Found problems: 179
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 $\angle NT D$
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 $ED_1 = EA_1$
A pentagon (not necessarily convex) has all sides of length $1$ and its product of cosine of any four angles equal to zero. Find all possible values of the area of such a pentagon.
In the convex pentagon $ ABCDE,$ the sides $ BC, CD, DE$ are equal. Moreover each diagonal of the pentagon is parallel to a side ($ AC$ is parallel to $ DE$, $ BD$ is parallel to $ AE$ etc.). Prove that $ ABCDE$ is a regular pentagon.
Given a convex pentagon $ABCDE$ in which $\angle ABC = \angle AED = 90^o$, $\angle BAC= \angle DAE$. Let $K$ be the midpoint of the side $CD$, and $P$ the intersection point of lines $AD$ and $BK$, $Q$ be the intersection point of lines $AC$ and $EK$. Prove that $BQ = PE$.
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 $\angle E$.
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 convex pentagon such that the triangles $ABC, BCD, CDE, DEA$ and $EAB$ have equal areas. Show that $(1/4)$ area $(ABCDE) <$ area $(ABC) < (1/3)$ area $(ABCDE)$.
A real number is written next to each vertex of a regular pentagon. All five numbers are different. A triple of vertices is called [b] successful[/b] if they form an isosceles triangle for which the number written on the top vertex is either larger than both numbers written on the base vertices, or smaller than both. Find the maximum possible number of successful triples.
In the convex pentagon $ ABCDE,$ the sides $ BC, CD, DE$ are equal. Moreover each diagonal of the pentagon is parallel to a side ($ AC$ is parallel to $ DE$, $ BD$ is parallel to $ AE$ etc.). Prove that $ ABCDE$ is a regular pentagon.
Different points $A_1,A_2,A_3,A_4,A_5$ lie on a circle so that $A_1A_2 = A_2A_3 = A_3A_4 =A_4A_5$. Let $A_6$ be the diametrically opposite point to $A_2$, and $A_7$ be the intersection of $A_1A_5$ and $A_3A_6$. Prove that the lines $A_1A_6$ and $A_4A_7$ are perpendicular
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.
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 $90^o$?
Convex pentagon $ABCDE$ satisfies $AB \parallel DE$, $BE \parallel CD$, $BC \parallel AE$, $AB = 30$, $BC = 18$, $CD = 17$, and $DE = 20$. Find its area.
[i] Proposed by Michael Tang [/i]
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.
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]
The surface of a soccer ball is made up of black pentagons and white hexagons together. On the sides of each pentagon are nothing but hexagons, while on the sides of each border of hexagons alternately pentagons and hexagons. Determine from this information about the soccer ball , the number of its pentagons and its hexagons.
Every two of five regular pentagons on the plane have a common point.
Is it true that some of these pentagons have a common point?
Let $ABCDE$ be a convex pentagon and area of $\vartriangle ABC =$ area of $\vartriangle BCD =$ area of $\vartriangle CDE=$ area of $\vartriangle DEA =$ area of $\vartriangle EAB$. Given that area of $\vartriangle ABCDE = 2$. Evaluate the area of area of $\vartriangle ABC$.
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 $AK = KL$.
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.
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]
About pentagon $ABCDE$ is known that angle $A$ and angle $C$ are right and that the sides $| AB | = 4$, $| BC | = 5$, $| CD | = 10$, $| DE | = 6$. Furthermore, the point $C'$ that appears by mirroring $C$ in the line $BD$, lies on the line segment $AE$. Find angle $E$.
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 $ADB$.
A regular pentagon $ABCDE$ is given. The point $X$ is on his circumcircle, on the arc $\overarc{AE}$. Prove that $|AX|+|CX|+|EX|=|BX|+|DX|$.
[u][b]Remark:[/b][/u] Here's a more general version of the problem: Prove that for any point $X$ in the plane, $|AX|+|CX|+|EX|\ge|BX|+|DX|$, with equality only on the arc $\overarc{AE}$.