Found problems: 179
Let $A$ be a vertex of a regular star-shaped pentagon, the angle at $A$ being less than $180^o$ and the broken line $AA_1BB_1CC_1DD_1EE_1$ being its contour. Lines $AB$ and $DE$ meet at $F$. Prove that polygon $ABB_1CC_1DED_1$ has the same area as the quadrilateral $AD_1EF$.
Note: A regular star pentagon is a figure formed along the diagonals of a regular pentagon.
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)
The figure $ABCDE$ is a convex pentagon. Find the sum $\angle DAC + \angle EBD +\angle ACE +\angle BDA + \angle CEB$?
Is there a convex pentagon in which each diagonal is equal to a side?
Pentagon $ABCDE$ is inscribed in a circle. Distances from point $E$ to lines $AB$ , $BC$ and $CD$ are equal to $a, b$ and $c$, respectively. Find the distance from point $E$ to line $AD$.
A regular pentagon of side length $1$ is given. Determine the smallest $r$ for which the pentagon can be covered by five discs of radius $r$ and justify your answer.
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 $\angle APQ$.
You have a paper pentagon, $ABCDE$, such that $AB = BC = 3$ cm, $CD = DE= 5$ cm, $EA = 4$ cm, $\angle ABC = 100^o$ ,$ \angle CDE = 80^o$. You have to divide the pentagon into four triangles, by three straight cuts, so that with the four triangles assemble a rectangle, without gaps or overlays. (The triangles can be rotated and / or turned around.)
Let $ABCDE$ be a regular pentagon. We drew two circles around $A$ and $B$ with radius $AB$. Let $F$ mark the intersection of the two circles that is inside the pentagon. Let $G$ mark the intersection of lines $EF$ and $AD$. What is the degree measure of angle $AGE$?
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.
The pentagon $ABCDE$ has an inscribed circle and the diagonals $AD$ and $CE$ intersect in its centre $O$. Prove that the segment $BO$ and the side $DE$ are perpendicular.
(Folklore)
The convex $n$-gon ($n\ge 5$) is cut along all its diagonals. Prove that there are at least a pair of parts with the different areas.
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.
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.
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.
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.
$(YUG 2)$ 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.
Given a convex pentagon $ABCDE$, prove that if an arbitrary point $M$ inside the pentagon is connected by lines with all the pentagon’s vertices, then either one or three or five of these lines cross the sides of the pentagon opposite the vertices they pass.
Note: In reality, we need to exclude the points of the diagonals, because that in this case the drawn lines can pass not through the internal points of the sides, but through the vertices. But if the drawn diagonals are not considered or counted twice (because they are drawn from two vertices), then the statement remains true.
Let $ABCDE$ be a regular pentagon, and $F$ some point on the arc $AB$ of the circumcircle of $ABCDE$. Show that
\[
\frac{FD}{FE + FC} = \frac{FB + FA}{FD} = \frac{-1 + \sqrt{5}}{2},
\]
and that $FD + FB + FA = FE + FC$.
Each diagonal of a convex pentagon is parallel to one side of the pentagon. Prove that the ratio of the length of a diagonal to that of its corresponding side is the same for all five diagonals, and compute this ratio.
Let $P$ be a convex pentagon with all sides equal. Prove that if two of the angles of $P$ add to $180^o$, then it is possible to cover the plane with $P$, without overlaps.
In a convex pentagon $ABCDE$, the area of the triangles $ABC, ABD, ACD$ and $ADE$ are equal and have the value $F$. What is the area of the triangle $BCE$ ?
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.
$ABCDE$ 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 $AX = AY$.
In a convex pentagon $ABCDE$ the following conditions hold : $AB \parallel CD$, $BC \parallel DE$, and $\angle BAE = \angle AED$. Prove that $AB + BC = CD + DE$
[i]Proposed by Anton Trygub[/i]