In triangle $ABC$, $AB=13$, $BC=14$, and $CA=15$. Distinct points $D$, $E$, and $F$ lie on segments $\overline{BC}$, $\overline{CA}$, and $\overline{DE}$, respectively, such that $\overline{AD}\perp\overline{BC}$, $\overline{DE}\perp\overline{AC}$, and $\overline{AF}\perp\overline{BF}$. The length of segment $\overline{DF}$ can be written as $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. What is $m+n$? ${ \textbf{(A)}\ 18\qquad\textbf{(B)}\ 21\qquad\textbf{(C)}\ 24\qquad\textbf{(D}}\ 27\qquad\textbf{(E)}\ 30 $

Say that a (nondegenerate) triangle is [i]funny[/i] if it satisfies the following condition: the altitude, median, and angle bisector drawn from one of the vertices divide the triangle into 4 non-overlapping triangles whose areas form (in some order) a 4-term arithmetic sequence. (One of these 4 triangles is allowed to be degenerate.) Find with proof all funny triangles.

The semicircle with centre $O$ and the diameter $AC$ is divided in two arcs $AB$ and $BC$ with ratio $1: 3$. $M$ is the midpoint of the radium $OC$. Let $T$ be the point of arc $BC$ such that the area of the cuadrylateral $OBTM$ is maximum. Find such area in fuction of the radium.

The circular base of a hemisphere of radius $2$ rests on the base of a square pyramid of height $6$. The hemisphere is tangent to the other four faces of the pyramid. What is the edge-length of the base of the pyramid? $ \textbf{(A)}\ 3\sqrt{2} \qquad \textbf{(B)}\ \frac{13}{3} \qquad \textbf{(C)}\ 4\sqrt{2} \qquad \textbf{(D)}\ 6 \qquad \textbf{(E)}\ \frac{13}{2} $

The incircle of a non-isosceles triangle $ABC$ with the center $I$ touches the sides $BC$ at $D$. let $X$ is a point on arc $BC$ from circumcircle of triangle $ABC$ such that if $E,F$ are feet of perpendicular from $X$ on $BI,CI$ and $M$ is midpoint of $EF$ we have $MB=MC$. prove that $\widehat{BAD}=\widehat{CAX}$

The number $\frac{1}{2}$ is written on a blackboard. For a real number $c$ with $0 < c < 1$, a [i]$c$-splay[/i] is an operation in which every number $x$ on the board is erased and replaced by the two numbers $cx$ and $1-c(1-x)$. A [i]splay-sequence[/i] $C = (c_1,c_2,c_3,c_4)$ is an application of a $c_i$-splay for $i=1,2,3,4$ in that order, and its [i]power[/i] is defined by $P(C) = c_1c_2c_3c_4$. Let $S$ be the set of splay-sequences which yield the numbers $\frac{1}{17}, \frac{2}{17}, \dots, \frac{16}{17}$ on the blackboard in some order. If $\sum_{C \in S} P(C) = \tfrac mn$ for relatively prime positive integers $m$ and $n$, compute $100m+n$. [i]Proposed by Lewis Chen[/i]

Given triangle $ ABC$ with sidelengths $ a,b,c$. Tangents to incircle of $ ABC$ that parallel with triangle's sides form three small triangle (each small triangle has 1 vertex of $ ABC$). Prove that the sum of area of incircles of these three small triangles and the area of incircle of triangle $ ABC$ is equal to $ \frac{\pi (a^{2}\plus{}b^{2}\plus{}c^{2})(b\plus{}c\minus{}a)(c\plus{}a\minus{}b)(a\plus{}b\minus{}c)}{(a\plus{}b\plus{}c)^{3}}$ (hmm,, looks familiar, isn't it? :wink: )

Let $\Gamma$ and $\Gamma_1$ be two circles internally tangent at $A$, with centers $O$ and $O_1$ and radii $r$ and $r_1$, respectively ($r>r_1$). $B$ is a point diametrically opposed to $A$ in $\Gamma$, and $C$ is a point on $\Gamma$ such that $BC$ is tangent to $\Gamma_1$ at $P$. Let $A'$ the midpoint of $BC$. Given that $O_1A'$ is parallel to $AP$, find the ratio $r/r_1$.

The geometric series $ a \plus{} ar \plus{} ar^{2} \plus{} ...$ has a sum of $ 7$, and the terms involving odd powers of $ r$ have a sum of $ 3$. What is $ a \plus{} r$? $ \textbf{(A)}\ \frac {4}{3}\qquad \textbf{(B)}\ \frac {12}{7}\qquad \textbf{(C)}\ \frac {3}{2}\qquad \textbf{(D)}\ \frac {7}{3}\qquad \textbf{(E)}\ \frac {5}{2}$

In the figure, what is the ratio of the area of the gray squares to the area of the white squares? [asy] size((70)); draw((10,0)--(0,10)--(-10,0)--(0,-10)--(10,0)); draw((-2.5,-7.5)--(7.5,2.5)); draw((-5,-5)--(5,5)); draw((-7.5,-2.5)--(2.5,7.5)); draw((-7.5,2.5)--(2.5,-7.5)); draw((-5,5)--(5,-5)); draw((-2.5,7.5)--(7.5,-2.5)); fill((-10,0)--(-7.5,2.5)--(-5,0)--(-7.5,-2.5)--cycle, gray); fill((-5,0)--(0,5)--(5,0)--(0,-5)--cycle, gray); fill((5,0)--(7.5,2.5)--(10,0)--(7.5,-2.5)--cycle, gray); [/asy] $ \textbf{(A)}\ 3:10 \qquad\textbf{(B)}\ 3:8 \qquad\textbf{(C)}\ 3:7 \qquad\textbf{(D)}\ 3:5 \qquad\textbf{(E)}\ 1:1 $

Let $c\neq 1$ be a positive rational number. Show that it is possible to partition $\mathbb{N}$, the set of positive integers, into two disjoint nonempty subsets $A,B$ so that $x/y\neq c$ holds whenever $x$ and $y$ lie both in $A$ or both in $B$.

Given a triangle $ABC$, let $D$ be the midpoint of the side $AC$ and let $M$ be the point that divides the segment $BD$ in the ratio $1/2$; that is, $MB/MD=1/2$. The rays $AM$ and $CM$ meet the sides $BC$ and $AB$ at points $E$ and $F$, respectively. Assume the two rays perpendicular: $AM\perp CM$. Show that the quadrangle $AFED$ is cyclic if and only if the median from $A$ in triangle $ABC$ meets the line $EF$ at a point situated on the circle $ABC$.

A paint brush is swept along both diagonals of a square to produce the symmetric painted area, as shown. Half the area of the square is painted. What is the ratio of the side length of the square to the brush width? [asy]unitsize(15mm); defaultpen(linewidth(.8pt)); path P=(-sqrt(2)/2,1)--(0,1-sqrt(2)/2)--(sqrt(2)/2,1); path Pc=(-sqrt(2)/2,1)--(0,1-sqrt(2)/2)--(sqrt(2)/2,1)--cycle; path S=(-1,-1)--(-1,1)--(1,1)--(1,-1)--cycle; fill(S,gray); for(int i=0;i<4;++i) { fill(rotate(90*i)*Pc,white); draw(rotate(90*i)*P); } draw(S);[/asy]$ \textbf{(A)}\ 2\sqrt {2} \plus{} 1 \qquad \textbf{(B)}\ 3\sqrt {2}\qquad \textbf{(C)}\ 2\sqrt {2} \plus{} 2 \qquad \textbf{(D)}\ 3\sqrt {2} \plus{} 1 \qquad \textbf{(E)}\ 3\sqrt {2} \plus{} 2$

Consider a triangle $ABC$, inscribed in $O$ and $\angle A < \angle B$. Some point $P$ outside the circle satisfies $$\angle A=\angle PBA =180^{\circ}- \angle PCB$$ Let $D$ be the intersection of line $PB$ and $O$(different from $B$), and $Q$ the intersection of the tangent line of $O$ passing through $A$ and line $CD$. Show that $CQ : AB=AQ^2:AD^2$.

In $\Delta ABC$, $AC=kAB$, with $k>1$. The internal angle bisector of $\angle BAC$ meets $BC$ at $D$. The circle with $AC$ as diameter cuts the extension of $AD$ at $E$. Express $\dfrac{AD}{AE}$ in terms of $k$.

Is it possible to divide $5$ apples of the same size equally between six children so that no apple will be cut into more than $3$ pieces? (You are allowed to cut an apple into any number of equal pieces).

In the figure, $\triangle ABC$ has $\angle A =45^{\circ}$ and $\angle B =30^{\circ}$. A line $DE$, with $D$ on $AB$ and $\angle ADE =60^{\circ}$, divides $\triangle ABC$ into two pieces of equal area. (Note: the figure may not be accurate; perhaps $E$ is on $CB$ instead of $AC$.) The ratio $\frac{AD}{AB}$ is [asy] size((220)); draw((0,0)--(20,0)--(7,6)--cycle); draw((6,6)--(10,-1)); label("A", (0,0), W); label("B", (20,0), E); label("C", (7,6), NE); label("D", (9.5,-1), W); label("E", (5.9, 6.1), SW); label("$45^{\circ}$", (2.5,.5)); label("$60^{\circ}$", (7.8,.5)); label("$30^{\circ}$", (16.5,.5)); [/asy] $ \textbf{(A)}\ \frac{1}{\sqrt{2}} \qquad\textbf{(B)}\ \frac{2}{2+\sqrt{2}} \qquad\textbf{(C)}\ \frac{1}{\sqrt{3}} \qquad\textbf{(D)}\ \frac{1}{\sqrt[3]{6}} \qquad\textbf{(E)}\ \frac{1}{\sqrt[4]{12}} $

The bisector of the interior angle at the vertex $B$ of the triangle $ABC$ and the perpendicular line on side $BC$ passing through the vertex $C$ intersects at $D$. Let $M$ and $N$ be the midpoints of the segments $BC$ and $BD$, respectively, with $N$ on the side $AC$. Find all possibilities of the angles of the triangles $ABC$, if it is known that $\frac{| AM |}{| BC |}=\frac{|CD|}{|BD|}$. .

In the adjoining figure triangle $ ABC$ is inscribed in a circle. Point $ D$ lies on $ \stackrel{\frown}{AC}$ with $ \stackrel{\frown}{DC} \equal{} 30^\circ$, and point $ G$ lies on $ \stackrel{\frown}{BA}$ with $ \stackrel{\frown}{BG}\, > \, \stackrel{\frown}{GA}$. Side $ AB$ and side $ AC$ each have length equal to the length of chord $ DG$, and $ \angle CAB \equal{} 30^\circ$. Chord $ DG$ intersects sides $ AC$ and $ AB$ at $ E$ and $ F$, respectively. The ratio of the area of $ \triangle AFE$ to the area of $ \triangle ABC$ is [asy] size(200); defaultpen(linewidth(.8pt)); pair C = origin; pair A = 2.5*dir(75); pair B = A + 2.5*dir(-75); path circ =circumcircle(A,B,C); pair D = waypoint(circ,(7/12)); pair G = waypoint(circ,(1/6)); pair E = intersectionpoint(D--G,A--C); pair F = intersectionpoint(A--B,D--G); label("$A$",A,N); label("$B$",B,SE); label("$C$",C,SW); label("$D$",D,SW); label("$G$",G,NE); label("$E$",E,NW); label("$F$",F,W); label("$30^\circ$",A,12S+E,fontsize(6pt)); draw(A--B--C--cycle); draw(circ); draw(Arc(A,0.25,-75,-105)); draw(D--G);[/asy]$ \textbf{(A)}\ \frac {2 \minus{} \sqrt {3}}{3}\qquad \textbf{(B)}\ \frac {2\sqrt {3} \minus{} 3}{3}\qquad \textbf{(C)}\ 7\sqrt {3} \minus{} 12\qquad \textbf{(D)}\ 3\sqrt {3} \minus{} 5\qquad$ $ \textbf{(E)}\ \frac {9 \minus{} 5\sqrt {3}}{3}$

In the diagram below $ \angle CAB, \angle CBD$, and $\angle CDE$ are all right angles with side lengths $AC = 3$, $BC = 5$, $BD = 12$, and $DE = 84$. The distance from point $E$ to the line $AB$ can be expressed as the ratio of two relatively prime positive integers, $m$ and $n$. Find $m + n$. [asy] size(300); defaultpen(linewidth(0.8)); draw(origin--(3,0)--(0,4)--cycle^^(0,4)--(6,8)--(3,0)--(30,-4)--(6,8)); label("$A$",origin,SW); label("$B$",(0,4),dir(160)); label("$C$",(3,0),S); label("$D$",(6,8),dir(80)); label("$E$",(30,-4),E);[/asy]

Older television screens have an aspect ratio of $ 4: 3$. That is, the ratio of the width to the height is $ 4: 3$. The aspect ratio of many movies is not $ 4: 3$, so they are sometimes shown on a television screen by 'letterboxing' - darkening strips of equal height at the top and bottom of the screen, as shown. Suppose a movie has an aspect ratio of $ 2: 1$ and is shown on an older television screen with a $ 27$-inch diagonal. What is the height, in inches, of each darkened strip? [asy]unitsize(1mm); defaultpen(linewidth(.8pt)); filldraw((0,0)--(21.6,0)--(21.6,2.7)--(0,2.7)--cycle,grey,black); filldraw((0,13.5)--(21.6,13.5)--(21.6,16.2)--(0,16.2)--cycle,grey,black); draw((0,2.7)--(0,13.5)); draw((21.6,2.7)--(21.6,13.5));[/asy]$ \textbf{(A)}\ 2 \qquad \textbf{(B)}\ 2.25 \qquad \textbf{(C)}\ 2.5 \qquad \textbf{(D)}\ 2.7 \qquad \textbf{(E)}\ 3$

A sphere is inscribed in a truncated right circular cone as shown. The volume of the truncated cone is twice that of the sphere. What is the ratio of the radius of the bottom base of the truncated cone to the radius of the top base of the truncated cone? [asy] real r=(3+sqrt(5))/2; real s=sqrt(r); real Brad=r; real brad=1; real Fht = 2*s; import graph3; import solids; currentprojection=orthographic(1,0,.2); currentlight=(10,10,5); revolution sph=sphere((0,0,Fht/2),Fht/2); //draw(surface(sph),green+white+opacity(0.5)); //triple f(pair t) {return (t.x*cos(t.y),t.x*sin(t.y),t.x^(1/n)*sin(t.y/n));} triple f(pair t) { triple v0 = Brad*(cos(t.x),sin(t.x),0); triple v1 = brad*(cos(t.x),sin(t.x),0)+(0,0,Fht); return (v0 + t.y*(v1-v0)); } triple g(pair t) { return (t.y*cos(t.x),t.y*sin(t.x),0); } surface sback=surface(f,(3pi/4,0),(7pi/4,1),80,2); surface sfront=surface(f,(7pi/4,0),(11pi/4,1),80,2); surface base = surface(g,(0,0),(2pi,Brad),80,2); draw(sback,rgb(0,1,0)); draw(sfront,rgb(.3,1,.3)); draw(base,rgb(.4,1,.4)); draw(surface(sph),rgb(.3,1,.3)); [/asy] $ \textbf {(A) } \dfrac {3}{2} \qquad \textbf {(B) } \dfrac {1+\sqrt{5}}{2} \qquad \textbf {(C) } \sqrt{3} \qquad \textbf {(D) } 2 \qquad \textbf {(E) } \dfrac {3+\sqrt{5}}{2} $

What is the smallest possible volume to surface ratio of a solid cone with height = $1$ unit?

In triangle $ABC$, $AB=13,$ $BC=15$ and $CA=17.$ Point $D$ is on $\overline{AB},$ $E$ is on $\overline{BC},$ and $F$ is on $\overline{CA}.$ Let $AD=p\cdot AB,$ $BE=q\cdot BC,$ and $CF=r\cdot CA,$ where $p,$ $q,$ and $r$ are positive and satisfy $p+q+r=2/3$ and $p^2+q^2+r^2=2/5.$ The ratio of the area of triangle $DEF$ to the area of triangle $ABC$ can be written in the form $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$

The vertices $D,E,F$ of an equilateral triangle lie on the sides $BC,CA,AB$ respectively of a triangle $ABC.$ If $a,b,c$ are the respective lengths of these sides, and $S$ the area of $ABC,$ prove that \[ DE \geq \frac{2 \cdot \sqrt{2} \cdot S}{\sqrt{a^2 + b^2 + c^2 + 4 \cdot \sqrt{3} \cdot S}}. \]