Olympiad problems

2020 Korea National Olympiad, 6

Tags: pentagon , geometry

Let $ABCDE$ be a convex pentagon such that quadrilateral $ABDE$ is a parallelogram and quadrilateral $BCDE$ is inscribed in a circle. The circle with center $C$ and radius $CD$ intersects the line $BD, DE$ at points $F, G(\neq D)$, and points $A, F, G$ is on line l. Let $H$ be the intersection point of line $l$ and segment $BC$. Consider the set of circle $\Omega$ satisfying the following condition. Circle $\Omega$ passes through $A, H$ and intersects the sides $AB, AE$ at point other than $A$. Let $P, Q(\neq A)$ be the intersection point of circle $\Omega$ and sides $AB, AE$. Prove that $AP+AQ$ is constant.

2018 Peru IMO TST, 3

Tags: geometry , pentagon , inscribed circle

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 Sharygin Geometry Olympiad, 7

Tags: geometric inequality , diagonal , convex , pentagon , geometry

A convex pentagon $P $ is divided by all its diagonals into ten triangles and one smaller pentagon $P'$. Let $N$ be the sum of areas of five triangles adjacent to the sides of $P$ decreased by the area of $P'$. The same operations are performed with the pentagon $P'$, let $N'$ be the similar difference calculated for this pentagon. Prove that $N > N'$. (A.Belov)

1997 Moldova Team Selection Test, 2

Tags: parallelogram , pentagon , ratio , geometry

In a convex pentagon every diagonal is parallel to one side. Show that the ratios between the lengths of diagonals and the sides parallel to them are equal and find their value.

2009 Brazil Team Selection Test, 1

Tags: right triangle , cyclic , pentagon , geometry , orthocenter

Let $A, B, C, D, E$ points in circle of radius r, in that order, such that $AC = BD = CE = r$. The points $H_1, H_2, H_3$ are the orthocenters of the triangles $ACD$, $BCD$ and $BCE$, respectively. Prove that $H_1H_2H_3$ is a right triangle .

1904 Eotvos Mathematical Competition, 1

Tags: regular , pentagon , geometry

Prove that, if a pentagon (five-sided polygon) inscribed in a circle has equal angles, then its sides are equal.

2013 Czech-Polish-Slovak Junior Match, 3

Tags: cyclic , geometry , pentagon , equal segments

The $ABCDE$ pentagon is inscribed in a circle and $AB = BC = CD$. Segments $AC$ and $BE$ intersect at $K$, and Segments $AD$ and $CE$ intersect at point$ L$. Prove that $AK = KL$.

1998 North Macedonia National Olympiad, 1

Tags: pentagon , angle chasing , equal segments , geometry

Let $ABCDE$ be a convex pentagon with $AB = BC =CA$ and $CD = DE = EC$. Let $T$ be the centroid of $\vartriangle ABC$, and $N$ be the midpoint of $AE$. Compute $\angle NT D$

2006 Sharygin Geometry Olympiad, 4

Tags: geometry , regular , square , pentagon

a) Given two squares $ABCD$ and $DEFG$, with point $E$ lying on the segment $CD$, and points$ F,G$ outside the square $ABCD$. Find the angle between lines $AE$ and $BF$. b) Two regular pentagons $OKLMN$ and $OPRST$ are given, and the point $P$ lies on the segment $ON$, and the points $R, S, T$ are outside the pentagon $OKLMN$. Find the angle between straight lines $KP$ and $MS$.

2018 Germany Team Selection Test, 2

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.

1966 Spain Mathematical Olympiad, 3

Tags: similarity , pentagon , geometry

Given a regular pentagon, consider the convex pentagon limited by its diagonals. You are asked to calculate: a) The similarity relation between the two convex pentagons. b) The relationship of their areas. c) The ratio of the homothety that transforms the first into the second.

2012 Korea Junior Math Olympiad, 2

Tags: concyclic , geometry , pentagon

A pentagon $ABCDE$ is inscribed in a circle $O$, and satisfies $\angle A = 90^o, AB = CD$. Let $F$ be a point on segment $AE$. Let $BF$ hit $O$ again at $J(\ne B)$, $CE \cap DJ = K$, $BD\cap FK = L$. Prove that $B,L,E,F$ are cyclic.

2008 India Regional Mathematical Olympiad, 6

Tags: geometry , convex quadrilateral , pentagon , trigonometry

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 $ABCDE$.

Durer Math Competition CD 1st Round - geometry, 2014.C4

Tags: area , pentagon , geometry

$ABCDE$ is a convex pentagon with $AB = CD = EA = 1$, $\angle ABC = \angle DEA = 90^o$, and $BC + DE = 1$. What is the area of the pentagon?

1996 Swedish Mathematical Competition, 4

Tags: geometry , increasing , cyclic , pentagon , angle

The angles at $A,B,C,D,E$ of a pentagon $ABCDE$ inscribed in a circle form an increasing sequence. Show that the angle at $C$ is greater than $\pi/2$, and that this lower bound cannot be improved.

1999 Ukraine Team Selection Test, 5

Tags: pentagon , equal angles , geometry

A convex pentagon $ABCDE$ with $DC = DE$ and $\angle DCB = \angle DEA = 90^o$ is given. Let $F$ be a point on the segment $AB$ such that $AF : BF = AE : BC$. Prove that $\angle FCE = \angle ADE$ and $\angle FEC = \angle BDC$.

1966 Kurschak Competition, 1

Tags: 3d geometry , pentagon , geometry

Can we arrange $5$ points in space to form a pentagon with equal sides such that the angle between each pair of adjacent edges is $90^o$?

Estonia Open Senior - geometry, 2020.1.5

Tags: regular , pentagon , concurrency , geometry

A circle $c$ with center $A$ passes through the vertices $B$ and $E$ of a regular pentagon $ABCDE$ . The line $BC$ intersects the circle $c$ for second time at point $F$. The point $G$ on the circle $c$ is chosen such that $| F B | = | FG |$ and $B \ne G$. Prove that the lines $AB, EF$ and $DG$ intersect at one point.

2010 Morocco TST, 4

Tags: circumcircle , pentagon , geometry

Let $ ABCDE$ be a convex pentagon such that \[ \angle BAC \equal{} \angle CAD \equal{} \angle DAE\qquad \text{and}\qquad \angle ABC \equal{} \angle ACD \equal{} \angle ADE. \]The diagonals $BD$ and $CE$ meet at $P$. Prove that the line $AP$ bisects the side $CD$. [i]Proposed by Zuming Feng, USA[/i]

2009 Belarus Team Selection Test, 2

Tags: equal segments , geometry , pentagon

Does there exist a convex pentagon $A_1A_2A_3A_4A_5$ and a point $X$ inside it such that $XA_i=A_{i+2}A_{i+3}$ for all $i=1,...,5$ (all indices are considered modulo $5$) ? I. Voronovich

2003 Belarusian National Olympiad, 8

Tags: geometry theorems , convex , pentagon , area

Given a convex pentagon $ABCDE$ with $AB=BC, CD=DE, \angle ABC=150^o, \angle CDE=30^o, BD=2$. Find the area of $ABCDE$. (I.Voronovich)

2020 Ukrainian Geometry Olympiad - December, 3

Tags: geometry , equal segments , pentagon , angle

About the pentagon $ABCDE$ we know that $AB = BC = CD = DE$, $\angle C = \angle D =108^o$, $\angle B = 96^o$. Find the value in degrees of $\angle E$.

1969 IMO Shortlist, 50

Tags: pentagon , construction , angle bisector , geometry

$(NET 5)$ The bisectors of the exterior angles of a pentagon $B_1B_2B_3B_4B_5$ form another pentagon $A_1A_2A_3A_4A_5.$ Construct $B_1B_2B_3B_4B_5$ from the given pentagon $A_1A_2A_3A_4A_5.$

2021 Saint Petersburg Mathematical Olympiad, 3

Tags: pentagon , geometry

Given a convex pentagon $ABCDE$, points $A_1, B_1, C_1, D_1, E_1$ are such that $$AA_1 \perp BE, BB_1 \perp AC, CC_1 \perp BD, DD_1 \perp CE, EE_1 \perp DA.$$ In addition, $AE_1 = AB_1, BC_1 = BA_1, CB_1 = CD_1$ and $DC_1 = DE_1$. Prove that $ED_1 = EA_1$

2019 Canadian Mathematical Olympiad Qualification, 6

Tags: concurrency , pentagon , geometry , perpendicular

Pentagon $ABCDE$ is given in the plane. Let the perpendicular from $A$ to line $CD$ be $F$, the perpendicular from $B$ to $DE$ be $G$, from $C$ to $EA$ be $H$, from $D$ to $AB$ be $I$,and from $E$ to $BC$ be $J$. Given that lines $AF,BG,CH$, and $DI$ concur, show that they also concur with line $EJ$.