An uncrossed belt is fitted without slack around two circular pulleys with radii of $14$ inches and $4$ inches. If the distance between the points of contact of the belt with the pulleys is $24$ inches, then the distance between the centers of the pulleys in inches is $\textbf{(A) }24\qquad\textbf{(B) }2\sqrt{119}\qquad\textbf{(C) }25\qquad\textbf{(D) }26\qquad \textbf{(E) }4\sqrt{35}$

Points $E$ and $F$ lie inside rectangle $ABCD$ with $AE=DE=BF=CF=EF$. If $AB=11$ and $BC=8$, find the area of the quadrilateral $AEFB$.

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 $A$.

The incircle of triangle $ABC$ touches sides $AB$;$BC$;$CA$ at $M$;$N$;$K$, respectively. The line through $A$ parallel to $NK$ meets $MN$ at $D$. The line through $A$ parallel to $MN$ meets $NK$ at $E$. Show that the line $DE$ bisects sides $AB$ and $AC$ of triangle $ABC$. [i]M. Sonkin[/i]

[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 $\textbf{(A) }30^\circ\qquad\textbf{(B) }45^\circ\qquad\textbf{(C) }60^\circ\qquad\textbf{(D) }90^\circ\qquad \textbf{(E) }120^\circ$

The shaded region below is called a shark's fin falcata, a figure studied by Leonardo da Vinci. It is bounded by the portion of the circle of radius $3$ and center $(0,0)$ that lies in the first quadrant, the portion of the circle with radius $\tfrac{3}{2}$ and center $(0,\tfrac{3}{2})$ that lies in the first quadrant, and the line segment from $(0,0)$ to $(3,0)$. What is the area of the shark's fin falcata? [asy] import cse5;pathpen=black;pointpen=black; size(1.5inch); D(MP("x",(3.5,0),S)--(0,0)--MP("\frac{3}{2}",(0,3/2),W)--MP("y",(0,3.5),W)); path P=(0,0)--MP("3",(3,0),S)..(3*dir(45))..MP("3",(0,3),W)--(0,3)..(3/2,3/2)..cycle; draw(P,linewidth(2)); fill(P,gray); [/asy] $\textbf{(A) } \dfrac{4\pi}{5} \qquad\textbf{(B) } \dfrac{9\pi}{8} \qquad\textbf{(C) } \dfrac{4\pi}{3} \qquad\textbf{(D) } \dfrac{7\pi}{5} \qquad\textbf{(E) } \dfrac{3\pi}{2} $

The diagram below shows four regular hexagons each with side length $1$ meter attached to the sides of a square. This figure is drawn onto a thin sheet of metal and cut out. The hexagons are then bent upward along the sides of the square so that $A_1$ meets $A_2$, $B_1$ meets $B_2$, $C_1$ meets $C_2$, and $D_1$ meets $D_2$. If the resulting dish is filled with water, the water will rise to the height of the corner where the $A_1$ and $A_2$ meet. there are relatively prime positive integers $m$ and $n$ so that the number of cubic meters of water the dish will hold is $\sqrt{\frac{m}{n}}$. Find $m+n$. [asy] /* File unicodetex not found. */ /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(7cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -4.3, xmax = 14.52, ymin = -8.3, ymax = 6.3; /* image dimensions */ draw((0,1)--(0,0)--(1,0)--(1,1)--cycle); draw((1,1)--(1,0)--(1.87,-0.5)--(2.73,0)--(2.73,1)--(1.87,1.5)--cycle); draw((0,1)--(1,1)--(1.5,1.87)--(1,2.73)--(0,2.73)--(-0.5,1.87)--cycle); draw((0,0)--(1,0)--(1.5,-0.87)--(1,-1.73)--(0,-1.73)--(-0.5,-0.87)--cycle); draw((0,1)--(0,0)--(-0.87,-0.5)--(-1.73,0)--(-1.73,1)--(-0.87,1.5)--cycle); /* draw figures */ draw((0,1)--(0,0)); draw((0,0)--(1,0)); draw((1,0)--(1,1)); draw((1,1)--(0,1)); draw((1,1)--(1,0)); draw((1,0)--(1.87,-0.5)); draw((1.87,-0.5)--(2.73,0)); draw((2.73,0)--(2.73,1)); draw((2.73,1)--(1.87,1.5)); draw((1.87,1.5)--(1,1)); draw((0,1)--(1,1)); draw((1,1)--(1.5,1.87)); draw((1.5,1.87)--(1,2.73)); draw((1,2.73)--(0,2.73)); draw((0,2.73)--(-0.5,1.87)); draw((-0.5,1.87)--(0,1)); /* dots and labels */ dot((1.87,-0.5),dotstyle); label("$C_1$", (1.72,-0.1), NE * labelscalefactor); dot((1.87,1.5),dotstyle); label("$B_2$", (1.76,1.04), NE * labelscalefactor); dot((1.5,1.87),dotstyle); label("$B_1$", (0.96,1.8), NE * labelscalefactor); dot((-0.5,1.87),dotstyle); label("$A_2$", (-0.26,1.78), NE * labelscalefactor); dot((-0.87,1.5),dotstyle); label("$A_1$", (-0.96,1.08), NE * labelscalefactor); dot((-0.87,-0.5),dotstyle); label("$D_2$", (-1.02,-0.18), NE * labelscalefactor); dot((-0.5,-0.87),dotstyle); label("$D_1$", (-0.22,-0.96), NE * labelscalefactor); dot((1.5,-0.87),dotstyle); label("$C_2$", (0.9,-0.94), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]

The figure $ABCD$ is bounded by a semicircle $CDA$ and a quarter circle $ABC$. Given that the distance from $A$ to $C$ is $18$, find the area of the figure. [asy] size(200); defaultpen(linewidth(0.8)); pair A=(-9,0),B=(0,9*sqrt(2)-9),C=(9,0),D=(0,9); dot(A^^B^^C^^D); draw(arc(origin,9,0,180)^^arc((0,-9),9*sqrt(2),45,135)); label("$A$",A,S); label("$B$",B,N); label("$C$",C,S); label("$D$",D,N); [/asy]

Point $F$ is taken on the extension of side $AD$ of parallelogram $ABCD$. $BF$ intersects diagonal $AC$ at $E$ and side $DC$ at $G$. If $EF = 32$ and $GF = 24$, then $BE$ equals: [asy] size(7cm); pair A = (0, 0), B = (7, 0), C = (10, 5), D = (3, 5), F = (5.7, 9.5); pair G = intersectionpoints(B--F, D--C)[0]; pair E = intersectionpoints(A--C, B--F)[0]; draw(A--D--C--B--cycle); draw(A--C); draw(D--F--B); label("$A$", A, SW); label("$B$", B, SE); label("$C$", C, NE); label("$D$", D, NW); label("$F$", F, N); label("$G$", G, NE); label("$E$", E, SE); //Credit to MSTang for the asymptote [/asy] $\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 8\qquad \textbf{(C)}\ 10 \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 16$

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] $\textbf{(A)}\ 12 \qquad \textbf{(B)}\ 14 \qquad \textbf{(C)}\ 15 \qquad \textbf{(D)}\ 16 \qquad \textbf{(E)}\ 20$

In $\triangle ABC$, $AB<AC$. $D$ and $P$ are the feet of the internal and external angle bisectors of $\angle BAC$, respectively. $M$ is the midpoint of segment $BC$, and $\omega$ is the circumcircle of $\triangle APD$. Suppose $Q$ is on the minor arc $AD$ of $\omega$ such that $MQ$ is tangent to $\omega$. $QB$ meets $\omega$ again at $R$, and the line through $R$ perpendicular to $BC$ meets $PQ$ at $S$. Prove $SD$ is tangent to the circumcircle of $\triangle QDM$. [i]Proposed by Ray Li[/i]

Let $O$ be the center of the circumcircle $\omega$ of an acute-angle triangle $ABC$. A circle $\omega_1$ with center $K$ passes through $A$, $O$, $C$ and intersects $AB$ at $M$ and $BC$ at $N$. Point $L$ is symmetric to $K$ with respect to line $NM$. Prove that $BL \perp AC$.

Let $ABC$ be a right angled triangle at $A$. Denote $D$ the foot of the altitude through $A$ and $O_1, O_2$ the incentres of triangles $ADB$ and $ADC$. The circle with centre $A$ and radius $AD$ cuts $AB$ in $K$ and $AC$ in $L$. Show that $O_1, O_2, K$ and $L$ are on a line.

Let be a twice-differentiable function $ f:(0,\infty )\longrightarrow\mathbb{R} $ that admits a polynomial function of degree $ 1 $ or $ 2, $ namely, $ \alpha :(0,\infty )\longrightarrow\mathbb{R} $ as its asymptote. Prove the following propositions: [b]a)[/b] $ f''>0\implies f-\alpha >0 $ [b]b)[/b] $ \text{supp} f''=(0,\infty )\wedge f-\alpha >0\implies f''=0 $

In $\bigtriangleup ABC$, $D$ is a point on side $\overline{AC}$ such that $BD=DC$ and $\angle BCD$ measures $70^\circ$. What is the degree measure of $\angle ADB$? [asy] size(300); defaultpen(linewidth(0.8)); pair A=(-1,0),C=(1,0),B=dir(40),D=origin; draw(A--B--C--A); draw(D--B); dot("$A$", A, SW); dot("$B$", B, NE); dot("$C$", C, SE); dot("$D$", D, S); label("$70^\circ$",C,2*dir(180-35)); [/asy] $\textbf{(A) }100\qquad\textbf{(B) }120\qquad\textbf{(C) }135\qquad\textbf{(D) }140\qquad \textbf{(E) }150$

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] $\textbf{(A)}\ EFA \qquad \textbf{(B)}\ EFC \qquad \textbf{(C)}\ ABM \qquad \textbf{(D)}\ ABU \qquad \textbf{(E)}\ FMC$