The dimensions of a rectangle $R$ are $a$ and $b$, $a < b$. It is required to obtain a rectangle with dimensions $x$ and $y$, $x < a$, $y < a$, so that its perimeter is one-third that of $R$, and its area is one-third that of $R$. The number of such (different) rectangles is: $\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1\qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ \text{infinitely many}$

Consider 2 concentric circle radii $ R$ and $ r$ ($ R > r$) with centre $ O.$ Fix $ P$ on the small circle and consider the variable chord $ PA$ of the small circle. Points $ B$ and $ C$ lie on the large circle; $ B,P,C$ are collinear and $ BC$ is perpendicular to $ AP.$ [b]i.)[/b] For which values of $ \angle OPA$ is the sum $ BC^2 \plus{} CA^2 \plus{} AB^2$ extremal? [b]ii.)[/b] What are the possible positions of the midpoints $ U$ of $ BA$ and $ V$ of $ AC$ as $ \angle OPA$ varies?

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 $P$ be an interior point of triangle $ABC$. The lines $AP$, $BP$ and $CP$ divide each of the three sides into two segments. If the so-obtained six segments all have distinct integer lengths, what is the minimum possible perimeter of $ABC$?

Suppose we climb a mountain that is a cone with radius $100$ and height $4$. We start at the bottom of the mountain (on the perimeter of the base of the cone), and our destination is the opposite side of the mountain, halfway up (height $z = 2$). Our climbing speed starts at $v_0=2$ but gets slower at a rate inversely proportional to the distance to the mountain top (so at height $z$ the speed $v$ is $(h-z)v_0/h$). Find the minimum time needed to get to the destination.

For any natural numbers $n$, ( $n \geq 3$), let $f(n)$ denote the number of congruent integer-sided triangles with perimeter $n$. Show that (i) $f(1999) > f (1996)$; (ii) $f(2000) = f(1997)$.

Let there be a triangle $ABC$ with orthocenter $H$. Let the lengths of the heights be $h_a, h_b, h_c$ from points $A, B$ and respectively $C$, and the semi-perimeter $p$ of triangle $ABC$. It is known that $AH \cdot h_a + BH \cdot h_b + CH \cdot h_c = \frac{2}{3} \cdot p^2$. Show that $ABC$ is equilateral.

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

Prove that a line that divides a triangle into two polygons of equal area and equal perimeter passes through the center of the circle inscribed in the triangle. Prove an analogous property for a polygon that has an inscribed circle.

Let $n\ge3$ be an integer. A circle is divided into $2n$ arcs by $2n$ points. Each arc has one of three possible lengths, and no two adjacent arcs have the same lengths. The $2n$ points are colored alternately red and blue. Prove that the $n$-gon with red vertices and the $n$-gon with blue vertices have the same perimeter and the same area.

Let $ABCD$ be the quadrilateral whose area is the largest among the quadrilaterals with given sides $a, b, c, d$, and let $PORS$ be the quadrilateral inscribed in $ABCD$ with the smallest perimeter. Find this perimeter.

Figures $I$, $II$, and $III$ are squares. The perimeter of $I$ is $12$ and the perimeter of $II$ is $24$. The perimeter of $III$ is [asy] draw((0,0)--(15,0)--(15,6)--(12,6)--(12,9)--(0,9)--cycle); draw((9,0)--(9,9)); draw((9,6)--(12,6)); label("$III$",(4.5,4),N); label("$II$",(12,2.5),N); label("$I$",(10.5,6.75),N); [/asy] $\text{(A)}\ 9 \qquad \text{(B)}\ 18 \qquad \text{(C)}\ 36 \qquad \text{(D)}\ 72 \qquad \text{(D)}\ 81$

A quadrilateral is given, in which the lengths of some two sides are equal to $1$ and $4$. Also, the diagonal of length $2$ divides it into two isosceles triangles. Find the perimeter of this quadrilateral.

Two points on the circumference of a circle of radius r are selected independently and at random. From each point a chord of length r is drawn in a clockwise direction. What is the probability that the two chords intersect? $ \textbf{(A)}\ \frac{1}{6}\qquad\textbf{(B)}\ \frac{1}{5}\qquad\textbf{(C)}\ \frac{1}{4}\qquad\textbf{(D)}\ \frac{1}{3}\qquad\textbf{(E)}\ \frac{1}{2} $

Suppose $F$ is a polygon with lattice vertices and sides parralell to x-axis and y-axis.Suppose $S(F),P(F)$ are area and perimeter of $F$. Find the smallest k that: $S(F) \leq k.P(F)^2$

Two tangents are drawn to a circle from an exterior point $A$; they touch the circle at points $B$ and $C$ respectively. A third tangent intersects segment $AB$ in $P$ and $AC$ in $R$, and touches the circle at $Q$. If $AB=20$, then the perimeter of triangle $APR$ is ${{ \textbf{(A)}\ 42\qquad\textbf{(B)}\ 40.5 \qquad\textbf{(C)}\ 40\qquad\textbf{(D)}\ 39\frac{7}{8} }\qquad\textbf{(E)}\ \text{not determined by the given information} } $

[u][b]The author of this posting is : Peter VDD[/b][/u] ____________________________________________________________________ most of us didn't really expect to get this, but here it goes (flanders mathematical olympiad 2004, today) triangle with sides 501m, 668m, 835m how many lines can be draws so that the line halves both area and circumference?

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)

A [i]stick[/i] is defined as a $1 \times k$ or $k\times 1$ rectangle for any integer $k \ge 1$. We wish to partition the cells of a $2022 \times 2022$ chessboard into $m$ non-overlapping sticks, such that any two of these $m$ sticks share at most one unit of perimeter. Determine the smallest $m$ for which this is possible. [i]Holden Mui[/i]

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.

In the adjoining diagram, $BO$ bisects $\angle CBA$, $CO$ bisects $\angle ACB$, and $MN$ is parallel to $BC$. If $AB=12$, $BC=24$, and $AC=18$, then the perimeter of $\triangle AMN$ is [asy] size(200); defaultpen(linewidth(0.7)+fontsize(10)); pair B=origin, C=(24,0), A=intersectionpoints(Circle(B,12), Circle(C,18))[0], O=incenter(A,B,C), M=intersectionpoint(A--B, O--O+40*dir(180)), N=intersectionpoint(A--C, O--O+40*dir(0)); draw(B--M--O--B--C--O--N--C^^N--A--M); label("$A$", A, dir(90)); label("$B$", B, dir(O--B)); label("$C$", C, dir(O--C)); label("$M$", M, dir(90)*dir(B--A)); label("$N$", N, dir(90)*dir(A--C)); label("$O$", O, dir(90));[/asy] $\textbf {(A) } 30 \qquad \textbf {(B) } 33 \qquad \textbf {(C) } 36 \qquad \textbf {(D) } 39 \qquad \textbf {(E) } 42$

City of Mar del Plata is a square shaped $WSEN$ land with $2(n + 1)$ streets that divides it into $n \times n$ blocks, where $n$ is an even number (the leading streets form the perimeter of the square). Each block has a dimension of $100 \times 100$ meters. All streets in Mar del Plata are one-way. The streets which are parallel and adjacent to each other are directed in opposite direction. Street $WS$ is driven in the direction from $W$ to $S$ and the street $WN$ travels from $W$ to $N$. A street cleaning car starts from point $W$. The driver wants to go to the point $E$ and in doing so, he must cross as much as possible roads. What is the length of the longest route he can go, if any $100$-meter stretch cannot be crossed more than once? (The figure shows a plan of the city for $n=6$ and one of the possible - but not the longest - routes of the street cleaning car. See http://goo.gl/maps/JAzD too.) [img]http://s14.postimg.org/avfg7ygb5/CPS_2012_P5.jpg[/img]

Nondegenerate $ \triangle ABC$ has integer side lengths, $ BD$ is an angle bisector, $ AD \equal{} 3$, and $ DC \equal{} 8$. What is the smallest possible value of the perimeter? $ \textbf{(A)}\ 30 \qquad \textbf{(B)}\ 33 \qquad \textbf{(C)}\ 35 \qquad \textbf{(D)}\ 36 \qquad \textbf{(E)}\ 37$

A quadrilateral is given, in which the lengths of some two sides are equal to $1$ and $4$. Also, the diagonal of length $2$ divides it into two isosceles triangles. Find the perimeter of this quadrilateral.

In a small town, there are $n \times n$ houses indexed by $(i, j)$ for $1 \leq i, j \leq n$ with $(1, 1)$ being the house at the top left corner, where $i$ and $j$ are the row and column indices, respectively. At time 0, a fire breaks out at the house indexed by $(1, c)$, where $c \leq \frac{n}{2}$. During each subsequent time interval $[t, t+1]$, the fire fighters defend a house which is not yet on fire while the fire spreads to all undefended [i]neighbors[/i] of each house which was on fire at time t. Once a house is defended, it remains so all the time. The process ends when the fire can no longer spread. At most how many houses can be saved by the fire fighters? A house indexed by $(i, j)$ is a [i]neighbor[/i] of a house indexed by $(k, l)$ if $|i - k| + |j - l|=1$.