Found problems: 179
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]
A regular star-shaped hexagon is split into $4$ parts. Construct from them a convex polygon.
Note: A regular six-pointed star is a figure that is obtained by combining a regular triangle and a triangle symmetrical to it relative to its center
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.
In convex pentagon $ABCDE, AC$ is parallel to $DE, AB$ is perpendicular to $AE$, and $BC$ is perpendicular to $CD$. If $H$ is the orthocentre of triangle $ABC$ and $M$ is the midpoint of segment $DE$, prove that $AD, CE$ and $HM$ are concurrent.
Given is a convex pentagon $ABCDE$ in which $\angle A = 60^o$, $\angle B = 100^o$, $\angle C = 140^o$.
Show that this pentagon can be placed in a circle with a radius of $\frac23 AD$.
To each vertex of a regular pentagon an integer is assigned, so that the sum of all five numbers is positive. If three consecutive vertices are assigned the numbers $x,y,z$ respectively, and $y<0$, then the following operation is allowed: $x,y,z$ are replaced by $x+y,-y,z+y$ respectively. Such an operation is performed repeatedly as long as at least one of the five numbers is negative. Determine whether this procedure necessarily comes to an end after a finite number of steps.
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.
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 picture shows $10$ equal regular pentagons where each two neighbouring pentagons have a common side. The smaller circle is tangent to one side of each pentagon and the larger circle passes through the opposite vertices of these sides. Find the area of the larger circle if the area of the smaller circle is $1$.
[img]https://cdn.artofproblemsolving.com/attachments/0/6/84fe98370868a5cf28d92d4b207ccb00e6eaa3.png[/img]
$ABCDE$ 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?
Consider the pentagon $ABCDE$ such that $AB = AE = x$, $AC = AD = y$, $\angle BAE = 90^o$ and $\angle ACB = \angle ADE = 135^o$. It is known that $C$ and $D$ are inside the triangle $BAE$. Determine the length of $CD$ in terms of $x$ and $y$.
$ABCDE$ 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 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]
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]
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.
$ABCBE$ 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]
The pentagon $ABCDE$ is inscribed in the circle. Line segments $AC$ and $BD$ intersect at point $K$. Line segment $CE$ touches the circumcircle of triangle $ABK$ at point $N$. Find the angle $CNK$ if $\angle ECD = 40^o.$
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 the convex pentagon $ABCDE$, $AE = AD$, $AB = AC$, and angle $CAD$ equals the sum of angles $AEB$ and $ABE$. Prove that segment $CD$ is double the length of median $AM$ of triangle $ABE$.
An approximate construction of a regular pentagon goes as follows. Inscribe an arbitrary convex pentagon $P_1P_2P_3P_4P_5$ in a circle. Now choose an arror bound $\epsilon > 0$ and apply the following procedure.
(a) Denote $P_0 = P_5$ and $P_6 = P_1$ and construct the midpoint $Q_i$ of the circular arc $P_{i-1}P_{i+1}$ containing $P_i$.
(b) Rename the vertices $Q_1,...,Q_5$ as $P_1,...,P_5$.
(c) Repeat this procedure until the difference between the lengths of the longest and the shortest among the arcs $P_iP_{i+1}$ is less than $\epsilon$.
Prove this procedure must end in a finite time for any choice of $\epsilon$ and the points $P_i$.
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 $\angle CBD= \angle ADF$.
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 $BDJK$.
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Does there exist a convex pentagon from which a similar pentagon can be cut off by a straight line?
(S Tokarev)
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 $AB^2 + CE^2 = BE^2 + CD^2$
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 $\angle FEC = \angle BDC$.