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)

Let there be a pentagon $ABCDE$ inscribed in a circle $O$. The tangent to $O$ at $E$ is parallel to $AD$. A point $F$ lies on $O$ and it is in the opposite side of $A$ with respect to $CD$, and satisfies $AB \cdot BC \cdot DF = AE \cdot ED \cdot CF$ and $\angle CFD = 2\angle BFE$. Prove that the tangent to $O$ at $B,E$ and line $AF$ concur at one point.

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.

The inner angles of the pentagon inscribed in the circle are equal to each other. Prove that this pentagon is regular.

A convex pentagon $ABCDE$ is given with $AB=BC$, $CD=DE$ and $\angle A=\angle C=\angle E>90^{\circ}$. Prove that the pentagon is circumscribed [I]Proposed by F. Baharev[/i]

There are $k$ points in the interior of a pentagon. Together with the vertices of the pentagon they form a $(k + 5)$-element set $M$. The area of the pentagon is defined by connecting lines between the points of $M$ into sub-areas in such a way that it is divided into sub-areas in such a way that no sub-areas have a point on their interior of $M$ and contains exactly three points of $M$ on the boundary of each part. None of the connecting lines has a point in common with any other connecting line or pentagon side, which does not belong to $M$. With such a decomposition of the pentagon, there can be an even number of connecting lines (including the pentagon sides) go out? The answer has to be justified.

You can inscribe a circle in the pentagon $ABCDE$. It is also known that $\angle ABC = \angle BAE = \angle CDE = 90^o$. Find the measure of the angle $ADB$.

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 be five nonzero complex numbers having the same absolute value and such that zero is equal to their sum, which is equal to the sum of their squares. Prove that the affixes of these numbers in the complex plane form a regular pentagon. [i]Daniel Jinga[/i]

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.

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.

$ABCDE$ is a regular pentagon. Two circles $C_1$ and $C_2$ are drawn through $B$ with centers $A$ and $C$ respectively. Let the other intersection of $C_1$ and $C_2$ be $P$. The circle with center $P$ which passes through $E$ and $D$ intersects $C_2$ at $X$ and $AE$ at $Y$. Prove that $AX = AY$.

Dinesh has several squares and regular pentagons, all with side length $ 1$. He wants to arrange the shapes alternately to form a closed loop (see diagram). How many pentagons would Dinesh need to do so? [img]https://cdn.artofproblemsolving.com/attachments/8/9/6345d7150298fe26cfcfba554656804ed25a6d.jpg[/img]

The point $M$ is the center of a regular pentagon $ABCDE$. The point $P$ is an inner point of the line segment $[DM]$. The circumscribed circle of triangle $\vartriangle ABP$ intersects the side $[AE]$ at point $Q$ (different from $A$). The perpendicular from $P$ on $CD$ intersects the side $[AE] $ at point $S$. Prove that $PS$ is the bisector of $\angle APQ$.

The pentagon $ABCDE$ is inscribed in the circle $\omega$. Let $T$ be the intersection point of the diagonals $BE$ and $AD$. A line is drawn through the point $T$ parallel to $CD$, which intersects $AB$ and $CE$ at points $X$ and $Y$, respectively. Prove that the circumscribed circle of the triangle $AXY$ is tangent to $\omega$.

The pentagon $ABCDE$ has $AB = BC, CD = DE, \angle ABC = 120^o, \angle CDE = 60^o$ and $BD = 2$. Calculate the area of the pentagon.

$ABCDE$ is a cyclic convex pentagon, and $AC = BD = CE$. $AC$ and $BD$ intersect at $X$, and $BD$ and $CE$ intersect at $Y$ . If $AX = 6$, $XY = 4$, and $Y E = 7$, then the area of pentagon $ABCDE$ can be written as $\frac{a\sqrt{b}}{c}$ , where $a$, $ b$, $c$ are integers, $c$ is positive, $b$ is square-free, and gcd$(a, c) = 1$. Find $100a + 10b + c$.

The figure below, composed of four regular pentagons with a side length of $1$, was glued in space as follows. First, it was folded along the broken sections, by combining the bold sections, and then formed in such a way that colored sections formed a square. Find the length of the segment $AB$ created in this way. [img]https://cdn.artofproblemsolving.com/attachments/0/7/bddad6449f74cbc7ea2623957ef05b3b0d2f19.png[/img]

The vertices of a convex pentagon $ABCDE$ lie on a circle $\gamma_1$. The diagonals $AC , CE, EB, BD$, and $DA$ are tangents to another circle $\gamma_2$ with the same centre as $\gamma_1$. (a) Show that all angles of the pentagon $ABCDE$ have the same size and that all edges of the pentagon have the same length. (b) What is the ratio of the radii of the circles $\gamma_1$ and $\gamma_2$? (The answer should be given in terms of integers, the four basic arithmetic operations and extraction of roots only.)

In a parallelogram $ABCD$ , $BD$ is the largest diagonal. By matching $B$ with $D$ by a bend, a regular pentagon is formed. Calculate the measures of the angles formed by the diagonal $BD$ with each of the sides of the parallelogram.

Let $ABCDE$ be a convex pentagon such that $\angle BAC = \angle CAD = \angle DAE$ and $\angle ABC = \angle ACD = \angle ADE$. Diagonals $BD$ and $CE$ meet at $P$. Prove that $AP$ bisects side $CD$.

Let $ABCDE$ be a convex pentagon with $\angle AEB=\angle BDC=90^o$ and line $AC$ bisects $\angle BAE$ and $\angle DCB$ internally. The circumcircle of $ABE$ intersects line $AC$ again at $P$. (a) Show that $P$ is the circumcenter of $BDE$. (b) Show that $A, C, D, E$ are concyclic.

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]

All sides of a convex pentagon are equal, and all angles are different. Prove that the maximum and minimum angles are adjacent to the same side of the pentagon.

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.