Found problems: 663
Let $s=\frac{AB+BC+AC}{2}$ be half-perimeter of triangle $ABC$. Let $L$ and $N$be a point's on ray's $AB$ and $CB$, for which $AL=CN=s$. Let $K$ is point, symmetric of point $B$ by circumcenter of $ABC$. Prove, that perpendicular from $K$ to $NL$ passes through incenter of $ABC$.
Solution for problem [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=125&t=365714&p=2011659#p2011659]here[/url]
Let $ABCD$ be a convex quadrilateral with perimeter $\tfrac{5}{2}$ and $AC=BD=1$. Determine the maximum possible area of $ABCD$.
A magician should determine the area of a hidden convex $ 2008$-gon $ A_{1}A_{2}\cdots A_{2008}$. In each step he chooses two points on the perimeter, whereas the chosen points can be vertices or points dividing selected sides in selected ratios. Then his helper divides the polygon into two parts by the line through these two points and announces the area of the smaller of the two parts. Show that the magician can find the area of the polygon in $ 2006$ steps.
Let $P$ be a simple polygon completely in $C$, a circle with radius $1$, such that $P$ does not pass through the center of $C$. The perimeter of $P$ is $36$. Prove that there is a radius of $C$ that intersects $P$ at least $6$ times, or there is a circle which is concentric with $C$ and have at least $6$ common points with $P$.
[i]Proposed by Seyed Reza Hosseini[/i]
$(BUL 3)$ 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.
Let $P$ be a point inside or outside (but not on) of a triangle $ABC$. Prove that $PA +PB +PC$ is greater than half of the perimeter of the triangle
In a triangle $ ABC $ that has perimeter $ P, $ prove that it's isosceles if and only if
$$ P^2+\sin^2 (\angle ABC-\angle BCA) =4\cdot AB\cdot AC\cdot\cos^2\frac{\angle ABC}{2}\cdot\cos^2\frac{\angle BCA}{2} . $$
Joe has a triangle with area $\sqrt{3}.$ What's the smallest perimeter it could have?
The incircle of the triangle $ABC$ touches its sides $AB,BC,CA$ at points $C_1,A_1,B_1$ respectively. If $r$ is the inradius of $\vartriangle ABC, P,P_1$ are the perimeters of $\vartriangle ABC, \vartriangle A_1B_1C_1$ respectively, prove that $P+P_1 \ge 9 \sqrt3 r$.
I. Voronovich
Let $\triangle{ABC}$ be a triangle with integer side lengths and the property that $\angle{B} = 2\angle{A}$. What is the least possible perimeter of such a triangle?
$
\textbf{(A) }13 \qquad
\textbf{(B) }14 \qquad
\textbf{(C) }15 \qquad
\textbf{(D) }16 \qquad
\textbf{(E) }17 \qquad
$
Let $T_1$ be a triangle with sides $2011, 2012,$ and $2013$. For $n \ge 1$, if $T_n=\triangle ABC$ and $D,E,$ and $F$ are the points of tangency of the incircle of $\triangle ABC$ to the sides $AB,BC$ and $AC$, respectively, then $T_{n+1}$ is a triangle with side lengths $AD,BE,$ and $CF$, if it exists. What is the perimeter of the last triangle in the sequence $(T_n)$?
$ \textbf{(A)}\ \frac{1509}{8} \qquad
\textbf{(B)}\ \frac{1509}{32} \qquad
\textbf{(C)}\ \frac{1509}{64} \qquad
\textbf{(D)}\ \frac{1509}{128} \qquad
\textbf{(E)}\ \frac{1509}{256} $
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 $\frac{S}{S'}\geq 3+2\sqrt{2}$.
The two squares shown share the same center $O$ and have sides of length 1. The length of $\overline{AB}$ is $43/99$ and the area of octagon $ABCDEFGH$ is $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$
[asy]
real alpha = 25;
pair W=dir(225), X=dir(315), Y=dir(45), Z=dir(135), O=origin;
pair w=dir(alpha)*W, x=dir(alpha)*X, y=dir(alpha)*Y, z=dir(alpha)*Z;
draw(W--X--Y--Z--cycle^^w--x--y--z--cycle);
pair A=intersectionpoint(Y--Z, y--z),
C=intersectionpoint(Y--X, y--x),
E=intersectionpoint(W--X, w--x),
G=intersectionpoint(W--Z, w--z),
B=intersectionpoint(Y--Z, y--x),
D=intersectionpoint(Y--X, w--x),
F=intersectionpoint(W--X, w--z),
H=intersectionpoint(W--Z, y--z);
dot(O);
label("$O$", O, SE);
label("$A$", A, dir(O--A));
label("$B$", B, dir(O--B));
label("$C$", C, dir(O--C));
label("$D$", D, dir(O--D));
label("$E$", E, dir(O--E));
label("$F$", F, dir(O--F));
label("$G$", G, dir(O--G));
label("$H$", H, dir(O--H));[/asy]
We are given a finite collection of segments in the plane, of total length 1. Prove that there exists a line $ l$ such that the sum of the lengths of the projections of the given segments to the line $ l$ is less than $ \frac{2}{\pi}.$
Let $O$ be a point in the plane and let $F$ be a (not necessary convex) polygon. Let $P$ be the perimeter of $F$, let $D$ be sum of the distances of the point $O$ from the vertices of $F$, and let $H$ be sum of the distances of the point $O$ from the lines that pass through the vertices of $F$. Show that
\[D^2-H^2 \geq \frac{P^2}{4}.\]
Let $ABCD$ be a convex quadrilateral. $AC$ and $BD$ meet at $P$, with $\angle APD=60^{\circ}$. Let $E,F,G$, and $H$ be the midpoints of $AB,BC,CD$ and $DA$ respectively. Find the greatest positive real number $k$ for which
\[EG+3HF\ge kd+(1-k)s \]
where $s$ is the semi-perimeter of the quadrilateral $ABCD$ and $d$ is the sum of the lengths of its diagonals. When does the equality hold?
Jorge's teacher asks him to plot all the ordered pairs $ (w, l)$ of positive integers for which $ w$ is the width and $ l$ is the length of a rectangle with area 12. What should his graph look like?
$ \textbf{(A)}$[asy]size(75);
draw((0,-1)--(0,13));
draw((-1,0)--(13,0));
dot((1,12));
dot((2,6));
dot((3,4));
dot((4,3));
dot((6,2));
dot((12,1));
label("$l$", (0,6), W);
label("$w$", (6,0), S);[/asy]
$ \textbf{(B)}$[asy]size(75);
draw((0,-1)--(0,13));
draw((-1,0)--(13,0));
dot((1,1));
dot((3,3));
dot((5,5));
dot((7,7));
dot((9,9));
dot((11,11));
label("$l$", (0,6), W);
label("$w$", (6,0), S);[/asy]
$ \textbf{(C)}$[asy]size(75);
draw((0,-1)--(0,13));
draw((-1,0)--(13,0));
dot((1,11));
dot((3,9));
dot((5,7));
dot((7,5));
dot((9,3));
dot((11,1));
label("$l$", (0,6), W);
label("$w$", (6,0), S);[/asy]
$ \textbf{(D)}$[asy]size(75);
draw((0,-1)--(0,13));
draw((-1,0)--(13,0));
dot((1,6));
dot((3,6));
dot((5,6));
dot((7,6));
dot((9,6));
dot((11,6));
label("$l$", (0,6), W);
label("$w$", (6,0), S);[/asy]
$ \textbf{(E)}$[asy]size(75);
draw((0,-1)--(0,13));
draw((-1,0)--(13,0));
dot((6,1));
dot((6,3));
dot((6,5));
dot((6,7));
dot((6,9));
dot((6,11));
label("$l$", (0,6), W);
label("$w$", (6,0), S);[/asy]
In the convex $ABCDEF$ (has all interior angles less than $180^o$) with the perimeter $s$ the triangles $ACE$ and $BDF$ have perimeters $u$ and $v$ respectively.
a) Show the inequalities $\frac{1}{2} \le \frac{s}{u+v}\le 1$
b) Check whether $1$ is replaced by a smaller number or $1/2$ by a larger number can the inequality remains valid for all convex hexagons.
In a right triangle, the length of one side is a prime and the lengths of the other
side and the hypotenuse are integral. The ratio of the triangle perimeter and the incircle diameter is also an integer. Find all possible side lengths of the triangle.
The number of scalene triangles having all sides of integral lengths, and perimeter less than $ 13$ is:
$ \textbf{(A)}\ 1 \qquad\textbf{(B)}\ 2 \qquad\textbf{(C)}\ 3 \qquad\textbf{(D)}\ 4 \qquad\textbf{(E)}\ 18$
An angle $< 45^o$ is given in the plane of the drawing. Furthermore, the projection $P_1$ of a point $P$ lying above the plane of the drawing and the distance from $P$ to $P_1$ are given. $P_1$ lies within the given angle. On the legs of the given angle, construct points $A$ and $B$, respectively, such that the triangle $PAB$ has a minimal perimeter.
On the sides of triangle $ABC$ take the points $M_1, N_1, P_1$ such that each line $MM_1, NN_1, PP_1$ divides the perimeter of $ABC$ in two equal parts ($M, N, P$ are respectively the midpoints of the sides $BC, CA, AB$).
[b]I.[/b] Prove that the lines $MM_1, NN_1, PP_1$ are concurrent at a point $K$.
[b]II.[/b] Prove that among the ratios $\frac{KA}{BC}, \frac{KB}{CA}, \frac{KC}{AB}$ there exist at least a ratio which is not less than $\frac{1}{\sqrt{3}}$.
The checkered plane is painted black and white, after a chessboard fashion. A polygon $\Pi$ of area $S$ and perimeter $P$ consists of some of these unit squares (i.e., its sides go along the borders of the squares).
Prove the polygon $\Pi$ contains not more than $\dfrac {S} {2} + \dfrac {P} {8}$, and not less than $\dfrac {S} {2} - \dfrac {P} {8}$ squares of a same color.
(Alexander Magazinov)
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$.
In the below figure, there is a regular hexagon and three squares whose sides are equal to $4$ cm. Let $M,N$, and $P$ be the centers of the squares. The perimeter of the triangle $MNP$ can be written in the form $a + b\sqrt3$ (cm), where $a, b$ are integers. Compute the value of $a + b$.
[img]https://cdn.artofproblemsolving.com/attachments/e/8/5996e994d4bbed8d3b3269d3e38fc2ec5d2f0b.png[/img]