Olympiad problems

2004 Estonia Team Selection Test, 6

Tags: polyhedron , pentagon , 3d geometry , hexagon , geometry , combinatorial geometry , combinatorics

Call a convex polyhedron a [i]footballoid [/i] if it has the following properties. (1) Any face is either a regular pentagon or a regular hexagon. (2) All neighbours of a pentagonal face are hexagonal (a [i]neighbour [/i] of a face is a face that has a common edge with it). Find all possibilities for the number of pentagonal and hexagonal faces of a footballoid.

1979 IMO Shortlist, 4

Tags: 3d geometry , prism , combinatorics , pentagon , geometry

We consider a prism which has the upper and inferior basis the pentagons: $A_{1}A_{2}A_{3}A_{4}A_{5}$ and $B_{1}B_{2}B_{3}B_{4}B_{5}$. Each of the sides of the two pentagons and the segments $A_{i}B_{j}$ with $i,j=1,\ldots,5$ is colored in red or blue. In every triangle which has all sides colored there exists one red side and one blue side. Prove that all the 10 sides of the two basis are colored in the same color.

1979 Bulgaria National Olympiad, Problem 5

Tags: geometry , triangle , pentagon

A convex pentagon $ABCDE$ satisfies $AB=BC=CA$ and $CD=DE=EC$. Let $S$ be the center of the equilateral triangle $ABC$ and $M$ and $N$ be the midpoints of $BD$ and $AE$, respectively. Prove that the triangles $SME$ and $SND$ are similar.

1997 Denmark MO - Mohr Contest, 3

Tags: reflection , pentagon , angle , geometry

About pentagon $ABCDE$ is known that angle $A$ and angle $C$ are right and that the sides $| AB | = 4$, $| BC | = 5$, $| CD | = 10$, $| DE | = 6$. Furthermore, the point $C'$ that appears by mirroring $C$ in the line $BD$, lies on the line segment $AE$. Find angle $E$.

2018 Brazil Team Selection Test, 1

Tags: inscribed circle , pentagon , geometry

Let $ABCDE$ be a convex pentagon such that $AB=BC=CD$, $\angle{EAB}=\angle{BCD}$, and $\angle{EDC}=\angle{CBA}$. Prove that the perpendicular line from $E$ to $BC$ and the line segments $AC$ and $BD$ are concurrent.

2016 NIMO Problems, 3

Tags: geometry , pentagon

Convex pentagon $ABCDE$ satisfies $AB \parallel DE$, $BE \parallel CD$, $BC \parallel AE$, $AB = 30$, $BC = 18$, $CD = 17$, and $DE = 20$. Find its area. [i] Proposed by Michael Tang [/i]

Brazil L2 Finals (OBM) - geometry, 2012.4

Tags: pentagon , angle chasing , region , equilateral , angle

The figure below shows a regular $ABCDE$ pentagon inscribed in an equilateral triangle $MNP$ . Determine the measure of the angle $CMD$. [img]http://4.bp.blogspot.com/-LLT7hB7QwiA/Xp9fXOsihLI/AAAAAAAAL14/5lPsjXeKfYwIr5DyRAKRy0TbrX_zx1xHQCK4BGAYYCw/s200/2012%2Bobm%2Bl2.png[/img]

2013 Sharygin Geometry Olympiad, 1

Tags: geometry , right angle , equal segments , pentagon

Let $ABCDE$ be a pentagon with right angles at vertices $B$ and $E$ and such that $AB = AE$ and $BC = CD = DE$. The diagonals $BD$ and $CE$ meet at point $F$. Prove that $FA = AB$.

2018 Germany Team Selection Test, 2

Tags: geometry , inscribed circle , pentagon

Let $ABCDE$ be a convex pentagon such that $AB=BC=CD$, $\angle{EAB}=\angle{BCD}$, and $\angle{EDC}=\angle{CBA}$. Prove that the perpendicular line from $E$ to $BC$ and the line segments $AC$ and $BD$ are concurrent.

2012 Bosnia and Herzegovina Junior BMO TST, 1

Tags: pentagon , circles , geometry

On circle $k$ there are clockwise points $A$, $B$, $C$, $D$ and $E$ such that $\angle ABE = \angle BEC = \angle ECD = 45^{\circ}$. Prove that $AB^2 + CE^2 = BE^2 + CD^2$

1969 All Soviet Union Mathematical Olympiad, 124

Tags: geometry , pentagon

Given a pentagon with all equal sides. a) Prove that there exist such a point on the maximal diagonal, that every side is seen from it inside a right angle. (side $AB$ is seen from the point $C$ inside an arbitrary angle that is greater or equal than $\angle ACB$) b) Prove that the circles constructed on its sides as on the diameters cannot cover the pentagon entirely.

Kvant 2019, M2583

Tags: pentagon , geometry

On the side $DE$ and on the diagonal $BE$ of the regular pentagon $ABCDE$ we consider the squares $DEFG$ and $BEHI$. [list=a] [*] Prove that $A,I,$ and $G$ are collinear. [*] Prove that on this line lies also the centre $O$ of the square $BDJK$. [/list]

2009 JBMO Shortlist, 4

Tags: trapezoid , pentagon , geometry

Let $ ABCDE$ be a convex pentagon such that $ AB\plus{}CD\equal{}BC\plus{}DE$ and $ k$ a circle with center on side $ AE$ that touches the sides $ AB$, $ BC$, $ CD$ and $ DE$ at points $ P$, $ Q$, $ R$ and $ S$ (different from vertices of the pentagon) respectively. Prove that lines $ PS$ and $ AE$ are parallel.

1969 IMO Longlists, 70

Tags: pentagon , euclidean distance , geometric inequality , geometry

$(YUG 2)$ A park has the shape of a convex pentagon of area $50000\sqrt{3} m^2$. A man standing at an interior point $O$ of the park notices that he stands at a distance of at most $200 m$ from each vertex of the pentagon. Prove that he stands at a distance of at least $100 m$ from each side of the pentagon.

1983 Bundeswettbewerb Mathematik, 1

Tags: 3d geometry , pentagon , hexagon , geometry

The surface of a soccer ball is made up of black pentagons and white hexagons together. On the sides of each pentagon are nothing but hexagons, while on the sides of each border of hexagons alternately pentagons and hexagons. Determine from this information about the soccer ball , the number of its pentagons and its hexagons.

2007 Alexandru Myller, 3

Tags: pentagon , geometry

The convex pentagon $ ABCDE $ has the following properties: $ \text{(i)} AB=BC $ $ \text{(ii)} \angle ABE+\angle CBD =\angle DBE $ $ \text{(iii)} \angle AEB +\angle BDC=180^{\circ} $ Prove that the orthocenter of $ BDE $ lies on $ AC. $

1993 Rioplatense Mathematical Olympiad, Level 3, 6

Tags: equal segments , geometry , pentagon , right triangle , angle

Let $ABCDE$ be pentagon such that $AE = ED$ and $BC = CD$. It is known that $\angle BAE + \angle EDC + \angle CB A = 360^o$ and that $P$ is the midpoint of $AB$. Show that the triangle $ECP$ is right.

2014 Contests, 3

Tags: convex , diagonal , pentagon , geometry

Is there a convex pentagon in which each diagonal is equal to a side?

2018 Auckland Mathematical Olympiad, 3

Tags: geometry , area , pentagon

Consider the pentagon below. Find its area. [img]https://cdn.artofproblemsolving.com/attachments/7/b/02ad3852b72682513cf62a170ed4aa45c23785.png[/img]

Estonia Open Junior - geometry, 2019.2.1

Tags: equilateral , angle , pentagon , geometry

A pentagon can be divided into equilateral triangles. Find all the possibilities that the sizes of the angles of this pentagon can be.

2018 Hanoi Open Mathematics Competitions, 7

Tags: pentagon , geometry

Suppose that $ABCDE$ is a convex pentagon with $\angle A = 90^o,\angle B = 105^o,\angle C = 90^o$ and $AB = 2,BC = CD = DE =\sqrt2$. If the length of $AE$ is $\sqrt{a }- b$ where $a, b$ are integers, what is the value of $a + b$?

2023 Novosibirsk Oral Olympiad in Geometry, 4

Tags: congruent triangles , congruence , pentagon , geometry

Inside the convex pentagon $ABCDE$, a point $O$ was chosen, and it turned out that all five triangles $AOB$, $BOC$, $COD$, $DOE$ and $EOA$ are congrunet to each other. Prove that these triangles are isosceles or right-angled.

Ukraine Correspondence MO - geometry, 2009.7

Tags: equal angles , geometry , pentagon

Let $ABCDE$ be a convex pentagon such that $AE\parallel BC$ and $\angle ADE = \angle BDC$. The diagonals $AC$ and $BE$ intersect at point $F$. Prove that $\angle CBD= \angle ADF$.

1979 IMO Longlists, 12

Tags: 3d geometry , prism , combinatorics , pentagon , geometry

We consider a prism which has the upper and inferior basis the pentagons: $A_{1}A_{2}A_{3}A_{4}A_{5}$ and $B_{1}B_{2}B_{3}B_{4}B_{5}$. Each of the sides of the two pentagons and the segments $A_{i}B_{j}$ with $i,j=1,\ldots,5$ is colored in red or blue. In every triangle which has all sides colored there exists one red side and one blue side. Prove that all the 10 sides of the two basis are colored in the same color.

2005 Oral Moscow Geometry Olympiad, 3

Tags: pentagon , geometry , combinatorial geometry

$ABCBE$ is a regular pentagon. Point $B'$ is symmetric to point $B$ wrt line $AC$ (see figure). Is it possible to pave the plane with pentagons equal to $AB'CBE$? (S. Markelov) [img]https://cdn.artofproblemsolving.com/attachments/9/2/cbb5756517e85e56c4a931e761a6b4da8fe547.png[/img]