A triangle with side lengths in the ratio $ 3: 4: 5$ is inscribed in a circle of radius $ 3.$ What is the area of the triangle? $ \textbf{(A)}\ 8.64 \qquad \textbf{(B)}\ 12 \qquad \textbf{(C)}\ 5\pi \qquad \textbf{(D)}\ 17.28 \qquad \textbf{(E)}\ 18$

Three circles, with radii $p$, $q$ and $r$ and centres $A$, $B$ and $C$ respectively, touch one another externally at points $D$, $E$ and $F$. Prove that the ratio of the areas of $\triangle DEF$ and $\triangle ABC$ equals \[\frac{2pqr}{(p+q)(q+r)(r+p)}.\]

[asy] draw(unitsquare);draw((0,0)--(.4,1)^^(0,.6)--(1,.2)); label("D",(0,1),NW);label("E",(.4,1),N);label("C",(1,1),NE); label("P",(0,.6),W);label("M",(.25,.55),E);label("Q",(1,.2),E); label("A",(0,0),SW);label("B",(1,0),SE); //Credit to Zimbalono for the diagram[/asy] Inside square $ABCD$ (See figure) with sides of length $12$ inches, segment $AE$ is drawn where $E$ is the point on $DC$ which is $5$ inches from $D$. The perpendicular bisector of $AE$ is drawn and intersects $AE$, $AD$, and $BC$ at points $M$, $P$, and $Q$ respectively. The ratio of segment $PM$ to $MQ$ is $\textbf{(A) }5:12\qquad\textbf{(B) }5:13\qquad\textbf{(C) }5:19\qquad\textbf{(D) }1:4\qquad \textbf{(E) }5:21$

A regular octahedron is formed by joining the centers of adjoining faces of a cube. The ratio of the volume of the octahedron to the volume of the cube is $ \textbf{(A)}\ \frac{\sqrt{3}}{12} \qquad\textbf{(B)}\ \frac{\sqrt{6}}{16} \qquad\textbf{(C)}\ \frac{1}{6} \qquad\textbf{(D)}\ \frac{\sqrt{2}}{8} \qquad\textbf{(E)}\ \frac{1}{4} $

Given is a triangle $ABC$ and points $D$ and $E$, respectively on $] BC [$ and $] AB [$. $F$ it is intersection of lines $AD$ and $CE$. We denote as $| CD | = a, | BD | = b, | DF | = c$ and $| AF | = d$. Determine the ratio $\frac{| BE |}{|AE |}$ in terms of $a, b, c$ and $d$ [img]https://cdn.artofproblemsolving.com/attachments/5/7/856c97045db2d9a26841ad00996a2b809addaa.png[/img]

In a rectangular plot of land, a man walks in a very peculiar fashion. Labeling the corners $ABCD$, he starts at $A$ and walks to $C$. Then, he walks to the midpoint of side $AD$, say $A_1$. Then, he walks to the midpoint of side $CD$ say $C_1$, and then the midpoint of $A_1D$ which is $A_2$. He continues in this fashion, indefinitely. The total length of his path if $AB=5$ and $BC=12$ is of the form $a + b\sqrt{c}$. Find $\displaystyle\frac{abc}{4}$.

$ \Delta ABC$ is a triangle such that $ AB \neq AC$. The incircle of $ \Delta ABC$ touches $ BC, CA, AB$ at $ D, E, F$ respectively. $ H$ is a point on the segment $ EF$ such that $ DH \bot EF$. Suppose $ AH \bot BC$, prove that $ H$ is the orthocentre of $ \Delta ABC$. Remark: the original question has missed the condition $ AB \neq AC$

Let $P(x)$ be a polynomial of degree $n \le 10$ with integral coefficients such that for every $k \in \{1, 2, \dots, 10\}$ there is an integer $m$ with $P(m) = k$. Furthermore, it is given that $|P(10) - P(0)| < 1000$. Prove that for every integer $k$ there is an integer $m$ such that $P(m) = k.$

In triangle $ABC$, the medians and bisectors corresponding to sides $BC$, $CA$, $AB$ are $m_a$, $m_b$, $m_c$ and $w_a$, $w_b$, $w_c$ respectively. $P=w_a \cap m_b$, $Q=w_b \cap m_c$, $R=w_c \cap m_a$. Denote the areas of triangle $ABC$ and $PQR$ by $F_1$ and $F_2$ respectively. Find the least positive constant $m$ such that $\frac{F_1}{F_2}<m$ holds for any $\triangle{ABC}$.

Points $A$, $B$, $C$, and $D$ lie on a circle such that chords $\overline{AC}$ and $\overline{BD}$ intersect at a point $E$ inside the circle. Suppose that $\angle ADE =\angle CBE = 75^\circ$, $BE=4$, and $DE=8$. The value of $AB^2$ can be written in the form $a+b\sqrt{c}$ for positive integers $a$, $b$, and $c$ such that $c$ is not divisible by the square of any prime. Find $a+b+c$. [i]Proposed by Tony Kim[/i]

In a geometric sequence of real numbers, the sum of the first two terms is 7, and the sum of the first 6 terms is 91. The sum of the first 4 terms is $\text{(A)}\ 28 \qquad \text{(B)}\ 32 \qquad \text{(C)}\ 35 \qquad \text{(D)}\ 49 \qquad \text{(E)}\ 84$

Larry and Julius are playing a game, taking turns throwing a ball at a bottle sitting on a ledge. Larry throws first. The winner is the first person to knock the bottle off the ledge. At each turn the probability that a player knocks the bottle off the ledge is $\frac{1}{2}$, independently of what has happened before. What is the probability that Larry wins the game? $\textbf{(A) }\frac{1}{2}\qquad\textbf{(B) }\frac{3}{5}\qquad\textbf{(C) }\frac{2}{3}\qquad\textbf{(D) }\frac{3}{4}\qquad\textbf{(E) }\frac{4}{5}$

A circle of radius 2 is cut into four congruent arcs. The four arcs are joined to form the star figure shown. What is the ratio of the area of the star figure to the area of the original circle? [asy] size(0,50); draw((-1,1)..(-2,2)..(-3,1)..(-2,0)..cycle); dot((-1,1)); dot((-2,2)); dot((-3,1)); dot((-2,0)); draw((1,0){up}..{left}(0,1)); dot((1,0)); dot((0,1)); draw((0,1){right}..{up}(1,2)); dot((1,2)); draw((1,2){down}..{right}(2,1)); dot((2,1)); draw((2,1){left}..{down}(1,0)); [/asy] $\textbf{(A)}\hspace{.05in}\dfrac{4-\pi}\pi \qquad \textbf{(B)}\hspace{.05in}\dfrac1\pi \qquad \textbf{(C)}\hspace{.05in}\dfrac{\sqrt2}{\pi} \qquad \textbf{(D)}\hspace{.05in}\dfrac{\pi-1}\pi \qquad \textbf{(E)}\hspace{.05in}\dfrac3\pi $

The diagram shows an octagon consisting of $10$ unit squares. The portion below $\overline{PQ}$ is a unit square and a triangle with base $5$. If $\overline{PQ}$ bisects the area of the octagon, what is the ratio $\frac{XQ}{QY}$? [asy] import graph; size(300); real lsf = 0.5; pen dp = linewidth(0.7) + fontsize(10); defaultpen(dp); pen ds = black; pen xdxdff = rgb(0.49,0.49,1); draw((0,0)--(6,0),linewidth(1.2pt)); draw((0,0)--(0,1),linewidth(1.2pt)); draw((0,1)--(1,1),linewidth(1.2pt)); draw((1,1)--(1,2),linewidth(1.2pt)); draw((1,2)--(5,2),linewidth(1.2pt)); draw((5,2)--(5,1),linewidth(1.2pt)); draw((5,1)--(6,1),linewidth(1.2pt)); draw((6,1)--(6,0),linewidth(1.2pt));draw((1,1)--(5,1),linewidth(1.2pt)); draw((1,1)--(1,0),linewidth(1.2pt));draw((2,2)--(2,0),linewidth(1.2pt)); draw((3,2)--(3,0),linewidth(1.2pt)); draw((4,2)--(4,0),linewidth(1.2pt)); draw((5,1)--(5,0),linewidth(1.2pt)); draw((0,0)--(5,1.5),linewidth(1.2pt)); dot((0,0),ds); label("$P$", (-0.23,-0.26),NE*lsf); dot((0,1),ds); dot((1,1),ds); dot((1,2),ds); dot((5,2),ds); label("$X$", (5.14,2.02),NE*lsf); dot((5,1),ds); label("$Y$", (5.12,1.14),NE*lsf); dot((6,1),ds); dot((6,0),ds); dot((1,0),ds); dot((2,0),ds); dot((3,0),ds); dot((4,0),ds); dot((5,0),ds); dot((2,2),ds); dot((3,2),ds); dot((4,2),ds); dot((5,1.5),ds); label("$Q$", (5.14,1.51),NE*lsf); clip((-4.19,-5.52)--(-4.19,6.5)--(10.08,6.5)--(10.08,-5.52)--cycle); [/asy] $\textbf{(A)}\ \frac 25 \qquad \textbf{(B)}\ \frac 12 \qquad \textbf{(C)}\ \frac 35 \qquad \textbf{(D)}\ \frac 23 \qquad \textbf{(E)}\ \frac 34$

A square is drawn inside a rectangle. The ratio of the width of the rectangle to a side of the square is $ 2: 1$. The ratio of the rectangle's length to its width is $ 2: 1$. What percent of the rectangle's area is inside the square? $ \textbf{(A)}\ 12.5 \qquad \textbf{(B)}\ 25 \qquad \textbf{(C)}\ 50 \qquad \textbf{(D)}\ 75 \qquad \textbf{(E)}\ 87.5$

Let $k$ and $a$ are positive constants. Denote by $V_1$ the volume of the solid generated by a rotation of the figure enclosed by the curve $C: y=\frac{x}{x+k}\ (x\geq 0)$, the line $x=a$ and the $x$-axis around the $x$-axis, and denote by $V_2$ that of the solid by a rotation of the figure enclosed by the curve $C$, the line $y=\frac{a}{a+k}$ and the $y$-axis around the $y$-axis. Find the ratio $\frac{V_2}{V_1}.$

Two geometric sequences $ a_1,a_2,a_3,\ldots$ and $b_1,b_2,b_3\ldots $have the same common ratio, with $a_1=27$,$b_1=99$, and $a_{15}=b_{11}$. Find $a_9.$

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 $\widehat{BAD}=\widehat{CAX}$

Let $ O$ be the circumcircle of a $ \Delta ABC$ and let $ I$ be its incenter, for a point $ P$ of the plane let $ f(P)$ be the point obtained by reflecting $ P'$ by the midpoint of $ OI$, with $ P'$ the homothety of $ P$ with center $ O$ and ratio $ \frac{R}{r}$ with $ r$ the inradii and $ R$ the circumradii,(understand it by $ \frac{OP}{OP'}\equal{}\frac{R}{r}$). Let $ A_1$, $ B_1$ and $ C_1$ the midpoints of $ BC$, $ AC$ and $ AB$, respectively. Show that the rays $ A_1f(A)$, $ B_1f(B)$ and $ C_1f(C)$ concur on the incircle.

Consider a cylinder and a cone with a common base such that the volume of the part of the cylinder enclosed in the cone equals the volume of the part of the cylinder outside the cone. Find the ratio of the height of the cone to the height of the cylinder.

Let $ABCD$ be a rectangle and $E$ the reflection of $A$ with respect to the diagonal $BD$. If $EB = EC$, what is the ratio $\frac{AD}{AB}$ ?

Let $ABCD$ be a quadrilateral with $\angle A + \angle C = 60^o$. If $AB \cdot CD = BC \cdot AD$, prove that $AB \cdot CD = AC \cdot BD$. Leonard Giugiuc

Two right circular cylinders have the same volume. The radius of the second cylinder is $10\%$ more than the radius of the first. What is the relationship between the heights of the two cylinders? $\textbf{(A) }\text{The second height is 10\% less than the first.}$ $\textbf{(B) }\text{The first height is 10\% more than the second.}$ $\textbf{(C) }\text{The second height is 21\% less than the first.}$ $\textbf{(D) }\text{The first height is 21\% more than the second.}$ $\textbf{(E) }\text{The second height is 80\% of the first.}$

Triangle $ABC$ is given with its centroid $G$ and cicumcentre $O$ is such that $GO$ is perpendicular to $AG$. Let $A'$ be the second intersection of $AG$ with circumcircle of triangle $ABC$. Let $D$ be the intersection of lines $CA'$ and $AB$ and $E$ the intersection of lines $BA'$ and $AC$. Prove that the circumcentre of triangle $ADE$ is on the circumcircle of triangle $ABC$.

In a triangle $ABC$, the angle bisector at $A,B,C$ meet the opposite sides at $A_1,B_1,C_1$, respectively. Prove that if the quadrilateral $BA_1B_1C_1$ is cyclic, then $$\frac{AC}{AB+BC}=\frac{AB}{AC+CB}+\frac{BC}{BA+AC}.$$