Let $ABCD$ be a convex quadrilateral with perimeter $\tfrac{5}{2}$ and $AC=BD=1$. Determine the maximum possible area of $ABCD$.

The apothem of a square having its area numerically equal to its perimeter is compared with the apothem of an equilateral triangle having its area numerically equal to its perimeter. The first apothem will be: $ \textbf{(A)}\ \text{equal to the second} \qquad\textbf{(B)}\ \frac {4}{3} \text{ times the second} \qquad\textbf{(C)}\ \frac {2}{\sqrt {3}} \text{ times the second} \\ \textbf{(D)}\ \frac {\sqrt {2}}{\sqrt {3}} \text{ times the second} \qquad\textbf{(E)}\ \text{indeterminately related to the second}$ [i][Note: The apothem of a regular polygon is a line segment from the center to the midpoint of one of its sides.][/i]

Let a regular polygon with $p$ vertices be given, where $p$ is an odd prime number. At every vertex there is one monkey. An owner of monkeys takes $p$ peanuts, goes along the perimeter of polygon clockwise and delivers to the monkeys by the following rule: Gives the first peanut for the leader, skips the two next vertices and gives the second peanut to the monkey at the next vertex; skip four next vertices gives the second peanut for the monkey at the next vertex ... after giving the $k$-th peanut, he skips the $2 \cdot k$ next vertices and gives $k+1$-th for the monkey at the next vertex. He does so until all $p$ peanuts are delivered. [b]I.[/b] How many monkeys are there which does not receive peanuts? [b]II.[/b] How many edges of polygon are there which satisfying condition: both two monkey at its vertex received peanut(s)?

A circle and a square have the same perimeter. Then: $ \textbf{(A)}\ \text{their areas are equal} \qquad\textbf{(B)}\ \text{the area of the circle is the greater}$ $ \textbf{(C)}\ \text{the area of the square is the greater}$ $ \textbf{(D)}\ \text{the area of the circle is } \pi \text{ times the area of the square} \\ \qquad\textbf{(E)}\ \text{none of these}$

Two right angled triangles are given, such that the incircle of the first one is equal to the circumcircle of the second one. Let $S$ (respectively $S'$) be the area of the first triangle (respectively of the second triangle). Prove that $\frac{S}{S'}\geq 3+2\sqrt{2}$.

A straight line passing through the center of the circumscribed circle and the intersection point of the heights of the non-equilateral triangle $ABC$ divides its perimeter and area in the same ratio.Find this ratio.

An equilateral triangle and a regular hexagon have equal perimeters. If the area of the triangle is 4, what is the area of the hexagon? $\textbf{(A)}\hspace{.05in}4 \qquad \textbf{(B)}\hspace{.05in}5 \qquad \textbf{(C)}\hspace{.05in}6 \qquad \textbf{(D)}\hspace{.05in}4\sqrt3 \qquad \textbf{(E)}\hspace{.05in}6\sqrt3 $

Let $ABCD$ be a square of side paper $10$ and $P$ a point on side $BC$. By folding the paper along the $AP$ line, point $B$ determines the point $Q$, as seen in the figure. The line $PQ$ cuts the side $CD$ at $R$. Calculate the perimeter of the triangle $ PCR$ [img]https://3.bp.blogspot.com/-ZSyCUznwutE/XNY7cz7reQI/AAAAAAAAKLc/XqgQnjm8DQYq6Q7fmCAKJwKt3ihoL8AuQCK4BGAYYCw/s400/may%2B2013%2Bl1.png[/img]

Given a line $p$ and a triangle $\Delta$ in the plane, construct an equilateral triangle one of whose vertices lies on the line $p$, while the other two halve the perimeter of $\Delta.$

In a perpendicular triangle the perimeter is 60 and the altitude on the hypotenuse is 12. Then, the length of the hypotenuse is $ \text{(A)}\ 24 \qquad \text{(B)}\ 25 \qquad \text{(C)}\ 26 \qquad \text{(D)}\ 27 \qquad \text{(E)}\ 28$

$S_1$ and $S_2$ are two circles that intersect at two different points $P$ and $Q$. Let $\ell_1$ and $\ell_2$ be two parallel lines such that $\ell_1$ passes through the point $P$ and intersects $S_1,S_2$ at $A_1,A_2$ respectively (both distinct from $P$), and $\ell_2$ passes through the point $Q$ and intersects $S_1,S_2$ at $B_1,B_2$ respectively (both distinct from $Q$). Show that the triangles $A_1QA_2$ and $B_1PB_2$ have the same perimeter.

Given two circles $C_1,C_2$ with common center, the radius of $C_2$ is twice the radius of $C_1$. Quadrilateral $A_1A_2A_3A_4$ is inscribed in $C_1$. The extension of $A_4A_1$ meets $C_2$ at $B_1$; the extension of $A_1A_2$ meets $C_2$ at $B_2$; the extension of $A_2A_3$ meets $C_2$ at $B_3$; the extension of $A_3A_4$ meets $C_2$ at $B_4$. Prove that $P(B_1B_2B_3B_4)\ge 2P(A_1A_2A_3A_4)$, and in what case the equality holds? ($P(X)$ denotes the perimeter of quadrilateral $X$)

Point $C$ lies inside the right angle $AOB$. Prove that the perimeter of triangle $ABC$ is greater than $2 OC$.

In the triangle $ABC$, the side $BC$ is equal to $a$. Point $F$ is midpoint of $AB$, $I$ is the point of intersection of the bisectors of triangle $ABC$. It turned out that $\angle AIF = \angle ACB$. Find the perimeter of the triangle $ABC$. (Grigory Filippovsky)

If $a$, $b$ and $c$ are sides of triangle which perimeter equals $1$, prove that: $a^2+b^2+c^2+4abc<\frac{1}{2}$

Three non-overlapping regular plane polygons, at least two of which are congruent, all have sides of length $ 1$. The polygons meet at a point $ A$ in such a way that the sum of the three interior angles at $ A$ is $ 360^\circ$. Thus the three polygons form a new polygon with $ A$ as an interior point. What is the largest possible perimeter that this polygon can have? $ \textbf{(A)}\ 12\qquad \textbf{(B)}\ 14\qquad \textbf{(C)}\ 18\qquad \textbf{(D)}\ 21\qquad \textbf{(E)}\ 24$

Let $T_m$ be a number of non-congruent triangles which perimeter is $m$ and all its sides are positive integers. Prove that: $a)$ $T_{1999} > T_{2000}$ $b)$ $T_{4n+1}=T_{4n-2}+n$, $(n \in \mathbb{N})$

Let $ ABC$ be an acute triangle with the incircle $ C(I,r)$ and the circumcircle $ C(O,R)$ . Denote $ D\in BC$ for which $ AD\perp BC$ and $ AD \equal{} h_a$ . Prove that $ DI^2 \equal{} (2R \minus{} h_a)(h_a \minus{} 2r)$ .

$(BUL 3)$ One hundred convex polygons are placed on a square with edge of length $38 cm.$ The area of each of the polygons is smaller than $\pi cm^2,$ and the perimeter of each of the polygons is smaller than $2\pi cm.$ Prove that there exists a disk with radius $1$ in the square that does not intersect any of the polygons.

On a plane, Bob chooses 3 points $A_0$, $B_0$, $C_0$ (not necessarily distinct) such that $A_0B_0+B_0C_0+C_0A_0=1$. Then he chooses points $A_1$, $B_1$, $C_1$ (not necessarily distinct) in such a way that $A_1B_1=A_0B_0$ and $B_1C_1=B_0C_0$. Next he chooses points $A_2$, $B_2$, $C_2$ as a permutation of points $A_1$, $B_1$, $C_1$. Finally, Bob chooses points $A_3$, $B_3$, $C_3$ (not necessarily distinct) in such a way that $A_3B_3=A_2B_2$ and $B_3C_3=B_2C_2$. What are the smallest and the greatest possible values of $A_3B_3+B_3C_3+C_3A_3$ Bob can obtain?

Determine all the quadrilaterals that can be divided by a diagonal into two triangles of equal area and equal perimeter.

Given a triangle $ABC$, the points $D$, $E$, and $F$ lie on the sides $BC$, $CA$, and $AB$, respectively, are such that $$DC + CE = EA + AF = FB + BD.$$ Prove that $$DE + EF + FD \ge \frac12 (AB + BC + CA).$$

In a triangle $ABC$, the excircle $\omega_a$ opposite $A$ touches $AB$ at $P$ and $AC$ at $Q$, while the excircle $\omega_b$ opposite $B$ touches $BA$ at $M$ and $BC$ at $N$. Let $K$ be the projection of $C$ onto $MN$ and let $L$ be the projection of $C$ onto $PQ$. Show that the quadrilateral $MKLP$ is cyclic. ([i]Bulgaria[/i])

An equilateral triangle is drawn with a side of length $ a$. A new equilateral triangle is formed by joining the midpoints of the sides of the first one. Then a third equilateral triangle is formed by joining the midpoints of the sides of the second; and so on forever. The limit of the sum of the perimeters of all the triangles thus drawn is: $ \textbf{(A)}\ \text{Infinite} \qquad\textbf{(B)}\ 5\frac {1}{4}a \qquad\textbf{(C)}\ 2a \qquad\textbf{(D)}\ 6a \qquad\textbf{(E)}\ 4\frac {1}{2}a$

Given an isosceles triangle $ABC$ with a vertex at the point $B$. Based on $AC$, an arbitrary point $D $ is selected, different from the vertices $A$ and $C $. On the line $AC $ select the point $E $ outside the segment $AC$, for which $AE = CD$. Prove that the perimeter $\Delta BDE$ is larger than the perimeter $\Delta ABC$.