A triangle $ABC$ is inscribed in a circle of radius $R.$ Let $BD$ and $CE$ be the bisectors of the angles $B$ and $C$ respectively and let the line $DE$ meet the arc $AB$ not containing $C$ at point $K.$ Let $A_1, B_1, C_1$ be the feet of perpendiculars from $K$ to $BC, AC, AB,$ and $x, y$ be the distances from $D$ and $E$ to $BC,$ respectively. (a) Express the lengths of $KA_1, KB_1, KC_1$ in terms of $x, y$ and the ratio $l = KD/ED.$ (b) Prove that $\frac{1}{KB}=\frac{1}{KA}+\frac{1}{KC}.$

Let $ k$ be the inscribed circle of non-isosceles triangle $ \triangle ABC$, which center is $ S$. Circle $ k$ touches sides $ BC,CA,AB$ in points $ P,Q,R$ respectively. Line $ QR$ intersects $ BC$ in point $ M$. Let a circle which contains points $ B$ and $ C$ touch $ k$ in point $ N$. Circumscribed circle of $ \triangle MNP$ intersects line $ AP$ in point $ L$, different from $ P$. Prove that points $ S,L$ and $ M$ are collinear.

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]

Let $AEF$ be a triangle with $EF = 20$ and $AE = AF = 21$. Let $B$ and $D$ be points chosen on segments $AE$ and $AF,$ respectively, such that $BD$ is parallel to $EF.$ Point $C$ is chosen in the interior of triangle $AEF$ such that $ABCD$ is cyclic. If $BC = 3$ and $CD = 4,$ then the ratio of areas $\tfrac{[ABCD]}{[AEF]}$ can be written as $\tfrac{a}{b}$ for relatively prime positive integers $a, b$. Compute $100a + b$.

Points $A_1,A_2,A_3,...,A_{12}$ are the vertices of a regular polygon in that order. The 12 diagonals $A_1A_6,A_2A_7,A_3A_8,...,A_{11}A_4,A_{12}A_5$ are marked, as in the picture below. Let $X$ be some point in the plane. From $X$, we draw perpendicular lines to all 12 marked diagonals. Let $B_1,B_2,B_3,...,B_{12}$ be the feet of the perpendiculars, so that $B_1$ lies on $A_1A_6$, $B_2$ lies on $A_2A_7$ and so on. Evaluate the ratio $\frac{XA_1+XA_2+\dots+XA_{12}}{B_1B_6+B_2B_7+\dots+B_{12}B_5}$. [img]https://i.imgur.com/DUuwFth.png[/img]

Consider the triangle $ABC$ and the points $D \in (BC)$ and $M \in (AD)$. Lines $BM$ and $AC$ meet at $E$, lines $CM$ and $AB$ meet at $F$, and lines $EF$ and $AD$ meet at $N$. Prove that $$\frac{AN}{DN}=\frac{1}{2}\cdot \frac{AM}{DM}$$

Pete has an instrument which can locate the midpoint of a line segment, and also the point which divides the line segment into two segments whose lengths are in a ratio of $n : (n + 1)$, where $n$ is any positive integer. Pete claims that with this instrument, he can locate the point which divides a line segment into two segments whose lengths are at any given rational ratio. Is Pete right?

Ana, Bob, and Cao bike at constant rates of $8.6$ meters per second, $6.2$ meters per second, and $5$ meters per second, respectively. They all begin biking at the same time from the northeast corner of a rectangular field whose longer side runs due west. Ana starts biking along the edge of the field, initially heading west, Bob starts biking along the edge of the field, initially heading south, and Cao bikes in a straight line across the field to a point D on the south edge of the field. Cao arrives at point D at the same time that Ana and Bob arrive at D for the first time. The ratio of the field's length to the field's width to the distance from point D to the southeast corner of the field can be represented as $p : q : r$, where $p$, $q$, and $r$ are positive integers with p and q relatively prime. Find $p + q + r$.

Let $D$ be the midpoint of side $BC$ of triangle $ABC$ and $E$ the midpoint of median $AD$. Line $BE$ intersects side $CA$ at $F$. Prove that the area of quadrilateral $CDEF$ is $\frac{5}{12}$ the area of triangle $ABC$.

Let $ABC$ be a triangle inscribed in circle $\omega$, and let the medians from $B$ and $C$ intersect $\omega$ at $D$ and $E$ respectively. Let $O_1$ be the center of the circle through $D$ tangent to $AC$ at $C$, and let $O_2$ be the center of the circle through $E$ tangent to $AB$ at $B$. Prove that $O_1$, $O_2$, and the nine-point center of $ABC$ are collinear. [i]Proposed by Michael Kural[/i]

Let $P$ be an interior point of the triangle $ABC$ which is not on the median belonging to $BC$ and satisfying $\angle CAP = \angle BCP. \: BP \cap CA = \{B'\} \: , \: CP \cap AB = \{C'\}$ and $Q$ is the second point of intersection of $AP$ and the circumcircle of $ABC. \: B'Q$ intersects $CC'$ at $R$ and $B'Q$ intersects the line through $P$ parallel to $AC$ at $S.$ Let $T$ be the point of intersection of lines $B'C'$ and $QB$ and $T$ be on the other side of $AB$ with respect to $C.$ Prove that \[\angle BAT = \angle BB'Q \: \Longleftrightarrow \: |SQ|=|RB'| \]

In trapezoid $ ABCD$ with $ \overline{BC}\parallel\overline{AD}$, let $ BC\equal{}1000$ and $ AD\equal{}2008$. Let $ \angle A\equal{}37^\circ$, $ \angle D\equal{}53^\circ$, and $ m$ and $ n$ be the midpoints of $ \overline{BC}$ and $ \overline{AD}$, respectively. Find the length $ MN$.

Assume every side length of a triangle $ABC$ is more than $2$ and two of its angles are given by $\angle ABC = 57^\circ$ and $ACB = 63^\circ$. Point $P$ is chosen on side $BC$ with $BP:PC = 2:1$. Points $M,N$ are chosen on sides $AB$ and $AC$, respectively so that $BM = 2$ and $CN = 1$. Let $Q$ be the point on segment $MN$ for which $MQ:QN = 2:1$. Find the value of $PQ$. Your answer must be in simplest form.

Three unit squares and two line segments connecting two pairs of vertices are shown. What is the area of $\triangle ABC$? [asy] size(200); defaultpen(linewidth(.6pt)+fontsize(12pt)); dotfactor=4; draw((0,0)--(0,2)); draw((0,0)--(1,0)); draw((1,0)--(1,2)); draw((0,1)--(2,1)); draw((0,0)--(1,2)); draw((0,2)--(2,1)); draw((0,2)--(2,2)); draw((2,1)--(2,2)); label("$A$",(0,2),NW); label("$B$",(1,2),N); label("$C$",(4/5,1.55),W); dot((0,2)); dot((1,2)); dot((4/5,1.6)); dot((2,1)); dot((0,0)); [/asy] $ \textbf{(A)}\ \frac{1}{6}\qquad\textbf{(B)}\ \frac{1}{5}\qquad\textbf{(C)}\ \frac{2}{9}\qquad\textbf{(D)}\ \frac{1}{3}\qquad\textbf{(E)}\ \frac{\sqrt2}{4} $

Suppose that $ ABCD$ is a convex quadrilateral. Let $ F \equal{} AB\cap CD$, $ E \equal{} AD\cap BC$ and $ T \equal{} AC\cap BD$. Suppose that $ A,B,T,E$ lie on a circle which intersects with $ EF$ at $ P$. Prove that if $ M$ is midpoint of $ AB$, then $ \angle APM \equal{} \angle BPT$.