$\triangle ABC$ has area $240$. Points $X, Y, Z$ lie on sides $AB$, $BC$, and $CA$, respectively. Given that $\frac{AX}{BX} = 3$, $\frac{BY}{CY} = 4$, and $\frac{CZ}{AZ} = 5$, find the area of $\triangle XYZ$. [asy] size(175); defaultpen(linewidth(0.8)); pair A=(0,15),B=(0,-5),C=(25,0.5),X=origin,Y=(4C+B)/5,Z=(5A+C)/6; draw(A--B--C--cycle^^X--Y--Z--cycle); label("$A$",A,N); label("$B$",B,S); label("$C$",C,E); label("$X$",X,W); label("$Y$",Y,S); label("$Z$",Z,NE);[/asy]

Three clever monkeys divide a pile of bananas. The first monkey takes some bananas from the pile, keeps three-fourths of them, and divides the rest equally between the other two. The second monkey takes some bananas from the pile, keeps one-fourth of them, and divides the rest equally between the other two. The third monkey takes the remaining bananas from the pile, keeps one-twelfth of them, and divides the rest equally between the other two. Given that each monkey receives a whole number of bananas whenever the bananas are divided, and the numbers of bananas the first, second, and third monkeys have at the end of the process are in the ratio $3: 2: 1$, what is the least possible total for the number of bananas?

Let $ r$ be the distance from the origion to a point $ P$ with coordinates $ x$ and $ y$. Designate the ratio $ \frac{y}{r}$ by $ s$ and the ratio $ \frac{x}{r}$ by $ c$. Then the values of $ s^2 \minus{} c^2$ are limited to the numbers: $ \textbf{(A)}\ \text{less than }{\minus{}1}\text{ are greater than }{\plus{}1}\text{, both excluded}\qquad\\ \textbf{(B)}\ \text{less than }{\minus{}1}\text{ are greater than }{\plus{}1}\text{, both included}\qquad \\ \textbf{(C)}\ \text{between }{\minus{}1}\text{ and }{\plus{}1}\text{, both excluded}\qquad \\ \textbf{(D)}\ \text{between }{\minus{}1}\text{ and }{\plus{}1}\text{, both included}\qquad \\ \textbf{(E)}\ {\minus{}1}\text{ and }{\plus{}1}\text{ only}$

Let $ABCD$ be a convex quadrilateral whose vertices do not lie on a circle. Let $A'B'C'D'$ be a quadrangle such that $A',B', C',D'$ are the centers of the circumcircles of triangles $BCD,ACD,ABD$, and $ABC$. We write $T (ABCD) = A'B'C'D'$. Let us define $A''B''C''D'' = T (A'B'C'D') = T (T (ABCD)).$ [b](a)[/b] Prove that $ABCD$ and $A''B''C''D''$ are similar. [b](b) [/b]The ratio of similitude depends on the size of the angles of $ABCD$. Determine this ratio.

Construct the $ \triangle ABC$, given $ h_a$, $ h_b$ (the altitudes from $ A$ and $ B$) and $ m_a$, the median from the vertex $ A$.

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 $k$.

Let $ ABC$ be an equilateral triangle. Let $ \Omega$ be its incircle (circle inscribed in the triangle) and let $ \omega$ be a circle tangent externally to $ \Omega$ as well as to sides $ AB$ and $ AC$. Determine the ratio of the radius of $ \Omega$ to the radius of $ \omega$.

A high school basketball game between the Raiders and Wildcats was tied at the end of the first quarter. The number of points scored by the Raiders in each of the four quarters formed an increasing geometric sequence, and the number of points scored by the Wildcats in each of the four quarters formed an increasing arithmetic sequence. At the end of the fourth quarter, the Raiders had won by one point. Neither team scored more than $ 100$ points. What was the total number of points scored by the two teams in the first half? $ \textbf{(A)}\ 30 \qquad \textbf{(B)}\ 31 \qquad \textbf{(C)}\ 32 \qquad \textbf{(D)}\ 33 \qquad \textbf{(E)}\ 34$

Let $ABC$ be a triangle and let $M,N,P$ be points on the line $BC$ such that $AM,AN,AP$ are the altitude, the angle bisector and the median of the triangle, respectively. It is known that $\frac{[AMP]}{[ABC]}=\frac{1}{4}$ and $\frac{[ANP]}{[ABC]}=1-\frac{\sqrt{3}}{2}$. Find the angles of triangle $ABC$.

In the triangle $ABC$, let $AM$ and $CN$ internal bisectors, with $M$ in $BC$ and $N$ in $AB$. Prove that if $$\frac{\angle BNM}{\angle MNC}=\frac{\angle BMN}{\angle NMA}$$ then $ABC$ is isosceles.

Let $ ABC$ a triangle and $ X$, $ Y$ and $ Z$ points at the segments $ BC$, $ AC$ and $ AB$, respectively.Let $ A'$, $ B'$ and $ C'$ the circuncenters of triangles $ AZY$,$ BXZ$,$ CYX$, respectively.Prove that $ 4(A'B'C')\geq(ABC)$ with equality if and only if $ AA'$, $ BB'$ and $ CC'$ are concurrents. Note: $ (XYZ)$ denotes the area of $ XYZ$

Let m be a positive integer. Prove that there exist infinitely many prime numbers p such that m+p^3 is composite.

Let $O$ be the circumcenter of the triangle $ABC$. The segment $XY$ is the diameter of the circumcircle perpendicular to $BC$ and it meets $BC$ at $M$. The point $X$ is closer to $M$ than $Y$ and $Z$ is the point on $MY$ such that $MZ = MX$. The point $W$ is the midpoint of $AZ$. a) Show that $W$ lies on the circle through the midpoints of the sides of $ABC$; b) Show that $MW$ is perpendicular to $AY$.

In triangle $ABC$ shown in the adjoining figure, $M$ is the midpoint of side $BC$, $AB=12$ and $AC=16$. Points $E$ and $F$ are taken on $AC$ and $AB$, respectively, and lines $EF$ and $AM$ intersect at $G$. If $AE=2AF$ then $\frac{EG}{GF}$ equals [asy] draw((0,0)--(12,0)--(14,7.75)--(0,0)); draw((0,0)--(13,3.875)); draw((5,0)--(8.75,4.84)); label("A", (0,0), S); label("B", (12,0), S); label("C", (14,7.75), E); label("E", (8.75,4.84), N); label("F", (5,0), S); label("M", (13,3.875), E); label("G", (7,1)); [/asy] $ \textbf{(A)}\ \frac{3}{2} \qquad\textbf{(B)}\ \frac{4}{3} \qquad\textbf{(C)}\ \frac{5}{4} \qquad\textbf{(D)}\ \frac{6}{5} \\ \qquad\textbf{(E)}\ \text{not enough information to solve the problem} $

Let $n$ be an integer greater than two, and let $A_1,A_2, \cdots , A_{2n}$ be pairwise distinct subsets of $\{1, 2, ,n\}$. Determine the maximum value of \[\sum_{i=1}^{2n} \dfrac{|A_i \cap A_{i+1}|}{|A_i| \cdot |A_{i+1}|}\] Where $A_{2n+1}=A_1$ and $|X|$ denote the number of elements in $X.$

Let $ABC$ be an acute-angled triangle. Let $D$ and $E$ be the feet of the altitudes on the sides $BC$ or $AC$. Points $F$ and $G$ are located on the lines $AD$ and $BE$ in such a way that$ \frac{AF}{FD}=\frac{BG}{GE}$. The line passing through $C$ and $F$ intersects $BE$ at point $H$, and the line passing through $C$ and $G$ intersects $AD$ at point $I$. Prove that points $F, G, H$ and $I$ lie on a circle. (Walther Janous)