcircle of triangle $ABC$ has centre $O$. $P$ is the midpoint of $\widehat{BAC}$ and $QP$ is the diameter. Let $I$ be the incentre of $\triangle ABC$ and let $D$ be the intersection of $PI$ and $BC$. The circumcircle of $\triangle AID$ and the extension of $PA$ meet at $F$. The point $E$ lies on the line segment $PD$ such that $DE=DQ$. Let $R,r$ be the radius of the inscribed circle and circumcircle of $\triangle ABC$, respectively. Show that if $\angle AEF=\angle APE$, then $\sin^2\angle BAC=\dfrac{2r}R$

2004 Thailand Mathematical Olympiad, 1

Tags: geometry

A $\vartriangle ABC$ is given with $\angle A = 70^o$. The angle bisectors of $\vartriangle ABC$ intersect at $I$. Suppose that $CA + AI=BC$. Find, with proof, the value of $\angle B$.

2019 Estonia Team Selection Test, 11

Tags: circumcircle , combinatorial geometry , perimeter , geometry

Given a circle $\omega$ with radius $1$. Let $T$ be a set of triangles good, if the following conditions apply: (a) the circumcircle of each triangle in the set $T$ is $\omega$; (b) The interior of any two triangles in the set $T$ has no common point. Find all positive real numbers $t$, for which for each positive integer $n$ there is a good set of $n$ triangles, where the perimeter of each triangle is greater than $t$.

1994 Chile National Olympiad, 2

Tags: geometry , rectangle , triangle

Show that it is possible to cut any triangle into several pieces, so that a rectangle is formed when they are joined together.

2006 All-Russian Olympiad Regional Round, 8.7

Tags: geometry , angle

Segment equal to median $AA_0$ of triangle $ABC$ is drawn from point $A_0$ perpendicular to side $BC$ to the outer side of the triangle. Let's denote the second end of the constructed segment as $A_1$. Points $B_1$ and $C_1$ are constructed similarly. Find the angles of triangle $A_1B_1C_1$ if the angles of the triangle $ABC$ are $30^o$, $30^o$ and $120^o$. [hide=original wording]Медиану AA0 треугольника ABC отложили от точки A0 перпендикулярно стороне BC во внешнюю сторону треугольника. Обозначим второй конец построенного отрезка через A1. Аналогично строятся точки B1 и C1. Найдите углы треугольника A1B1C1, если углы треугольника ABC равны 30^o, 30^o и 120^o.[/hide]

2018 USAMTS Problems, 3:

Tags: cyclic quadrilateral , geometry

Cyclic quadrilateral $ABCD$ has $AC\perp BD$, $AB+CD=12$, and $BC+AD=13$. FInd the greatest possible area of $ABCD$.

1949-56 Chisinau City MO, 31

Tags: geometry , locus , chord , midpoint

Find the locus of the points that are the midpoints of the chords of the secant to the given circle and passing through a given point.

1998 All-Russian Olympiad, 8

Tags: geometry , rectangle , rotation , vector , function , induction , quadratic

A figure $\Phi$ composed of unit squares has the following property: if the squares of an $m \times n$ rectangle ($m,n$ are fixed) are filled with numbers whose sum is positive, the figure $\Phi$ can be placed within the rectangle (possibly after being rotated) so that the sum of the covered numbers is also positive. Prove that a number of such figures can be put on the $m\times n$ rectangle so that each square is covered by the same number of figures.

2019 Stars of Mathematics, 2

Tags: geometry

Let $A$ and $C$ be two points on a circle $X$ so that $AC$ is not diameter and $P$ a segment point on $AC$ different from its middle. The circles $c_1$ and $c_2$, inner tangents in $A$, respectively $C$, to circle $X$, pass through the point $P$ ¸ and intersect a second time at point $Q$. The line $PQ$ intersects the circle $X$ in points $B$ and $D$. The circle $c_1$ intersects the segments $AB$ and $AD$ in $K$, respectively $N$, and circle $c_2$ intersects segments $CB$ and ¸ $CD $ in $L$, respectively $M$. Show that: a) the $KLMN$ quadrilateral is isosceles trapezoid; b) $Q$ is the middle of the segment $BD$. Proposed by Thanos Kalogerakis

2021 Bosnia and Herzegovina Junior BMO TST, 3

Tags: geometry , right angle , equal angles

In the convex quadrilateral $ABCD$, $AD = BD$ and $\angle ACD = 3 \angle BAC$. Let $M$ be the midpoint of side $AD$. If the lines $CM$ and $AB$ are parallel, prove that the angle $\angle ACB$ is right.

2006 Petru Moroșan-Trident, 3

Tags: geometry , orthocenter

In an acute-angled triangle $ ABC $ consider $ A_1,B_1,C_1 $ to be the symmetric points of the orthocenter of $ ABC $ to the sides $ BC,AC,AB, $ respectively. Show that if the centroids of the triangles $ ABC,A_1B_1C_1 $ are the same, then $ ABC $ is equilateral. [i]Carmen Botea[/i]

2017 Korea Winter Program Practice Test, 4

Tags: number theory , geometry

For a point $P$ on the plane, denote by $\lVert P \rVert$ the distance to its nearest lattice point. Prove that there exists a real number $L > 0$ satisfying the following condition: For every $\ell > L$, there exists an equilateral triangle $ABC$ with side-length $\ell$ and $\lVert A \rVert, \lVert B \rVert, \lVert C \rVert < 10^{-2017}$.

2024 Euler Olympiad, Round 1, 3

Tags: geometry

In a convex trapezoid $ABCD$, side $AD$ is twice the length of the other sides. Let $E$ and $F$ be points on segments $AC$ and $BD$, respectively, such that $\angle BEC = 70^\circ$ and $\angle BFC = 80^\circ$. Determine the ratio of the areas of quadrilaterals $BEFC$ and $ABCD$. [i]Proposed by Zaza Meliqidze, Georgia [/i]

2022 BMT, 12

Tags: geometry

Parallelograms $ABGF$, $CDGB$ and $EFGD$ are drawn so that $ABCDEF$ is a convex hexagon, as shown. If $\angle ABG = 53^o$ and $\angle CDG = 56^o$, what is the measure of $\angle EFG$, in degrees? [img]https://cdn.artofproblemsolving.com/attachments/9/f/79d163662e02bc40d2636a76b73f632e59d584.png[/img]

Fractal Edition 1, P4

Tags: geometry

In triangle $ ABC $, $ D $, $ E $, and $ F $ are the feet of the perpendiculars from the vertices $ A $, $ B $, and $ C $, respectively. The parallel to $ EF $ through $ D $ intersects $ AB $ at $ P_B $ and $ AC $ at $ P_C $. Let $ X $ be the intersection of $ EF $ and $ BC $. Prove that the circumcircle of triangle $ P_B P_C X $ passes through the midpoint of side $ BC $.

1974 IMO, 2

Tags: trigonometry , circumcircle , geometry , triangle , trigonometric inequality

Let $ABC$ be a triangle. Prove that there exists a point $D$ on the side $AB$ of the triangle $ABC$, such that $CD$ is the geometric mean of $AD$ and $DB$, iff the triangle $ABC$ satisfies the inequality $\sin A\sin B\le\sin^2\frac{C}{2}$. [hide="Comment"][i]Alternative formulation, from IMO ShortList 1974, Finland 2:[/i] We consider a triangle $ABC$. Prove that: $\sin(A) \sin(B) \leq \sin^2 \left( \frac{C}{2} \right)$ is a necessary and sufficient condition for the existence of a point $D$ on the segment $AB$ so that $CD$ is the geometrical mean of $AD$ and $BD$.[/hide]

2019 Stanford Mathematics Tournament, 10

Tags: geometry

Let $ABC$ be an acute triangle with $BC = 48$. Let $M$ be the midpoint of $BC$, and let $D$ and $E$ be the feet of the altitudes drawn from $B$ and $C$ to $AC$ and $AB$ respectively. Let $P$ be the intersection between the line through $A$ parallel to $BC$ and line $DE$. If $AP = 10$, compute the length of $PM$,

2016 Costa Rica - Final Round, G3

Tags: collinear , geometry , rectangle

Let the $JHIZ$ be a rectangle and let $A$ and $C$ be points on the sides $ZI$ and $ZJ$, respectively. The perpendicular from $A$ on $CH$ intersects line $HI$ at point $X$ and perpendicular from $C$ on $AH$ intersects line $HJ$ at point $Y$. Show that points $X, Y$, and $Z$ are collinear.

2021 Moldova Team Selection Test, 11

Tags: geometry

In a convex quadrilateral $ABCD$ the angles $BAD$ and $BCD$ are equal. Points $M$ and $N$ lie on the sides $(AB)$ and $(BC)$ such that the lines $MN$ and $AD$ are parallel and $MN=2AD$. The point $H$ is the orthocenter of the triangle $ABC$ and the point $K$ is the midpoint of $MN$. Prove that the lines $KH$ and $CD$ are perpendicular.

1912 Eotvos Mathematical Competition, 3

Tags: geometry , perpendicular

Prove that the diagonals of a quadrilateral are perpendicular if and only if the sum of the squares of one pair of opposite sides equals that of the other.

1999 Harvard-MIT Mathematics Tournament, 7

Tags: calculus , geometry , derivative

If a right triangle is drawn in a semicircle of radius $1/2$ with one leg (not the hypotenuse) along the diameter, what is the triangle's maximum possible area?

2018 Stanford Mathematics Tournament, 9

Tags: geometry

Let $ABCD$ be a cyclic quadrilateral with $3AB = 2AD$ and $BC = CD$. The diagonals $AC$ and $BD$ intersect at point $X$. Let $E$ be a point on $AD$ such that $DE = AB$ and $Y$ be the point of intersection of lines $AC$ and $BE$. If the area of triangle $ABY$ is $5$, then what is the area of quadrilateral $DEY X$?

2013 Rioplatense Mathematical Olympiad, Level 3, 2

Tags: analytic geometry , graphing lines , slope , similar triangles , geometry

Let $ABCD$ be a square, and let $E$ and $F$ be points in $AB$ and $BC$ respectively such that $BE=BF$. In the triangle $EBC$, let N be the foot of the altitude relative to $EC$. Let $G$ be the intersection between $AD$ and the extension of the previously mentioned altitude. $FG$ and $EC$ intersect at point $P$, and the lines $NF$ and $DC$ intersect at point $T$. Prove that the line $DP$ is perpendicular to the line $BT$.

2014 Bundeswettbewerb Mathematik, 3

Tags: geometry , parallelogram

A regular hexagon with side length $1$ is given. Using a ruler construct points in such a way that among the given and constructed points there are two such points that the distance between them is $\sqrt7$. Notes: ''Using a ruler construct points $\ldots$'' means: Newly constructed points arise only as the intersection of straight lines connecting two points that are given or already constructed. In particular, no length can be measured by the ruler.

2017 Princeton University Math Competition, A3

Tags: geometry

Triangle $ABC$ has incenter $I$. The line through $I$ perpendicular to $AI$ meets the circumcircle of $ABC$ at points $P$ and $Q$, where $P$ and $B$ are on the same side of $AI$. Let $X$ be the point such that $PX$ // $CI$ and $QX$ // $BI$. Show that $P B, QC$, and $IX$ intersect at a common point.