Let $ ABC$ be a triangle and $ K$ and $ L$ be two points on $ (AB)$, $ (AC)$ such that $ BK \equal{} CL$ and let $ P \equal{} CK\cap BL$. Let the parallel through $ P$ to the interior angle bisector of $ \angle BAC$ intersect $ AC$ in $ M$. Prove that $ CM \equal{} AB$.

What minimum number of colours is sufficient to colour all positive real numbers so that every two numbers whose ratio is 4 or 8 have different colours?

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).

Through an internal point $O$ of $\Delta ABC$ one draws 3 lines, parallel to each of the sides, intersecting in the points shown on the picture. [img]https://cdn.artofproblemsolving.com/attachments/e/3/03d4d1bb61eda8b4a72ff84466d700de47c147.png[/img] Find the value of $\frac{|AF|}{|AB|}+\frac{|BE|}{|BC|}+\frac{|CN|}{|CA|}$.

On average, for every 4 sports cars sold at the local dealership, 7 sedans are sold. The dealership predicts that it will sell 28 sports cars next month. How many sedans does it expect to sell? $ \textbf{(A)}\ 7 \qquad \textbf{(B)}\ 32 \qquad \textbf{(C)}\ 35 \qquad \textbf{(D)}\ 49 \qquad \textbf{(E)}\ 112$

$ABCDEF$ is a convex hexagon, such that in it $AC \parallel DF$, $BD \parallel AE$ and $CE \parallel BF$. Prove that \[AB^2+CD^2+EF^2=BC^2+DE^2+AF^2.\] [i]N. Sedrakyan[/i]

Each point of the plane is coloured in one of two colours. Given an odd integer number $n \geq 3$, prove that there exist (at least) two similar triangles whose similitude ratio is $n$, each of which has a monochromatic vertex-set. [i]Vasile Pop[/i]

Amy painted a dart board over a square clock face using the "hour positions" as boundaries. [See figure.] If $t$ is the area of one of the eight triangular regions such as that between $12$ o'clock and $1$ o'clock, and $q$ is the area of one of the four corner quadrilaterals such as that between $1$ o'clock and $2$ o'clock, then $\frac{q}{t}=$ [asy] size((80)); draw((0,0)--(4,0)--(4,4)--(0,4)--(0,0)--(.9,0)--(3.1,4)--(.9,4)--(3.1,0)--(2,0)--(2,4)); draw((0,3.1)--(4,.9)--(4,3.1)--(0,.9)--(0,2)--(4,2)); [/asy] $ \textbf{(A)}\ 2\sqrt{3}-2 \qquad\textbf{(B)}\ \frac{3}{2} \qquad\textbf{(C)}\ \frac{\sqrt{5}+1}{2} \qquad\textbf{(D)}\ \sqrt{3} \qquad\textbf{(E)}\ 2 $

Let $ ABC$ be a right triangle with $ \angle A \equal{} 90^\circ$. Let $ D$ be the midpoint of $ AB$ and let $ E$ be a point on segment $ AC$ such that $ AD \equal{} AE$. Let $ BE$ meet $ CD$ at $ F$. If $ \angle BFC \equal{} 135^\circ$, determine $ BC / AB$.

The bisector of $\angle{BAD}$ in the parallellogram $ABCD$ intersects the lines $BC$ and $CD$ at the points $K$ and $L$ respectively. Prove that the centre of the circle passing through the points $C,\ K$ and $L$ lies on the circle passing through the points $B,\ C$ and $D$.

Consider an isosceles triangle $ABC$, where $AB = AC$. $D$ is a point on the $AC$ side and $P$ a point on the segment $BD$ so that the angle $\angle APC = 90^o$ and $ \angle ABP = \angle BCP $. Determine the ratio $AD: DC$.

Triangle $ ABC$ has $ AC\equal{}3$, $ BC\equal{}4$, and $ AB\equal{}5$. Point $ D$ is on $ \overline{AB}$, and $ \overline{CD}$ bisects the right angle. The inscribed circles of $ \triangle ADC$ and $ \triangle BCD$ have radii $ r_a$ and $ r_b$, respectively. What is $ r_a/r_b$? $ \textbf{(A)}\ \frac{1}{28}\left(10\minus{}\sqrt{2}\right) \qquad \textbf{(B)}\ \frac{3}{56}\left(10\minus{}\sqrt{2}\right) \qquad \textbf{(C)}\ \frac{1}{14}\left(10\minus{}\sqrt{2}\right) \qquad \textbf{(D)}\ \frac{5}{56}\left(10\minus{}\sqrt{2}\right) \\ \textbf{(E)}\ \frac{3}{28}\left(10\minus{}\sqrt{2}\right)$

In a triangle $ABC$, points $D$ and $E$ are on segments $BC$ and $AC$ such that $BD=3DC$ and $AE=4EC$. Point $P$ is on line $ED$ such that $D$ is the midpoint of segment $EP$. Lines $AP$ and $BC$ intersect at point $S$. Find the ratio $BS/SD$.

The altitudes of a tetrahedron intersect in a point. Prove that this point, the foot of one of the altitudes, and the points dividing the other three altitudes in the ratio $2 : 1$ (measuring from the vertices) lie on a sphere. [i]D. Tereshin[/i]

Let $ABC$ be a triangle. For a point $P$ in its interior, we draw the threee lines through $P$ parallel to the sides of the triangle. This partitions $ABC$ in three triangles and three quadrilaterals. Let $V_A$ be the area of the quadrilateral which has $A$ as one vertex. Let $D_A$ be the area of the triangle which has a part of $BC$ as one of its sides. Define $V_B, D_B$ and $V_C, D_C$ similarly. Determine all possible values of $\frac{D_A}{V_A}+\frac{D_B}{V_B}+\frac{D_C}{V_C}$, as $P$ varies in the interior of the triangle.

In a triangle $ABC$, the bisector of the angle at $A$ of a triangle $ABC$ intersects the segment $BC$ and the circumcircle of $ABC$ at points $A_1$ and $A_2$, respectively. Points $B_1,B_2,C_1,C_2$ are analogously defined. Prove that $$\frac{A_1A_2}{BA_2+CA_2}+\frac{B_1B_2}{CB_2+AB_2}+\frac{C_1C_2}{AC_2+BC_2}\ge\frac34.$$

$ I_a$ is the excenter of the triangle $ ABC$ with respect to $ A$, and $ AI_a$ intersects the circumcircle of $ ABC$ at $ T$. Let $ X$ be a point on $ TI_a$ such that $ XI_a^2\equal{}XA.XT$. Draw a perpendicular line from $ X$ to $ BC$ so that it intersects $ BC$ in $ A'$. Define $ B'$ and $ C'$ in the same way. Prove that $ AA'$, $ BB'$ and $ CC'$ are concurrent.

Brenda and Sally run in opposite directions on a circular track, starting at diametrically opposite points. They first meet after Brenda has run $ 100$ meters. They next meet after Sally has run $ 150$ meters past their first meeting point. Each girl runs at a constant speed. What is the length of the track in meters? $ \textbf{(A)}\ 250 \qquad \textbf{(B)}\ 300 \qquad \textbf{(C)}\ 350 \qquad \textbf{(D)}\ 400\qquad \textbf{(E)}\ 500$

In this diagram a scheme is indicated for associating all the points of segment $ \overline{AB}$ with those of segment $ \overline{A'B'}$, and reciprocally. To described this association scheme analytically, let $ x$ be the distance from a point $ P$ on $ \overline{AB}$ to $ D$ and let $ y$ be the distance from the associated point $ P'$ of $ \overline{A'B'}$ to $ D'$. Then for any pair of associated points, if $ x \equal{} a,\, x \plus{} y$ equals: [asy]defaultpen(linewidth(.8pt)); unitsize(.8cm); pair D= (0,9); pair E = origin; pair A = (3,9); pair P = (3.6,9); pair B = (4,9); pair F = (1,0); pair G = (2.6,0); pair H = (5,0); dot((0,0));dot((1,0));dot((2,0));dot((3,0));dot((4,0));dot((5,0)); dot((0,9));dot((1,9));dot((2,9));dot((3,9));dot((4,9));dot((5,9)); draw((D+(0,0.5))--(0,-0.5)); draw(A--H); draw(P--G); draw(B--F); draw(F--H); draw(A--B); label("$D$",D,NW); label("$D'$",E,NW); label("0",(0,0),SE); label("1",(1,0),SE); label("2",(2,0),SE); label("3",(3,0),SE); label("4",(4,0),SE); label("5",(5,0),SE); label("0",(0,9),SE); label("1",(1,9),SE); label("2",(2,9),SE); label("3",(3,9),SW); label("4",(4,9),SE); label("5",(5,9),SE); label("$B'$",F,NW); label("$P'$",G,S); label("$A'$",H,NE); label("$A$",A,NW); label("$P$",P,N); label("$B$",B,NE);[/asy] $ \textbf{(A)}\ 13a\qquad \textbf{(B)}\ 17a \minus{} 51\qquad \textbf{(C)}\ 17 \minus{} 3a\qquad \textbf{(D)}\ \frac {17 \minus{} 3a}{4}\qquad \textbf{(E)}\ 12a \minus{} 34$

Let $ABC$ be a triangle such that $\angle BAC = 90^\circ$ and $AB < AC$. We divide the interior of the triangle into the following six regions: \begin{align*} S_1=\text{set of all points }\mathit{P}\text{ inside }\triangle ABC\text{ such that }PA<PB<PC \\ S_2=\text{set of all points }\mathit{P}\text{ inside }\triangle ABC\text{ such that }PA<PC<PB \\ S_3=\text{set of all points }\mathit{P}\text{ inside }\triangle ABC\text{ such that }PB<PA<PC \\ S_4=\text{set of all points }\mathit{P}\text{ inside }\triangle ABC\text{ such that }PB<PC<PA \\ S_5=\text{set of all points }\mathit{P}\text{ inside }\triangle ABC\text{ such that }PC<PA<PB \\ S_6=\text{set of all points }\mathit{P}\text{ inside }\triangle ABC\text{ such that }PC<PB<PA\end{align*} Suppose that the ratio of the area of the largest region to the area of the smallest non-empty region is $49 : 1$. Determine the ratio $AC : AB$.

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$

Define the infinite products \[ A = \prod\limits_{i=2}^{\infty} \left(1-\frac{1}{n^3}\right) \text{ and } B = \prod\limits_{i=1}^{\infty}\left(1+\frac{1}{n(n+1)}\right). \] If $\tfrac{A}{B} = \tfrac{m}{n}$ where $m,n$ are relatively prime positive integers, determine $100m+n$. [i]Proposed by Lewis Chen[/i]

Two parallelograms $ABCD$ and $A'B'C'D'$ are given in space. Points $A'',B'',C'',D''$ divide the segments $AA',BB',CC',DD'$ in the same ratio. What can be said about the quadrilateral $A''B''C''D''$?

In triangle $ABC$, $BC = 23$, $CA = 27$, and $AB = 30$. Points $V$ and $W$ are on $\overline{AC}$ with $V$ on $\overline{AW}$, points $X$ and $Y$ are on $\overline{BC}$ with $X$ on $\overline{CY}$, and points $Z$ and $U$ are on $\overline{AB}$ with $Z$ on $\overline{BU}$. In addition, the points are positioned so that $\overline{UV} \parallel \overline{BC}$, $\overline{WX} \parallel \overline{AB}$, and $\overline{YZ} \parallel \overline{CA}$. Right angle folds are then made along $\overline{UV}$, $\overline{WX}$, and $\overline{YZ}$. The resulting figure is placed on a level floor to make a table with triangular legs. Let $h$ be the maximum possible height of a table constructed from triangle $ABC$ whose top is parallel to the floor. Then $h$ can be written in the form $\tfrac{k \sqrt{m}}{n}$, where $k$ and $n$ are relatively prime positive integers and $m$ is a positive integer that is not divisible by the square of any prime. Find $k + m + n$. [asy] unitsize(1 cm); pair translate; pair[] A, B, C, U, V, W, X, Y, Z; A[0] = (1.5,2.8); B[0] = (3.2,0); C[0] = (0,0); U[0] = (0.69*A[0] + 0.31*B[0]); V[0] = (0.69*A[0] + 0.31*C[0]); W[0] = (0.69*C[0] + 0.31*A[0]); X[0] = (0.69*C[0] + 0.31*B[0]); Y[0] = (0.69*B[0] + 0.31*C[0]); Z[0] = (0.69*B[0] + 0.31*A[0]); translate = (7,0); A[1] = (1.3,1.1) + translate; B[1] = (2.4,-0.7) + translate; C[1] = (0.6,-0.7) + translate; U[1] = U[0] + translate; V[1] = V[0] + translate; W[1] = W[0] + translate; X[1] = X[0] + translate; Y[1] = Y[0] + translate; Z[1] = Z[0] + translate; draw (A[0]--B[0]--C[0]--cycle); draw (U[0]--V[0],dashed); draw (W[0]--X[0],dashed); draw (Y[0]--Z[0],dashed); draw (U[1]--V[1]--W[1]--X[1]--Y[1]--Z[1]--cycle); draw (U[1]--A[1]--V[1],dashed); draw (W[1]--C[1]--X[1]); draw (Y[1]--B[1]--Z[1]); dot("$A$",A[0],N); dot("$B$",B[0],SE); dot("$C$",C[0],SW); dot("$U$",U[0],NE); dot("$V$",V[0],NW); dot("$W$",W[0],NW); dot("$X$",X[0],S); dot("$Y$",Y[0],S); dot("$Z$",Z[0],NE); dot(A[1]); dot(B[1]); dot(C[1]); dot("$U$",U[1],NE); dot("$V$",V[1],NW); dot("$W$",W[1],NW); dot("$X$",X[1],dir(-70)); dot("$Y$",Y[1],dir(250)); dot("$Z$",Z[1],NE); [/asy]

Let $ABC$ be a triangle with $AB>AC$. Let $D$ be the foot of the internal angle bisector of $A$. Points $F$ and $E$ are on $AC,AB$ respectively such that $B,C,F,E$ are concyclic. Prove that the circumcentre of $DEF$ is the incentre of $ABC$ if and only if $BE+CF=BC$.