There are several discs whose radii are no more that $1$, and whose centers all lie on a segment with length ${l}$. Prove that the union of all the discs has a perimeter not exceeding $4l+8$. [i]Proposed by Morteza Saghafian - Iran[/i]

How many noncongruent integer-sided triangles with positive area and perimeter less than $15$ are neither equilateral, isosceles, nor right triangles? $\textbf{(A) }3\qquad\textbf{(B) }4\qquad\textbf{(C) }5\qquad\textbf{(D) }6\qquad\textbf{(E) }7$

A polygon $A$ that doesn't intersect itself and has perimeter $p$ is called [b]Rotund[/b] if for each two points $x,y$ on the sides of this polygon which their distance on the plane is less than $1$ their distance on the polygon is at most $\frac{p}{4}$. (Distance on the polygon is the length of smaller path between two points on the polygon) Now we shall prove that we can fit a circle with radius $\frac{1}{4}$ in any rotund polygon. The mathematicians of two planets earth and Tarator have two different approaches to prove the statement. In both approaches by "inner chord" we mean a segment with both endpoints on the polygon, and "diagonal" is an inner chord with vertices of the polygon as the endpoints. [b]Earth approach: Maximal Chord[/b] We know the fact that for every polygon, there exists an inner chord $xy$ with a length of at most 1 such that for any inner chord $x'y'$ with length of at most 1 the distance on the polygon of $x,y$ is more than the distance on the polygon of $x',y'$. This chord is called the [b]maximal chord[/b]. On the rotund polygon $A_0$ there's two different situations for maximal chord: (a) Prove that if the length of the maximal chord is exactly $1$, then a semicircle with diameter maximal chord fits completely inside $A_0$, so we can fit a circle with radius $\frac{1}{4}$ in $A_0$. (b) Prove that if the length of the maximal chord is less than one we still can fit a circle with radius $\frac{1}{4}$ in $A_0$. [b]Tarator approach: Triangulation[/b] Statement 1: For any polygon that the length of all sides is less than one and no circle with radius $\frac{1}{4}$ fits completely inside it, there exists a triangulation of it using diagonals such that no diagonal with length more than $1$ appears in the triangulation. Statement 2: For any polygon that no circle with radius $\frac{1}{4}$ fits completely inside it, can be divided into triangles that their sides are inner chords with length of at most 1. The mathematicians of planet Tarator proved that if the second statement is true, for each rotund polygon there exists a circle with radius $\frac{1}{4}$ that fits completely inside it. (c) Prove that if the second statement is true, then for each rotund polygon there exists a circle with radius $\frac{1}{4}$ that fits completely inside it. They found out that if the first statement is true then the second statement is also true, so they put a bounty of a doogh on proving the first statement. A young earth mathematician named J.N., found a counterexample for statement 1, thus receiving the bounty. (d) Find a 1392-gon that is counterexample for statement 1. But the Tarators are not disappointed and they are still trying to prove the second statement. (e) (Extra points) Prove or disprove the second statement. Time allowed for this problem was 150 minutes.

Four rectangular strips each measuring $4$ by $16$ inches are laid out with two vertical strips crossing two horizontal strips forming a single polygon which looks like a tic-tack-toe pattern. What is the perimeter of this polygon? [asy] size(100); draw((1,0)--(2,0)--(2,1)--(3,1)--(3,0)--(4,0)--(4,1)--(5,1)--(5,2)--(4,2)--(4,3)--(5,3)--(5,4)--(4,4)--(4,5)--(3,5)--(3,4)--(2,4)--(2,5)--(1,5)--(1,4)--(0,4)--(0,3)--(1,3)--(1,2)--(0,2)--(0,1)--(1,1)--(1,0)); draw((2,2)--(2,3)--(3,3)--(3,2)--cycle); [/asy]

Let U be the union of a finite number (not necessarily connected and not necessarily disjoint) of closed unit squares lying in the plane. Can the quotient of the perimeter and area of U be arbitrarily large? @below: i think "single" means "connected".

The perimeter of a semicircular region, measured in centimeters, is numerically equal to its area, measured in square centimeters. The radius of the semicircle, measured in centimeters, is $\text{(A)} \pi \qquad \text{(B)} \frac{2}{\pi} \qquad \text{(C)} 1 \qquad \text{(D)} \frac{1}{2} \qquad \text{(E)} \frac{4}{\pi} + 2$

Triangle $ ABC$ has side lengths $ AB \equal{} 5$, $ BC \equal{} 6$, and $ AC \equal{} 7$. Two bugs start simultaneously from $ A$ and crawl along the sides of the triangle in opposite directions at the same speed. They meet at point $ D$. What is $ BD$? $ \textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 5$

Let $k$ be a positive integer. In a circle with radius $1$, finitely many chords are drawn. You know that every diameter of the circle intersects at most $k$ of these chords. Prove that the sum of the lengths of all these chords is less than $k \cdot \pi$.

Let $C_1$ and $C_2$ be tangent circles internally at point $A$, with $C_2$ inside of $C_1$. Let $BC$ be a chord of $C_1$ that is tangent to $C_2$. Prove that the ratio between the length $BC$ and the perimeter of the triangle $ABC$ is constant, that is, it does not depend of the selection of the chord $BC$ that is chosen to construct the trangle.

Let $a, b, c$ be sides of a triangle whose perimeter does not exceed $2 \cdot \pi.$, Prove that $\sin a, \sin b, \sin c$ are sides of a triangle.

Let us divide by straight lines a quadrangle of unit area into $ n$ subpolygons and draw a circle into each subpolygon. Show that the sum of the perimeters of the circles is at most $ \pi \sqrt{n}$ (the lines are not allowed to cut the interior of a subpolygon). [i]G. and L. Fejes-Toth[/i]

Let us consider all rectangles with sides of length $a,b$ both of which are whole numbers. Do more of these rectangles have perimeter $2000$ or perimeter $2002$?

On the plane are $n$ paper disks of radius $1$ whose boundaries all pass through a certain point, which lies inside the region covered by the disks. Find the perimeter of this region. (P Kozhevnikov)

Given is a triangle $ABC$, $\angle C = 60^o$, $R$ the midpoint of side $AB$. There exist a point $P$ on the line $BC$ and a point $Q$ on the line $AC$ such that the perimeter of the triangle $PQR$ is minimal. a) Prove that and also indicate how the points $P$ and $Q$ can be constructed. b) If $AB = c$, $AC = b$, $BC = a$, then prove that the perimeter of the triangle $PQR$ equals $\frac12\sqrt{3c^2+6ab}$ .

Let $ E$ be the intersection of the diagonals of the convex quadrilateral $ ABCD$. The perimeters of $ \triangle AEB$, $ \triangle BEC$, $ \triangle CED$, and $ \triangle DEA$ are all same. If inradii of $ \triangle AEB$, $ \triangle BEC$, $ \triangle CED$ are $ 3,4,6$, respectively, then inradius of $ \triangle DEA$ will be ? $\textbf{(A)}\ \frac {9}{2} \qquad\textbf{(B)}\ \frac {7}{2} \qquad\textbf{(C)}\ \frac {13}{3} \qquad\textbf{(D)}\ 5 \qquad\textbf{(E)}\ \text{None}$

A large square is divided into a small square surrounded by four congruent rectangles as shown. The perimeter of each of the congruent rectangles is 14. What is the area of the large square? [asy]unitsize(3mm); defaultpen(linewidth(.8pt)); draw((0,0)--(7,0)--(7,7)--(0,7)--cycle); draw((1,0)--(1,6)); draw((7,1)--(1,1)); draw((6,7)--(6,1)); draw((0,6)--(6,6));[/asy]$ \textbf{(A)}\ \ 49 \qquad \textbf{(B)}\ \ 64 \qquad \textbf{(C)}\ \ 100 \qquad \textbf{(D)}\ \ 121 \qquad \textbf{(E)}\ \ 196$

a) Each point, on the perimeter of a square, is colored either red, or blue. Show that, there is a right-angled triangle where all the corners are on the square of the square and so that all the corners are on points of the same color. b) Show that, it is possible to color each point on the perimeter of one square, red, white, or blue so that, there is not a right-angled triangle where all the three corners are at points of same color.

Inside a rectangle is inscribed a quadrilateral, which has a vertex on each side of the rectangle. Prove that the perimeter of the inscribed quadrilateral is not smaller than double the length of a diagonal of the rectangle. (V. V . Proizvolov , Moscow)

Find three isosceles triangles, no two of which are congruent, with integer sides, such that each triangle's area is numerically equal to $6$ times its perimeter.

The square $ABCD$ has to be decomposed into $n$ triangles (which are not overlapping) and which have all angles acute. Find the smallest integer $n$ for which there exist a solution of that problem and for such $n$ construct at least one decomposition. Answer whether it is possible to ask moreover that (at least) one of these triangles has the perimeter less than an arbitrarily given positive number.

Given an angle less than $180^o$, and a point $M$ outside the angle. Draw a line through $M$ so that the triangle, whose vertices are the vertex of the angle and the intersection points of its legs with the line drawn, has a given perimeter.

For some positive integers $p$, there is a quadrilateral $ABCD$ with positive integer side lengths, perimeter $p$, right angles at $B$ and $C$, $AB=2$, and $CD=AD$. How many different values of $p<2015$ are possible? $\textbf{(A) }30\qquad\textbf{(B) }31\qquad\textbf{(C) }61\qquad\textbf{(D) }62\qquad\textbf{(E) }63$

Points $M$ and $N$ are the midpoints of sides $PA$ and $PB$ of $\triangle PAB$. As $P$ moves along a line that is parallel to side $AB$, how many of the four quantities listed below change? $\mathrm{(A)}\ \text{the length of the segment} MN$ $\mathrm{(B)}\ \text{the perimeter of }\triangle PAB$ $\mathrm{(C)}\ \text{ the area of }\triangle PAB$ $\mathrm{(D)}\ \text{ the area of trapezoid} ABNM$ [asy] draw((2,0)--(8,0)--(6,4)--cycle); draw((4,2)--(7,2)); draw((1,4)--(9,4),Arrows); label("$A$",(2,0),SW); label("$B$",(8,0),SE); label("$M$",(4,2),W); label("$N$",(7,2),E); label("$P$",(6,4),N);[/asy] $\mathrm{(A)}\ 0 \qquad\mathrm{(B)}\ 1 \qquad\mathrm{(C)}\ 2 \qquad\mathrm{(D)}\ 3 \qquad\mathrm{(E)}\ 4$

What is the area of the regular hexagon with perimeter $60$?

Unit square $ABCD$ is divided into $10^{12}$ smaller squares (not necessarily equal). Prove that the sum of perimeters of all the smaller squares having common points with diagonal $AC$ does not exceed 1500. [i]Proposed by A. Kanel-Belov[/i]