The incircle of $ABC$ touches $BC$, $CA$, $AB$ at $K$, $L$, $M$ respectively. The line through $A$ parallel to $LK$ meets $MK$ at $P$, and the line through $A$ parallel to $MK$ meets $LK$ at $Q$. Show that the line $PQ$ bisects $AB$ and bisects $AC$.

If each of two intersecting lines intersects a hyperbola and neither line is tangent to the hyperbola, then the possible number of points of intersection with the hyperbola is: $ \textbf{(A)}\ 2 \qquad\textbf{(B)}\ 2\text{ or }3 \qquad\textbf{(C)}\ 2\text{ or }4 \qquad\textbf{(D)}\ 3\text{ or }4 \qquad\textbf{(E)}\ 2,3,\text{ or }4$

Let $ABC$ be an acute triangle, and let $X$ be a variable interior point on the minor arc $BC$ of its circumcircle. Let $P$ and $Q$ be the feet of the perpendiculars from $X$ to lines $CA$ and $CB$, respectively. Let $R$ be the intersection of line $PQ$ and the perpendicular from $B$ to $AC$. Let $\ell$ be the line through $P$ parallel to $XR$. Prove that as $X$ varies along minor arc $BC$, the line $\ell$ always passes through a fixed point. (Specifically: prove that there is a point $F$, determined by triangle $ABC$, such that no matter where $X$ is on arc $BC$, line $\ell$ passes through $F$.) [i]Robert Simson et al.[/i]

On a long straight stretch of one-way single-lane highway, cars all travel at the same speed and all obey the safety rule: the distance from the back of the car ahead to the front of the car behind is exactly one car length for each 15 kilometers per hour of speed or fraction thereof (Thus the front of a car traveling 52 kilometers per hour will be four car lengths behind the back of the car in front of it.) A photoelectric eye by the side of the road counts the number of cars that pass in one hour. Assuming that each car is 4 meters long and that the cars can travel at any speed, let $ M$ be the maximum whole number of cars that can pass the photoelectric eye in one hour. Find the quotient when $ M$ is divided by 10.

A cylinder radius $12$ and a cylinder radius $36$ are held tangent to each other with a tight band. The length of the band is $m\sqrt{k}+n\pi$ where $m$, $k$, and $n$ are positive integers, and $k$ is not divisible by the square of any prime. Find $m + k + n$. [asy] size(150); real t=0.3; void cyl(pair x, real r, real h) { pair xx=(x.x,t*x.y); path B=ellipse(xx,r,t*r), T=ellipse((x.x,t*x.y+h),r,t*r), S=xx+(r,0)--xx+(r,h)--(xx+(-r,h))--xx-(r,0); unfill(S--cycle); draw(S); unfill(B); draw(B); unfill(T); draw(T); } real h=8, R=3,r=1.2; pair X=(0,0), Y=(R+r)*dir(-50); cyl(X,R,h); draw(shift((0,5))*yscale(t)*arc(X,R,180,360)); cyl(Y,r,h); void str (pair x, pair y, real R, real r, real h, real w) { real u=(angle(y-x)+asin((R-r)/(R+r)))*180/pi+270; path P=yscale(t)*(arc(x,R,180,u)--arc(y,r,u,360)); path Q=shift((0,h))*P--shift((0,h+w))*reverse(P)--cycle; fill(Q,grey);draw(Q); } str(X,Y,R,r,3.5,1.5);[/asy]

The magnitudes of the sides of triangle $ABC$ are $a$, $b$, and $c$, as shown, with $c\le b\le a$. Through interior point $P$ and the vertices $A$, $B$, $C$, lines are drawn meeting the opposite sides in $A'$, $B'$, $C'$, respectively. Let $s=AA'+BB'+CC'$. Then, for all positions of point $P$, $s$ is less than: $\textbf{(A) }2a+b\qquad\textbf{(B) }2a+c\qquad\textbf{(C) }2b+c\qquad\textbf{(D) }a+2b\qquad \textbf{(E) }$ $a+b+c$ [asy] import math; defaultpen(fontsize(11pt)); pair A = (0,0), B = (1,3), C = (5,0), P = (1.5,1); pair X = extension(B,C,A,P), Y = extension(A,C,B,P), Z = extension(A,B,C,P); draw(A--B--C--cycle); draw(A--X); draw(B--Y); draw(C--Z); dot(P); dot(A); dot(B); dot(C); label("$A$",A,dir(210)); label("$B$",B,dir(90)); label("$C$",C,dir(-30)); label("$A'$",X,dir(-100)); label("$B'$",Y,dir(65)); label("$C'$",Z,dir(20)); label("$P$",P,dir(70)); label("$a$",X,dir(80)); label("$b$",Y,dir(-90)); label("$c$",Z,dir(110)); //Credit to bobthesmartypants for the diagram [/asy]

Point $F$ is taken in side $AD$ of square $ABCD$. At $C$ a perpendicular is drawn to $CF$, meeting $AB$ extended at $E$. The area of $ABCD$ is $256$ square inches and the area of triangle $CEF$ is $200$ square inches. Then the number of inches in $BE$ is: [asy] size(6cm); pair A = (0, 0), B = (1, 0), C = (1, 1), D = (0, 1), E = (1.3, 0), F = (0, 0.7); draw(A--B--C--D--cycle); draw(F--C--E--B); label("$A$", A, SW); label("$B$", B, S); label("$C$", C, N); label("$D$", D, NW); label("$E$", E, SE); label("$F$", F, W); //Credit to MSTang for the asymptote [/asy] $\textbf{(A)}\ 12 \qquad \textbf{(B)}\ 14 \qquad \textbf{(C)}\ 15 \qquad \textbf{(D)}\ 16 \qquad \textbf{(E)}\ 20$

On square $ABCD,$ points $E,F,G,$ and $H$ lie on sides $\overline{AB},\overline{BC},\overline{CD},$ and $\overline{DA},$ respectively, so that $\overline{EG} \perp \overline{FH}$ and $EG=FH = 34.$ Segments $\overline{EG}$ and $\overline{FH}$ intersect at a point $P,$ and the areas of the quadrilaterals $AEPH, BFPE, CGPF,$ and $DHPG$ are in the ratio $269:275:405:411.$ Find the area of square $ABCD$. [asy] size(200); defaultpen(linewidth(0.8)+fontsize(10.6)); pair A = (0,sqrt(850)); pair B = (0,0); pair C = (sqrt(850),0); pair D = (sqrt(850),sqrt(850)); draw(A--B--C--D--cycle); dotfactor = 3; dot("$A$",A,dir(135)); dot("$B$",B,dir(215)); dot("$C$",C,dir(305)); dot("$D$",D,dir(45)); pair H = ((2sqrt(850)-sqrt(120))/6,sqrt(850)); pair F = ((2sqrt(850)+sqrt(306)+7)/6,0); dot("$H$",H,dir(90)); dot("$F$",F,dir(270)); draw(H--F); pair E = (0,(sqrt(850)-6)/2); pair G = (sqrt(850),(sqrt(850)+sqrt(100))/2); dot("$E$",E,dir(180)); dot("$G$",G,dir(0)); draw(E--G); pair P = extension(H,F,E,G); dot("$P$",P,dir(60)); label("$w$", (H+E)/2,fontsize(15)); label("$x$", (E+F)/2,fontsize(15)); label("$y$", (G+F)/2,fontsize(15)); label("$z$", (H+G)/2,fontsize(15)); label("$w:x:y:z=269:275:405:411$",(sqrt(850)/2,-4.5),fontsize(11)); [/asy]

In a circle of radius 42, two chords of length 78 intersect at a point whose distance from the center is 18. The two chords divide the interior of the circle into four regions. Two of these regions are bordered by segments of unequal lenghts, and the area of either of them can be expressed uniquley in the form $m\pi-n\sqrt{d},$ where $m, n,$ and $d$ are positive integers and $d$ is not divisible by the square of any prime number. Find $m+n+d.$

In a plane we are given two distinct points $A,B$ and two lines $a, b$ passing through $B$ and $A$ respectively $(a \ni B, b \ni A)$ such that the line $AB$ is equally inclined to a and b. Find the locus of points $M$ in the plane such that the product of distances from $M$ to $A$ and a equals the product of distances from $M$ to $B$ and $b$ (i.e., $MA \cdot MA' = MB \cdot MB'$, where $A'$ and $B'$ are the feet of the perpendiculars from $M$ to $a$ and $b$ respectively).

Let $ABC$ be a triangle. Let $A_1$, $A_2$ be points on $AB$ and $AC$ respectively such that $A_1A_2 \parallel BC$ and the circumcircle of $\triangle AA_1A_2$ is tangent to $BC$ at $A_3$. Define $B_3$, $C_3$ similarly. Prove that $AA_3$, $BB_3$, and $CC_3$ are concurrent. [i]Carl Lian.[/i]

Let $ABC$ be a triangle with $\angle ABC $ as the largest angle. Let $R$ be its circumcenter. Let the circumcircle of triangle $ARB$ cut $AC$ again at $X$. Prove that $RX$ is perpendicular to $BC$.

The set of points satisfying the pair of inequalities $y>2x$ and $y>4-x$ is contained entirely in quadrants: ${{ \textbf{(A)}\ \text{I and II} \qquad\textbf{(B)}\ \text{II and III} \qquad\textbf{(C)}\ \text{I and III} \qquad\textbf{(D)}\ \text{III and IV} }\qquad\textbf{(E)}\ \text{I and IV} } $

Triangle $ABC$ has $AC = 3$, $BC = 5$, $AB = 7$. A circle is drawn internally tangent to the circumcircle of $ABC$ at $C$, and tangent to $AB$. Let $D$ be its point of tangency with $AB$. Find $BD - DA$. [asy] /* File unicodetex not found. */ /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(6cm); real labelscalefactor = 2.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -4.5, xmax = 7.01, ymin = -3, ymax = 8.02; /* image dimensions */ /* draw figures */ draw(circle((1.37,2.54), 5.17)); draw((-2.62,-0.76)--(-3.53,4.2)); draw((-3.53,4.2)--(5.6,-0.44)); draw((5.6,-0.44)--(-2.62,-0.76)); draw(circle((-0.9,0.48), 2.12)); /* dots and labels */ dot((-2.62,-0.76),dotstyle); label("$C$", (-2.46,-0.51), SW * labelscalefactor); dot((-3.53,4.2),dotstyle); label("$A$", (-3.36,4.46), NW * labelscalefactor); dot((5.6,-0.44),dotstyle); label("$B$", (5.77,-0.17), SE * labelscalefactor); dot((0.08,2.37),dotstyle); label("$D$", (0.24,2.61), SW * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); label("$7$",(-3.36,4.46)--(5.77,-0.17), NE * labelscalefactor); label("$3$",(-3.36,4.46)--(-2.46,-0.51),SW * labelscalefactor); label("$5$",(-2.46,-0.51)--(5.77,-0.17), SE * labelscalefactor); /* end of picture */ [/asy]

Let $\Gamma$ be a circumference and $A$ a point outside it. Let $B$ and $C$ be points in $\Gamma$ such that $AB$ and $AC$ are tangent to $\Gamma$. Let $P$ be a point in $\Gamma$. Let $D$, $E$ and $F$ be points in $BC$, $AC$ and $AB$ respectively, such that $PD \perp BC$, $PE \perp AC$, and $PF \perp AB$. Show that $PD^2 = PE \cdot PF$

Let $ ABC$ be a triangle. Point $ D$ lies on its sideline $ BC$ such that $ \angle CAD \equal{} \angle CBA.$ Circle $ (O)$ passing through $ B,D$ intersects $ AB,AD$ at $ E,F$, respectively. $ BF$ meets $ DE$ at $ G$.Denote by$ M$ the midpoint of $ AG.$ Show that $ CM\perp AO.$

[asy] import cse5; size(180); pathpen=black; pair A1=(0,0), A2=(1,0), A3=(0.5,sqrt(3)/2); D(MP("A_1",A1)--MP("A_2",A2)--MP("A_3",A3,N)--cycle); pair A4=(A1+A2)/2, A5 = (A3+A2)/2, A6 = (A4+A3)/2; D(MP("A_4",A4,S)--MP("A_6",A6,W)--A3); D(A6--MP("A_5",A5,NE)--A4); //Credit to chezbgone2 for the diagram[/asy] If $\triangle A_1A_2A_3$ is equilateral and $A_{n+3}$ is the midpoint of line segment $A_nA_{n+1}$ for all positive integers $n$, then the measure of $\measuredangle A_{44}A_{45}A_{43}$ equals $\textbf{(A) }30^\circ\qquad\textbf{(B) }45^\circ\qquad\textbf{(C) }60^\circ\qquad\textbf{(D) }90^\circ\qquad \textbf{(E) }120^\circ$

A rectangular board of 8 columns has squared numbered beginning in the upper left corner and moving left to right so row one is numbered 1 through 8, row two is 9 through 16, and so on. A student shades square 1, then skips one square and shades square 3, skips two squares and shades square 6, skips 3 squares and shades square 10, and continues in this way until there is at least one shaded square in each column. What is the number of the shaded square that first achieves this result? [asy] unitsize(20); for(int a = 0; a < 10; ++a) { draw((0,a)--(8,a)); } for (int b = 0; b < 9; ++b) { draw((b,0)--(b,9)); } draw((0,0)--(0,-.5)); draw((1,0)--(1,-1.5)); draw((.5,-1)--(1.5,-1)); draw((2,0)--(2,-.5)); draw((4,0)--(4,-.5)); draw((5,0)--(5,-1.5)); draw((4.5,-1)--(5.5,-1)); draw((6,0)--(6,-.5)); draw((8,0)--(8,-.5)); fill((0,8)--(1,8)--(1,9)--(0,9)--cycle,black); fill((2,8)--(3,8)--(3,9)--(2,9)--cycle,black); fill((5,8)--(6,8)--(6,9)--(5,9)--cycle,black); fill((1,7)--(2,7)--(2,8)--(1,8)--cycle,black); fill((6,7)--(7,7)--(7,8)--(6,8)--cycle,black); label("$2$",(1.5,8.2),N); label("$4$",(3.5,8.2),N); label("$5$",(4.5,8.2),N); label("$7$",(6.5,8.2),N); label("$8$",(7.5,8.2),N); label("$9$",(0.5,7.2),N); label("$11$",(2.5,7.2),N); label("$12$",(3.5,7.2),N); label("$13$",(4.5,7.2),N); label("$14$",(5.5,7.2),N); label("$16$",(7.5,7.2),N); [/asy] $\text{(A)}\ 36 \qquad \text{(B)}\ 64 \qquad \text{(C)}\ 78 \qquad \text{(D)}\ 91 \qquad \text{(E)}\ 120$

Let $ K $ be the midpoint of the side $ AB $ of a given triangle $ ABC $. Let $ L $ and $ M$ be points on the sides $ AC $ and $ BC$, respectively, such that $ \angle CLK = \angle KMC $. Prove that the perpendiculars to the sides $ AB, AC, $ and $ BC $ passing through $ K,L, $ and $M$, respectively, are concurrent.

Let $X$ be an arbitrary point inside the circumcircle of a triangle $ABC$. The lines $BX$ and $CX$ meet the circumcircle in points $K$ and $L$ respectively. The line $LK$ intersects $BA$ and $AC$ at points $E$ and $F$ respectively. Find the locus of points $X$ such that the circumcircles of triangles $AFK$ and $AEL$ touch.

Let $T_1$ and $T_2$ be the points of tangency of the excircles of a triangle $ABC$ with its sides $BC$ and $AC$ respectively. It is known that the reflection of the incenter of $ABC$ across the midpoint of $AB$ lies on the circumcircle of triangle $CT_1T_2$. Find $\angle BCA$.

This problem is easy but nobody solved it. point $A$ moves in a line with speed $v$ and $B$ moves also with speed $v'$ that at every time the direction of move of $B$ goes from $A$.We know $v \geq v'$.If we know the point of beginning of path of $A$, then $B$ must be where at first that $B$ can catch $A$.

Let $ABCDE$ be a pentagon inscribed in a circle such that $AB=CD=3$, $BC=DE=10$, and $AE=14$. The sum of the lengths of all diagonals of $ABCDE$ is equal to $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. What is $m+n$? $\textbf{(A) }129\qquad \textbf{(B) }247\qquad \textbf{(C) }353\qquad \textbf{(D) }391\qquad \textbf{(E) }421\qquad$

The incircle of triangle $ABC$ touches the side $AB$ at point $C'$; the incircle of triangle $ACC'$ touches the sides $AB$ and $AC$ at points $C_1, B_1$; the incircle of triangle $BCC'$ touches the sides $AB$ and $BC$ at points $C_2$, $A_2$. Prove that the lines $B_1C_1$, $A_2C_2$, and $CC'$ concur.

Chord $EF$ is the perpendicular bisector of chord $BC$, intersecting it in $M$. Between $B$ and $M$ point $U$ is taken, and $EU$ extended meets the circle in $A$. Then, for any selection of $U$, as described, triangle $EUM$ is similar to triangle: [asy] pair B = (-0.866, -0.5); pair C = (0.866, -0.5); pair E = (0, -1); pair F = (0, 1); pair M = midpoint(B--C); pair A = (-0.99, -0.141); pair U = intersectionpoints(A--E, B--C)[0]; draw(B--C); draw(F--E--A); draw(unitcircle); label("$B$", B, SW); label("$C$", C, SE); label("$A$", A, W); label("$E$", E, S); label("$U$", U, NE); label("$M$", M, NE); label("$F$", F, N); //Credit to MSTang for the asymptote [/asy] $\textbf{(A)}\ EFA \qquad \textbf{(B)}\ EFC \qquad \textbf{(C)}\ ABM \qquad \textbf{(D)}\ ABU \qquad \textbf{(E)}\ FMC$