Prove: If $x, y, z$ are the lengths of the angle bisectors of a triangle with perimeter 6, than we have: \[\frac{1}{x^2} + \frac{1}{y^2} + \frac{1}{z^2} \geq 1.\]

Let two ants stand on the perimeter of a regular $2019$-gon of unit side length. One of them stands on a vertex and the other one is on the midpoint of the opposite side. They start walking along the perimeter at the same speed counterclockwise. The locus of their midpoints traces out a figure $P$ in the plane with $N$ corners. Let the area enclosed by the convex hull of $P$ be $\tfrac{A}{B}\tfrac{\sin^m\left(\tfrac{\pi}{4038}\right)}{\tan\left(\tfrac{\pi}{2019}\right)}$, where $A$ and $B$ are coprime positive integers, and $m$ is the smallest possible positive integer such that this formula holds. Find $A+B+m+N$. [i]Note:[/i] The [i]convex hull[/i] of a figure $P$ is the convex polygon of smallest area which contains $P$.

A Pythagorean triangle is a right triangle whose three sides are integers. The best known example is the triangle with rectangular sides $3$ and $4$ and hypotenuse $5$. Determine all Pythagorean triangles whose area is twice the perimeter.

Let ABC be a triangle and $M$ be an interior point. Prove that \[ \min\{MA,MB,MC\}+MA+MB+MC<AB+AC+BC.\]

In a triangle with integer side lengths, one side is three times as long as a second side, and the length of the third side is 15. What is the greatest possible perimeter of the triangle? $ \textbf{(A) } 43 \qquad \textbf{(B) } 44 \qquad \textbf{(C) } 45 \qquad \textbf{(D) } 46 \qquad \textbf{(E) } 47$

In quadrilateral $ABCD, \angle B$ is a right angle, diagonal $\overline{AC}$ is perpendicular to $\overline{CD},$ $AB=18, BC=21,$ and $CD=14.$ Find the perimeter of $ABCD$.

A convex quadrilateral $ ABCD$ with area $ 2002$ contains a point $ P$ in its interior such that $ PA \equal{} 24$, $ PB \equal{} 32$, $ PC \equal{} 28$, and $ PD \equal{} 45$. FInd the perimeter of $ ABCD$. $ \textbf{(A)}\ 4\sqrt {2002}\qquad \textbf{(B)}\ 2\sqrt {8465}\qquad \textbf{(C)}\ 2\left(48 \plus{} \sqrt {2002}\right)$ $ \textbf{(D)}\ 2\sqrt {8633}\qquad \textbf{(E)}\ 4\left(36 \plus{} \sqrt {113}\right)$

An $ABC$ triangle divide by a $d$ line such that, new two pieces' areas and perimeters are equal. Prove that $ABC$'s incenter lies $d$

The floor of a rectangular room is covered with square tiles. The room is 10 tiles long and 5 tiles wide. The number of tiles that touch the walls of the room is $\textbf{(A)}\ 26 \qquad \textbf{(B)}\ 30 \qquad \textbf{(C)}\ 34 \qquad \textbf{(D)}\ 46 \qquad \textbf{(E)}\ 50$

Given a tetrahedron $ABCD$, $E$, $F$, $G$, are on the respectively on the segments $AB$, $AC$ and $AD$. Prove that: i) area $EFG \leq$ max{area $ABC$,area $ABD$,area $ACD$,area $BCD$}. ii) The same as above replacing "area" for "perimeter".

$A_1A_2A_3A_4$ is a tangential quadrilateral with perimeter $p_1$ and sum of the diagonals $k_1$ .$B_1B_2B_3B_4$ is a tangential quadrilateral with perimeter $p_2$ and sum of the diagonals $k_2$ .Prove that $A_1A_2A_3A_4$ and $B_1B_2B_3B_4$ are congruent squares if $$ p_1^2+p_2^2=(k_1+k_2)^2 $$

The point $A_1$ on the perimeter of a convex quadrilateral $ABCD$ is such that the line $AA_1$ divides the quadrilateral into two parts of equal area. The points $B_1$, $C_1$, $D_1$ are defined similarly. Prove that the area of the quadrilateral $A_1B_1C_1D_1$ is greater than a quarter of the area of $ABCD$. [i]L. Emelyanov [/i]

Let $n$ be a natural number. Given a chessboard sized $m \times n$. The sides of the small squares of chessboard are not on the perimeter of the chessboard will be colored so that each small square has exactly two sides colored. Prove that a coloring like that is possible if and only if $m \cdot n$ is even.

10 vertices of a regular 100-gon are coloured red and ten other (distinct) vertices are coloured blue. Prove that there is at least one connection edge (segment) of two red which is as long as the connection edge of two blue points. [hide="Hint"]Possible approaches are pigeon hole principle, proof by contradiction, consider turns (bijective congruent mappings) which maps red in blue points. [/hide]

[b]i)[/b] Let $0<a<b$.Prove that amongst all triangles having base $a$ and perimeter $a+b$ the triangle having two sides(other than the base) equal to $\frac {b}{2}$ has the maximum area. [b]ii)[/b]Using $i)$ or otherwise, prove that amongst all quadrilateral having give perimeter the square has the maximum area.

Vitya cut the chessboard along the borders of the cells into pieces of the same perimeter. It turned out that not all of the received parts are equal. What is the largest possible number of parts that Vitya could get?

Let $ABC$ be an equilateral triangle and $n{}, n>1$ an integer. Let $S{}$ be the set of the $n-1$ lines parallel with $BC$ that cut $ABC$ in $n{}$ figures with equal areas and $S^{'}$ be the set of the $n-1$ lines parallel with $BC$ that cut $ABC$ in $n{}$ figures with equal perimeters. Show that $S{}$ and $S^{'}$ are disjunctive.

A right-angled triangle has perimeter $60$ and the altitude of the hypotenuse has a length $12$. Determine the lengths of the sides.

Let $AB$ be diameter of a circle $\omega$ and let $C$ be a point on $\omega$, different from $A$ and $B$. The perpendicular from $C$ intersects $AB$ at $D$ and $\omega$ at $E(\neq C)$. The circle with centre at $C$ and radius $CD$ intersects $\omega$ at $P$ and $Q$. If the perimeter of the triangle $PEQ$ is $24$, find the length of the side $PQ$

In order for Mateen to walk a kilometer ($1000$m) in his rectangular backyard, he must walk the length $25$ times or walk its perimeter $10$ times. What is the area of Mateen's backyard in square meters? $\text{(A)}\ 40 \qquad \text{(B)}\ 200 \qquad \text{(C)}\ 400 \qquad \text{(D)}\ 500 \qquad \text{(E)}\ 1000$

In right triangle $ ABC$, $ BC \equal{} 5$, $ AC \equal{} 12$, and $ AM \equal{} x$; $ \overline{MN} \perp \overline{AC}$, $ \overline{NP} \perp \overline{BC}$; $ N$ is on $ AB$. If $ y \equal{} MN \plus{} NP$, one-half the perimeter of rectangle $ MCPN$, then: [asy]defaultpen(linewidth(.8pt)); unitsize(2cm); pair A = origin; pair M = (1,0); pair C = (2,0); pair P = (2,0.5); pair B = (2,1); pair Q = (1,0.5); draw(A--B--C--cycle); draw(M--Q--P); label("$A$",A,SW); label("$M$",M,S); label("$C$",C,SE); label("$P$",P,E); label("$B$",B,NE); label("$N$",Q,NW);[/asy]$ \textbf{(A)}\ y \equal{} \frac {1}{2}(5 \plus{} 12) \qquad \textbf{(B)}\ y \equal{} \frac {5x}{12} \plus{} \frac {12}{5}\qquad \textbf{(C)}\ y \equal{} \frac {144 \minus{} 7x}{12}\qquad$ $ \textbf{(D)}\ y \equal{} 12\qquad \qquad\quad\,\, \textbf{(E)}\ y \equal{} \frac {5x}{12} \plus{} 6$

Out of all the nondegenerate triangles with positive integer sides and perimeter $100$, find the one with the smallest area.

The perimeter of triangle $APM$ is $152,$ and the angle $PAM$ is a right angle. A circle of radius $19$ with center $O$ on $\overline{AP}$ is drawn so that it is tangent to $\overline{AM}$ and $\overline{PM}.$ Given that $OP=m/n,$ where $m$ and $n$ are relatively prime positive integers, find $m+n.$

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.

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.