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: 179

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 .
A regular dodecagon $A_1A_2...A_{12}$ is inscribed in a circle with a diameter of $20$ cm . Calculate the perimeter of the pentagon $A_1A_3A_6A_8A_{11}$. (Alexey Panasenko)
The pentagon $ABCDE$ below is such that the quadrilateral $ABCD$ is a square and $BC=DE$. What is the measure of the angle $\angle AEC$?
As shown in figure , in the pentagon $ABCDE$, $BC = DE$, $CD \parallel BE$, $AB>AE$. If $\angle BAC = \angle DAE$ and $\frac{AB}{BD}=\frac{AE}{ED}$. Prove that $AC$ bisects the line segment $BE$. [img]https://cdn.artofproblemsolving.com/attachments/3/2/5ce44f1e091786b865ae4319bda56c3ddbb8d7.png[/img]
The diagonals of a convex pentagon divide each of its interior angles into three equal parts. Does it follow that the pentagon is regular?
Prove that, if a pentagon (five-sided polygon) inscribed in a circle has equal angles, then its sides are equal.
$ABCDE$ is a pentagon whose vertices lie on circle $\omega$ where $\angle DAB=90^{\circ}$. Let $EB$ and $AC$ intersect at $F$, $EC$ meet $BD$ at $G$. $M$ is the midpoint of arc $AB$ on $\omega$, not containing $C$. If $FG\parallel DE\parallel CM$ holds, then what is the value of $\frac{|GE|}{|GD|}$?
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)
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]
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.
Let $ABCDE$ be a cyclic pentagon such that the diagonals $AC$ and $AD$ intersect $BE$ at $P$ and $Q$, respectively, with $BP \cdot QE = PQ^2$. Prove that $BC \cdot DE = CD \cdot PQ$.
We call a pentagon [i]distinguished [/i] if either all side lengths or all angles are equal. We call it [i]very distinguished[/i] if in addition two of the other parts are equal. i.e. $5$ sides and $2$ angles or $2$ sides and $5$ angles.Show that every very distinguished pentagon has an axis of symmetry.
Consider the pentagon below. Find its area. [img]https://cdn.artofproblemsolving.com/attachments/7/b/02ad3852b72682513cf62a170ed4aa45c23785.png[/img]
The pentagon $A_1A_2A_3A_4A_5$ contains bisectors $\ell_1$, $\ell_2$, $...$, $\ell_5$ of angles $\angle A_1$, $\angle A_2$, $ ...$ , $\angle A_5$ respectively. Bisectors $\ell_1$ and $\ell_2$ intersect at point $B_1$, $\ell_2$ and $\ell_3$ - at point $B_2$, etc., $\ell_5$ and $\ell_1$ intersect at point $B_5$. Can the pentagon $B_1B_2B_3B_4B_5$ be convex?
All the points situated more close than $1$ cm to ALL the vertices of the regular pentagon with $1$ cm side, are deleted from that pentagon. Find the area of the remained figure.
The symbol of the Olympiad shows $5$ regular hexagons with side $a$, located inside a regular hexagon with side $b$. Find ratio $\frac{a}{b}$. [img]https://1.bp.blogspot.com/-OwyAl75LwiM/YIsThl3SG6I/AAAAAAAANS0/LwHEsAfyZMcqVIS8h_jr_n46OcMJaSTgQCLcBGAsYHQ/s0/2019%2BUkraine%2Bcorrespondence%2B5-12%2Bp8.png[/img]
$(NET 5)$ 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 $A_1A_2A_3A_4A_5.$
A rectangle of paper of $3$ cm by $9$ cm is folded along a straight line, making two opposite vertices coincide. In this way a pentagon is formed. Calculate it's area.
In a convex pentagon, let the perpendicular line from a vertex to the opposite side be called an altitude. Prove that if four of the altitudes are concurrent at a point then the fifth altitude also passes through this point.
Let $ABCDE$ be a convex pentagon such that $AB = BC = CD$ and $\angle BDE = \angle EAC = 30 ^{\circ}$. Find the possible values of $\angle BEC$. [i]Proposed by Josef Tkadlec (Czech Republic)[/i]
A pentagon can be divided into equilateral triangles. Find all the possibilities that the sizes of the angles of this pentagon can be.
The vertices of the pentagon $ABCDE$ are on a circle, and the points $H_1, H_2, H_3,H_4$ are the orthocenters of the triangles $ABC, ABE, ACD, ADE$ respectively . Prove that the quadrilateral determined by the four orthocenters is square if and only if $BE \parallel CD$ and the distance between them is $\frac{BE + CD}{2}$.
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 $EJ$.
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.
Given a convex pentagon $ABCDE$ with $\angle ABC= \angle ADE$ and $\angle AEC= \angle ADB$ . Prove that $\angle BAC = \angle DAE$ .