Found problems: 663
A quadrilateral has all integer sides lengths, a perimeter of $26$, and one side of length $4$. What is the greatest possible length of one side of this quadrilateral?
$\textbf{(A)}~9\qquad\textbf{(B)}~10\qquad\textbf{(C)}~11\qquad\textbf{(D)}~12\qquad\textbf{(E)}~13$
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]
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]
$\mathrm{(A)}\ 0 \qquad\mathrm{(B)}\ 1 \qquad\mathrm{(C)}\ 2 \qquad\mathrm{(D)}\ 3 \qquad\mathrm{(E)}\ 4$
The touching point of the excircle with the side of a triangle and the base of the altitude to this side are symmetric wrt the base of the corresponding bisector. Prove that this side is equal to one third of the perimeter.
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.
Given is a regular polygon. Volodya wants to mark $k$ points on its perimeter so that any another regular polygon (maybe having a different number of sides) doesn’t contain all marked points on its perimeter. Find the minimal $k$ sufficient for any given polygon.
Prove that if triangle $T_1$ contains triangle $T_2$, then the perimeter of triangle $T_1$ is not less than the perimeter of triangle $T_2$.
A square piece of paper, 4 inches on a side, is folded in half vertically. Both layers are then cut in half parallel to the fold. Three new rectangles are formed, a large one and two small ones. What is the ratio of the perimeter of one of the small rectangles to the perimeter of the large rectangle?
[asy]
draw((0,8)--(0,0)--(4,0)--(4,8)--(0,8)--(3.5,8.5)--(3.5,8));
draw((2,-1)--(2,9),dashed);[/asy]
$ \text{(A)}\ \frac{1}{3}\qquad\text{(B)}\ \frac{1}{2}\qquad\text{(C)}\ \frac{3}{4}\qquad\text{(D)}\ \frac{4}{5}\qquad\text{(E)}\ \frac{5}{6} $
As shown in the following figure, a heart is a shape consist of three semicircles with diameters $AB$, $BC$ and $AC$ such that $B$ is midpoint of the segment $AC$. A heart $\omega$ is given. Call a pair $(P, P')$ bisector if $P$ and $P'$ lie on $\omega$ and bisect its perimeter. Let $(P, P')$ and $(Q,Q')$ be bisector pairs. Tangents at points $P, P', Q$, and $Q'$ to $\omega$ construct a convex quadrilateral $XYZT$. If the quadrilateral $XYZT$ is inscribed in a circle, find the angle between lines $PP'$ and $QQ'$.
[img]https://cdn.artofproblemsolving.com/attachments/3/c/8216889594bbb504372d8cddfac73b9f56e74c.png[/img]
[i]Proposed by Mahdi Etesamifard - Iran[/i]
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 $k \cdot \pi$.
A triangle is circumscribed about a circle of radius $r$ inches. If the perimeter of the triangle is $P$ inches and the area is $K$ square inches, then $\frac{P}{K}$ is:
$ \text{(A)}\text{independent of the value of} \; r\qquad\text{(B)}\ \frac{\sqrt{2}}{r}\qquad\text{(C)}\ \frac{2}{\sqrt{r}}\qquad\text{(D)}\ \frac{2}{r}\qquad\text{(E)}\ \frac{r}{2} $
The diagonals of a convex quadrilateral $ABCD$ are perpendicular to each other and intersect at the point $O$. The sum of the inradii of triangles $AOB$ and $COD$ is equal to the sum of the inradii of triangles $BOC$ and $DOA$.
$(a)$ Prove that $ABCD$ has an incircle.
$(b)$ Prove that $ABCD$ is symmetric about one of its diagonals.
The cube $ABCDA' B' C' D' $has of length a. Consider the points $K \in [AB], L \in [CC' ], M \in [D'A']$.
a) Show that $\sqrt3 KL \ge KB + BC + CL$
b) Show that the perimeter of triangle $KLM$ is strictly greater than $2a\sqrt3$.
In the plane, consider a point $ X$ and a polygon $ \mathcal{F}$ (which is not necessarily convex). Let $ p$ denote the perimeter of $ \mathcal{F}$, let $ d$ be the sum of the distances from the point $ X$ to the vertices of $ \mathcal{F}$, and let $ h$ be the sum of the distances from the point $ X$ to the sidelines of $ \mathcal{F}$. Prove that $ d^2 \minus{} h^2\geq\frac {p^2}{4}.$
Out of all the nondegenerate triangles with positive integer sides and perimeter $100$, find the one with the smallest area.
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.
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} $
The diagonal of square I is $a+b$. The perimeter of square II with [i]twice[/i] the area of I is:
$ \textbf{(A)}\ (a+b)^2\qquad\textbf{(B)}\ \sqrt{2}(a+b)^2\qquad\textbf{(C)}\ 2(a+b)\qquad\textbf{(D)}\ \sqrt{8}(a+b) \qquad$
$\textbf{(E)}\ 4(a+b) $
$ABCD$ is a trapezoid with $AB$ parallel to $CD$. The external bisectors of the angles at $ B$ and $C$ intersect at $ P$. The external bisectors of the angles at $ A$ and $D$ intersect at $Q$. Show that the length of $PQ$ is equal to half the perimeter of the trapezoid $ABCD$.
What is the smallest whole number larger than the perimeter of any triangle with a side of length $ 5$ and a side of length $19$?
$\textbf{(A) }24\qquad\textbf{(B) }29\qquad\textbf{(C) }43\qquad\textbf{(D) }48\qquad \textbf{(E) }57$
Determine the greatest possible value of the constant $c$ that satisfies the following condition: for any convex heptagon the sum of the lengthes of all it’s diagonals is greater than $cP$, where $P$ is the perimeter of the heptagon.
(I. Zhuk)
Let $ d$ be the sum of the lengths of all the diagonals of a plane convex polygon with $ n$ vertices (where $ n>3$). Let $ p$ be its perimeter. Prove that:
\[ n\minus{}3<{2d\over p}<\Bigl[{n\over2}\Bigr]\cdot\Bigl[{n\plus{}1\over 2}\Bigr]\minus{}2,\]
where $ [x]$ denotes the greatest integer not exceeding $ x$.
Given a triangle $ABC$, consider the semicircle with diameter $\overline{EF}$ on $\overline{BC}$ tangent to $\overline{AB}$ and $\overline{AC}$. If $BE=1$, $EF=24$, and $FC=3$, find the perimeter of $\triangle{ABC}$.
[i]Proposed by Ray Li[/i]
All sides and diagonals of a convex $n$-gon, $n\ge 3$, are coloured one of two colours. Show that there exist $\left[\frac{n+1}{3}\right]$ pairwise disjoint monochromatic segments.
[i](Two segments are disjoint if they do not share an endpoint or an interior point).[/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)?