The lengths of the sides of an acute angled triangle are successive integers. Prove that the altitude to the second longest side divides this side into two segments whose difference in length equals $4$.

Given triangle $ ABC$ and two points $ X$, $ Y$ not lying on its circumcircle. Let $ A_1$, $ B_1$, $ C_1$ be the projections of $ X$ to $ BC$, $ CA$, $ AB$, and $ A_2$, $ B_2$, $ C_2$ be the projections of $ Y$. Prove that the perpendiculars from $ A_1$, $ B_1$, $ C_1$ to $ B_2C_2$, $ C_2A_2$, $ A_2B_2$, respectively, concur if and only if line $ XY$ passes through the circumcenter of $ ABC$.

If the circumradius of a regular $n$-gon is $1$ and the ratio of its perimeter over its area is $\dfrac{4\sqrt 3}{3}$, what is $n$? $ \textbf{(A)}\ 3 \qquad\textbf{(B)}\ 4 \qquad\textbf{(C)}\ 5 \qquad\textbf{(D)}\ 6 \qquad\textbf{(E)}\ 8 $

In $\triangle ABC$, three lines are drawn parallel to side $BC$ dividing the altitude of the triangle into four equal parts. If the area of the second largest part is $35$, what is the area of the whole $\triangle ABC$? [asy] defaultpen(linewidth(0.7)); size(120); pair B = (0,0), C = (1,0), A = (0.7,1); pair[] AB, AC; draw(A--B--C--cycle); for(int i = 1; i < 4; ++i) { AB.push((i*A + (4-i)*B)/4); AC.push((i*A + (4-i)*C)/4); draw(AB[i-1] -- AC[i-1]); } filldraw(AB[1]--AB[0]--AC[0]--AC[1]--cycle, gray(0.7)); label("$A$",A,N); label("$B$",B,S); label("$C$",C,S);[/asy]

Consider $\vartriangle A_0B_0C_0$ and points $C_1, A_1, B_1$ on its sides $A_0B_0, B_0C_0, C_0A_0$, points $C_2, A_2,B_2$ on the sides $A_1B_1, B_1C_1, C_1A_1$ of $\vartriangle A_1B_1C_1$, respectively, etc., so that $$\frac{A_0B_1}{B_1C_0}= \frac{B_0C_1}{C_1A_0}= \frac{C_0A_1}{A_1B_0}= k, \frac{A_1B_2}{B_2C_1}= \frac{B_1C_2}{C_2A_1}= \frac{C_1A_2}{A_2B_1}= \frac{1}{k^2}$$ and, in general, $$\frac{A_nB_{n+1}}{B_{n+1}C_n}= \frac{B_nC_{n+1}}{C_{n+1}A_n}= \frac{C_nA_{n+1}}{A_{n+1}B_n} =k^{2n}$$ for $n$ even , $\frac{1}{k^{2n}}$ for $n$ odd. Prove that $\vartriangle ABC$ formed by lines $A_0A_1, B_0B_1, C_0C_1$ is contained in $\vartriangle A_nB_nC_n$ for any $n$.

Let $ABC$ be a triangle with $AB = 13$, $BC = 14$, and $CA = 15$. Let $\Gamma$ be the circumcircle of $ABC$, let $O$ be its circumcenter, and let $M$ be the midpoint of minor arc $BC$. Circle $\omega_1$ is internally tangent to $\Gamma$ at $A$, and circle $\omega_2$, centered at $M$, is externally tangent to $\omega_1$ at a point $T$. Ray $AT$ meets segment $BC$ at point $S$, such that $BS - CS = \dfrac4{15}$. Find the radius of $\omega_2$

The expression $16^n+4^n+1$ is equiavalent to the expression $(2^{p(n)}-1)/(2^{q(n)}-1)$ for all positive integers $n>1$ where $p(n)$ and $q(n)$ are functions and $\tfrac{p(n)}{q(n)}$ is constant. Find $p(2006)-q(2006)$.

In isosceles triangle $ABC$ ($AC=BC$), angle bisector $\angle BAC$ and altitude $CD$ from point $C$ intersect at point $O$, such that $CO=3 \cdot OD$. In which ratio does altitude from point $A$ on side $BC$ divide altitude $CD$ of triangle $ABC$

A triangle has vertices $P=(-8,5)$, $Q=(-15,-19)$, and $R=(1,-7)$. The equation of the bisector of $\angle P$ can be written in the form $ax+2y+c=0$. Find $a+c$.

The sides of triangle $BAC$ are in the ratio $2: 3: 4$. $BD$ is the angle-bisector drawn to the shortest side $AC$, dividing it into segments $AD$ and $CD$. If the length of $AC$ is $10$, then the length of the longer segment of $AC$ is: $\text{(A)} \ 3\frac12 \qquad \text{(B)} \ 5 \qquad \text{(C)} \ 5\frac57 \qquad \text{(D)} \ 6 \qquad \text{(E)} \ 7\frac12$

In rectangle $ABCD$, $AB=12$ and $BC=10$. Points $E$ and $F$ lie inside rectangle $ABCD$ so that $BE=9$, $DF=8$, $\overline{BE} \parallel \overline{DF}$, $\overline{EF} \parallel \overline{AB}$, and line $BE$ intersects segment $\overline{AD}$. The length $EF$ can be expressed in the form $m\sqrt{n}-p$, where $m,n,$ and $p$ are positive integers and $n$ is not divisible by the square of any prime. Find $m+n+p$.

Squares $ABCD$, $EFGH$, and $GHIJ$ are equal in area. Points $C$ and $D$ are the midpoints of sides $IH$ ad $HE$, respectively. What is the ratio of the area of the shaded pentagon $AJICB$ to the sum of the areas of the three squares? [asy] pair A,B,C,D,E,F,G,H,I,J; A = (0.5,2); B = (1.5,2); C = (1.5,1); D = (0.5,1); E = (0,1); F = (0,0); G = (1,0); H = (1,1); I = (2,1); J = (2,0); draw(A--B); draw(C--B); draw(D--A); draw(F--E); draw(I--J); draw(J--F); draw(G--H); draw(A--J); filldraw(A--B--C--I--J--cycle,grey); draw(E--I); dot("$A$", A, NW); dot("$B$", B, NE); dot("$C$", C, NE); dot("$D$", D, NW); dot("$E$", E, NW); dot("$F$", F, SW); dot("$G$", G, S); dot("$H$", H, N); dot("$I$", I, NE); dot("$J$", J, SE);[/asy] $\textbf{(A)}\ \frac14 \qquad \textbf{(B)}\ \frac7{24} \qquad \textbf{(C)}\ \frac13 \qquad \textbf{(D)}\ \frac38 \qquad \textbf{(E)}\ \frac5{12}$

The difference of two angles of a triangle is greater than $90^{\circ}$. Prove that the ratio of its circumradius and inradius is greater than $4$.

Line segment $AM=d>0$ is given in the plane. Furthermore, a positive number $v$ is given. Construct a right triangle $ABC$ with hypotenuse $AB$, altitude to the hypotenuse of the length $v$ and the leg $BC$ being divided by $M$ in ration $MB/MC=2/3$. Discuss conditions of solvability in terms of $d, v$.

Let $D$ be a point in the interior of $\bigtriangleup ABC$ such that $AB = ab$, $AC = ac$, $AD = ad$, $BC = bc$, $BD = bd$ and $CD = cd$. Prove that $\angle ABD + \angle ACD = \frac{\pi}{3}$.

If the radius of a circle is increased by $ 1$ unit, the ratio of the new circumference to the new diameter is: $ \textbf{(A)}\ \pi \plus{} 2 \qquad \textbf{(B)}\ \frac{2 \pi \plus{} 1}{2} \qquad \textbf{(C)}\ \pi \qquad \textbf{(D)}\ \frac{2 \pi \minus{} 1}{2} \qquad \textbf{(E)}\ \pi \minus{} 2$