This website contains problems from math contests. Problems and corresponding tags were obtained from the Art of Problem Solving website.

Tags were heavily modified to better represent problems.

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Found problems: 65

2013 Sharygin Geometry Olympiad, 4

The diagonals of a convex quadrilateral $ABCD$ meet at point $L$. The orthocenter $H$ of the triangle $LAB$ and the circumcenters $O_1, O_2$, and $O_3$ of the triangles $LBC, LCD$, and $LDA$ were marked. Then the whole configuration except for points $H, O_1, O_2$, and $O_3$ was erased. Restore it using a compass and a ruler.

1984 IMO, 1

Let $ABCD$ be a convex quadrilateral with the line $CD$ being tangent to the circle on diameter $AB$. Prove that the line $AB$ is tangent to the circle on diameter $CD$ if and only if the lines $BC$ and $AD$ are parallel.

1984 Brazil National Olympiad, 5

$ABCD$ is any convex quadrilateral. Squares center $E, F, G, H$ are constructed on the outside of the edges $AB, BC, CD$ and $DA$ respectively. Show that $EG$ and $FH$ are equal and perpendicular.

1997 Spain Mathematical Olympiad, 5

Prove that in every convex quadrilateral of area $1$, the sum of the lengths of the sides and diagonals is not smaller than $2(2+\sqrt2)$.

1984 IMO Shortlist, 14

Let $ABCD$ be a convex quadrilateral with the line $CD$ being tangent to the circle on diameter $AB$. Prove that the line $AB$ is tangent to the circle on diameter $CD$ if and only if the lines $BC$ and $AD$ are parallel.

2010 Junior Balkan Team Selection Tests - Romania, 2

Let $ABCD$ be a convex quadrilateral with $\angle BCD= 120^o, \angle {CBA} = 45^o, \angle {CBD} = 15^o$ and $\angle {CAB} = 90^o$. Show that $AB = AD$.

2021 Brazil National Olympiad, 1

Let \(ABCD\) be a convex quadrilateral in the plane and let \(O_{A}, O_{B}, O_{C}\) and \(O_{D}\) be the circumcenters of the triangles \(BCD, CDA, DAB\) and \(ABC\), respectively. Suppose these four circumcenters are distinct points. Prove that these points are not on a same circle.

2021 Iranian Combinatorics Olympiad, P6

Let $\mathcal{P}$ be a convex polygon and $\textbf{T}$ be a triangle with vertices among the vertices of $\mathcal{P}$. By removing $\textbf{T}$ from $\mathcal{P}$, we end up with $0, 1, 2,$ or $3$ smaller polygons (possibly with shared vertices) which we call the effect of $\textbf{T}$. A triangulation of $P$ is a way of dissecting it into some triangles using some non-intersecting diagonals. We call a triangulation of $\mathcal{P}$ $\underline{\text{beautiful}}$, if for each of its triangles, the effect of this triangle contains exactly one polygon with an odd number of vertices. Prove that a triangulation of $\mathcal{P}$ is beautiful if and only if we can remove some of its diagonals and end up with all regions as quadrilaterals.

2009 May Olympiad, 2

Let $ABCD$ be a convex quadrilateral such that the triangle $ABD$ is equilateral and the triangle $BCD$ is isosceles, with $\angle C = 90^o$. If $E$ is the midpoint of the side $AD$, determine the measure of the angle $\angle CED$.

2000 Greece JBMO TST, 2

Let $ABCD$ be a convex quadrilateral with $AB=CD$. From a random point $P$ of it's diagonal $BD$, we draw a line parallel to $AB$ that intersects $AD$ at point $M$ and a line parallel to $CD$ that intersects $BC$ at point $N$. Prove that: a) The sum $PM+PN$ is constant, independent of the position of $P$ on the diagonal $BD$. b) $MN\le BD$. When the equality holds?

2010 Silk Road, 1

In a convex quadrilateral it is known $ABCD$ that $\angle ADB + \angle ACB = \angle CAB + \angle DBA = 30^{\circ}$ and $AD = BC$. Prove that from the lengths $DB$, $CA$ and $DC$, you can make a right triangle.

1991 Mexico National Olympiad, 4

The diagonals $AC$ and $BD$ of a convex quarilateral $ABCD$ are orthogonal. Let $M,N,R,S$ be the midpoints of the sides $AB,BC,CD$ and $DA$ respectively, and let $W,X,Y,Z$ be the projections of the points $M,N,R$ and $S$ on the lines $CD,DA,AB$ and $BC$, respectively. Prove that the points $M,N,R,S,W,X,Y$ and $Z$ lie on a circle.

1989 Bundeswettbewerb Mathematik, 3

A convex polygon is divided into finitely many quadrilaterals. Prove that at least one of these quadrilaterals must also be convex.

1969 IMO Longlists, 45

Given $n>4$ points in the plane, no three collinear. Prove that there are at least $\frac{(n-3)(n-4)}{2}$ convex quadrilaterals with vertices amongst the $n$ points.

1969 IMO, 5

Given $n>4$ points in the plane, no three collinear. Prove that there are at least $\frac{(n-3)(n-4)}{2}$ convex quadrilaterals with vertices amongst the $n$ points.

2013 Turkmenistan National Math Olympiad, 4

Let $ ABCD$ be a convex quadrilateral such that the sides $ AB, AD, BC$ satisfy $ AB \equal{} AD \plus{} BC.$ There exists a point $ P$ inside the quadrilateral at a distance $ h$ from the line $ CD$ such that $ AP \equal{} h \plus{} AD$ and $ BP \equal{} h \plus{} BC.$ Show that: \[ \frac {1}{\sqrt {h}} \geq \frac {1}{\sqrt {AD}} \plus{} \frac {1}{\sqrt {BC}} \]

1969 IMO Longlists, 16

$(CZS 5)$ A convex quadrilateral $ABCD$ with sides $AB = a, BC = b, CD = c, DA = d$ and angles $\alpha = \angle DAB, \beta = \angle ABC, \gamma = \angle BCD,$ and $\delta = \angle CDA$ is given. Let $s = \frac{a + b + c +d}{2}$ and $P$ be the area of the quadrilateral. Prove that $P^2 = (s - a)(s - b)(s - c)(s - d) - abcd \cos^2\frac{\alpha +\gamma}{2}$

2008 India Regional Mathematical Olympiad, 6

Let $BCDK$ be a convex quadrilateral such that $BC=BK$ and $DC=DK$. $A$ and $E$ are points such that $ABCDE$ is a convex pentagon such that $AB=BC$ and $DE=DC$ and $K$ lies in the interior of the pentagon $ABCDE$. If $\angle ABC=120^{\circ}$ and $\angle CDE=60^{\circ}$ and $BD=2$ then determine area of the pentagon $ABCDE$.

1969 IMO Shortlist, 68

$(USS 5)$ Given $5$ points in the plane, no three of which are collinear, prove that we can choose $4$ points among them that form a convex quadrilateral.

1994 Mexico National Olympiad, 5

$ABCD$ is a convex quadrilateral. Take the $12$ points which are the feet of the altitudes in the triangles $ABC, BCD, CDA, DAB$. Show that at least one of these points must lie on the sides of $ABCD$.

1977 IMO Longlists, 22

Let $S$ be a convex quadrilateral $ABCD$ and $O$ a point inside it. The feet of the perpendiculars from $O$ to $AB, BC, CD, DA$ are $A_1, B_1, C_1, D_1$ respectively. The feet of the perpendiculars from $O$ to the sides of $S_i$, the quadrilateral $A_iB_iC_iD_i$, are $A_{i+1}B_{i+1}C_{i+1}D_{i+1}$, where $i = 1, 2, 3.$ Prove that $S_4$ is similar to S.

2008 IMO Shortlist, 7

Let $ ABCD$ be a convex quadrilateral with $ BA\neq BC$. Denote the incircles of triangles $ ABC$ and $ ADC$ by $ \omega_{1}$ and $ \omega_{2}$ respectively. Suppose that there exists a circle $ \omega$ tangent to ray $ BA$ beyond $ A$ and to the ray $ BC$ beyond $ C$, which is also tangent to the lines $ AD$ and $ CD$. Prove that the common external tangents to $ \omega_{1}$ and $\omega_{2}$ intersect on $ \omega$. [i]Author: Vladimir Shmarov, Russia[/i]

2000 IMO Shortlist, 6

Let $ ABCD$ be a convex quadrilateral. The perpendicular bisectors of its sides $ AB$ and $ CD$ meet at $ Y$. Denote by $ X$ a point inside the quadrilateral $ ABCD$ such that $ \measuredangle ADX \equal{} \measuredangle BCX < 90^{\circ}$ and $ \measuredangle DAX \equal{} \measuredangle CBX < 90^{\circ}$. Show that $ \measuredangle AYB \equal{} 2\cdot\measuredangle ADX$.

1992 IMO Longlists, 13

Let $ABCD$ be a convex quadrilateral such that $AC = BD$. Equilateral triangles are constructed on the sides of the quadrilateral. Let $O_1,O_2,O_3,O_4$ be the centers of the triangles constructed on $AB,BC,CD,DA$ respectively. Show that $O_1O_3$ is perpendicular to $O_2O_4.$

2006 Spain Mathematical Olympiad, 3

The diagonals $AC$ and $BD$ of a convex quadrilateral $ABCD$ intersect at $E$. Denotes by $S_1,S_2$ and $S$ the areas of the triangles $ABE$, $CDE$ and the quadrilateral $ABCD$ respectively. Prove that $\sqrt{S_1}+\sqrt{S_2}\le \sqrt{S}$ . When equality is reached?