Let $ ABCD$ be a rhombus with $ AC\equal{}16$ and $ BD\equal{}30$. Let $ N$ be a point on $ \overline{AB}$, and let $ P$ and $ Q$ be the feet of the perpendiculars from $ N$ to $ \overline{AC}$ and $ \overline{BD}$, respectively. Which of the following is closest to the minimum possible value of $ PQ$? [asy]unitsize(2.5cm); defaultpen(linewidth(.8pt)+fontsize(8pt)); pair D=(0,0), C=dir(0), A=dir(aSin(240/289)), B=shift(A)*C; pair Np=waypoint(B--A,0.6), P=foot(Np,A,C), Q=foot(Np,B,D); draw(A--B--C--D--cycle); draw(A--C); draw(B--D); draw(Np--Q); draw(Np--P); label("$D$",D,SW); label("$C$",C,SE); label("$B$",B,NE); label("$A$",A,NW); label("$N$",Np,N); label("$P$",P,SW); label("$Q$",Q,SSE); draw(rightanglemark(Np,P,C,2)); draw(rightanglemark(Np,Q,D,2));[/asy]$ \textbf{(A)}\ 6.5 \qquad \textbf{(B)}\ 6.75 \qquad \textbf{(C)}\ 7 \qquad \textbf{(D)}\ 7.25 \qquad \textbf{(E)}\ 7.5$

Let $ABC$ be an acute triangle with ${AB\neq AC}$. The incircle ${\omega}$ of the triangle touches the sides ${BC, CA}$ and ${AB}$ at ${D, E}$ and ${F}$, respectively. The perpendicular line erected at ${C}$ onto ${BC}$ meets ${EF}$ at ${M}$, and similarly the perpendicular line erected at ${B}$ onto ${BC}$ meets ${EF}$ at ${N}$. The line ${DM}$ meets ${\omega}$ again in ${P}$, and the line ${DN}$ meets ${\omega}$ again at ${Q}$. Prove that ${DP=DQ}$. Ruben Dario & Leo Giugiuc (Romania)

Let $ABC$ be a triangle and let $P$ be a point not lying on any of the three lines $AB$, $BC$, or $CA$. Distinct points $D$, $E$, and $F$ lie on lines $BC$, $AC$, and $AB$, respectively, such that $\overline{DE}\parallel \overline{CP}$ and $\overline{DF}\parallel \overline{BP}$. Show that there exists a point $Q$ on the circumcircle of $\triangle AEF$ such that $\triangle BAQ$ is similar to $\triangle PAC$. [i]Andrew Gu[/i]

Let $ABC$ be an arbitrary triangle. On the perpendicular bisector of $AB$, there is a point $P$ inside of triangle $ABC$. On the sides $BC$ and $CA$, triangles $BQC$ and $CRA$ are placed externally. These triangles satisfy $\vartriangle BPA \sim \vartriangle BQC \sim \vartriangle CRA$. (So $Q$ and $A$ lie on opposite sides of $BC$, and $R$ and $B$ lie on opposite sides of $AC$.) Show that the points $P, Q, C$ and $R$ form a parallelogram.

Trapezoid $ABCD$ has $\overline{AB} \parallel \overline{CD}$, $BC = CD = 43$, and $\overline{AD} \perp \overline{BD}$. Let $O$ be the intersection of the diagonals $\overline{AC}$ and $\overline{BD}$, and let $P$ be the midpoint of $\overline{BD}$. GIven that $OP = 11$, the length $AD$ can be written in the form $m\sqrt{n}$, where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime. What is $m + n$? $\textbf{(A)}\: 65\qquad\textbf{(B)}\: 132\qquad\textbf{(C)}\: 157\qquad\textbf{(D)}\: 194\qquad\textbf{(E)}\: 215$

Let $P$ be a point in the plane of a triangle $ABC$, different from its circumcenter. Prove that the triangle whose vertices are the projections of $P$ on the perpendicular bisectors of the sides of $ABC$, is similar to $ABC$.

Let $k$ be a circle with diameter $AB$. A point $C$ is chosen inside the segment $AB$ and a point $D$ is chosen on $k$ such that $BCD$ is an acute-angled triangle, with circumcentre denoted by $O$. Let $E$ be the intersection of the circle $k$ and the line $BO$ (different from $B$). Show that the triangles $BCD$ and $ECA$ are similar.

Let $O$ be the circumcenter of triangle $ABC$. Arbitrary points $M$ and $N$ lie on the sides $AC$ and $BC$, respectively. Points $P$ and $Q$ lie in the same half-plane as point $C$ with respect to the line $MN$, and satisfy $\triangle CMN \sim \triangle PAN \sim \triangle QMB$ (in this exact order). Prove that $OP=OQ$. [i]Proposed by Medeubek Kungozhin, Kazakhstan[/i]

In acute $\triangle ABC,$ let $D$ be the foot of perpendicular from $A$ on $BC$. Consider points $K, L, M$ on segment $AD$ such that $AK= KL= LM= MD$. Suppose the sum of the areas of the shaded region equals the sum of the areas of the unshaded regions in the following picture. Prove that $BD= DC$. [img]http://s27.postimg.org/a0d0plr4z/Untitled.png[/img]

Through a point on the hypotenuse of a right triangle, lines are drawn parallel to the legs of the triangle so that the trangle is divided into a square and two smaller right triangles. The area of one of the two small right triangles is $m$ times the area of the square. The ratio of the area of the other small right triangle to the area of the square is $\text{(A)}\ \frac1{2m+1}\qquad\text{(B)}\ m \qquad\text{(C)}\ 1-m\qquad\text{(D)}\ \frac1{4m} \qquad\text{(E)}\ \frac1{8m^2}$

For an olympiad geometry problem, Tina wants to draw an acute triangle whose angles each measure a multiple of $10^{\circ}$. She doesn't want her triangle to have any special properties, so none of the angles can measure $30^{\circ}$ or $60^{\circ}$, and the triangle should definitely not be isosceles. How many different triangles can Tina draw? (Similar triangles are considered the same.) [i]Proposed by Evan Chen[/i]

Let $\triangle ABC$ be a triangle with circumcircle $\Omega$ and let $N$ be the midpoint of the major arc $\widehat{BC}$. The incircle $\omega$ of $\triangle ABC$ is tangent to $AC$ and $AB$ at points $E$ and $F$ respectively. Suppose point $X$ is placed on the same side of $EF$ as $A$ such that $\triangle XEF\sim\triangle ABC$. Let $NX$ intersect $BC$ at a point $P$. Given that $AB=15$, $BC=16$, and $CA=17$, compute $\tfrac{PX}{XN}$.

Let $ABCD$ be a cyclic quadrilateral. Prove that there exists a point $X$ on segment $\overline{BD}$ such that $\angle BAC=\angle XAD$ and $\angle BCA=\angle XCD$ if and only if there exists a point $Y$ on segment $\overline{AC}$ such that $\angle CBD=\angle YBA$ and $\angle CDB=\angle YDA$.

Let $ABCD$ be a square and $X$ a point such that $A$ and $X$ are on opposite sides of $CD$. The lines $AX$ and $BX$ intersect $CD$ in $Y$ and $Z$ respectively. If the area of $ABCD$ is $1$ and the area of $XYZ$ is $\frac{2}{3}$, determine the length of $YZ$

Triangle $ABC$ has side-lengths $AB=12$, $BC=24$, and $AC=18$. The line through the incenter of $\triangle ABC$ parallel to $\overline{BC}$ intersects $\overline{AB}$ at $M$ and $\overline{AC}$ at $N$. What is the perimeter of $\triangle AMN$? $ \textbf{(A)}\ 27 \qquad \textbf{(B)}\ 30 \qquad \textbf{(C)}\ 33 \qquad \textbf{(D)}\ 36 \qquad \textbf{(E)}\ 42 $

Triangles $ABC$ and $ABD$ are isosceles with $AB =AC = BD$, and $BD$ intersects $AC$ at $E$. If $BD$ is perpendicular to $AC$, then $\angle C + \angle D$ is [asy] size(130); defaultpen(linewidth(0.8) + fontsize(11pt)); pair A, B, C, D, E; real angle = 70; B = origin; A = dir(angle); D = dir(90-angle); C = rotate(2*(90-angle), A) * B; draw(A--B--C--cycle); draw(B--D--A); E = extension(B, D, C, A); draw(rightanglemark(B, E, A, 1.5)); label("$A$", A, dir(90)); label("$B$", B, dir(210)); label("$C$", C, dir(330)); label("$D$", D, dir(0)); label("$E$", E, 1.5*dir(340)); [/asy] $\textbf{(A)}\ 115^\circ \qquad \textbf{(B)}\ 120^\circ \qquad \textbf{(C)}\ 130^\circ \qquad \textbf{(D)}\ 135^\circ \qquad \textbf{(E)}\ \text{not uniquely determined}$

A street has parallel curbs $ 40$ feet apart. A crosswalk bounded by two parallel stripes crosses the street at an angle. The length of the curb between the stripes is $ 15$ feet and each stripe is $ 50$ feet long. Find the distance, in feet, between the stripes. $ \textbf{(A)}\ 9 \qquad \textbf{(B)}\ 10 \qquad \textbf{(C)}\ 12 \qquad \textbf{(D)}\ 15 \qquad \textbf{(E)}\ 25$

The side lengths $a,$ $b,$ $c$ of a triangle $ABC$ form an arithmetical progression (such that $b-a=c-b$). The side lengths $a_{1},$ $b_{1},$ $c_{1}$ of a triangle $A_{1}B_{1}C_{1}$ also form an arithmetical progression (with $b_{1}-a_{1}=c_{1}-b_{1}$). [Hereby, $a=BC,$ $b=CA,$ $c=AB, $ $a_{1}=B_{1}C_{1},$ $b_{1}=C_{1}A_{1},$ $c_{1}=A_{1}B_{1}.$] Moreover, we know that $\measuredangle CAB=\measuredangle C_{1}A_{1}B_{1}.$ Show that triangles $ABC$ and $A_{1}B_{1}C_{1}$ are similar.

Let $ P$ be an interior point of an acute angled triangle $ ABC$. The lines $ AP,BP,CP$ meet $ BC,CA,AB$ at points $ D,E,F$ respectively. Given that triangle $ \triangle DEF$ and $ \triangle ABC$ are similar, prove that $ P$ is the centroid of $ \triangle ABC$.

The point $P$, inside the triangle $[ABC]$, lies on the perpendicular bisector of $[AB]$. $Q$ and $R$ points, exterior to the triangle, they are such that $ [BPA], [BQC]$ and $[CRA]$ are similar triangles. Shows that $[PQCR]$ is a parallelogram. [img]https://cdn.artofproblemsolving.com/attachments/f/5/6e036b127f8a013794b8246cbb1544e7280d4a.png[/img]

Let $A_1$, $A_2$, $A_3$ be three points in the plane, and for convenience, let $A_4= A_1$, $A_5 = A_2$. For $n = 1$, $2$, and $3$, suppose that $B_n$ is the midpoint of $A_n A_{n+1}$, and suppose that $C_n$ is the midpoint of $A_n B_n$. Suppose that $A_n C_{n+1}$ and $B_n A_{n+2}$ meet at $D_n$, and that $A_n B_{n+1}$ and $C_n A_{n+2}$ meet at $E_n$. Calculate the ratio of the area of triangle $D_1 D_2 D_3$ to the area of triangle $E_1 E_2 E_3$.

9.4, 10.3 Let $I$ be an incenter of $ABC$ ($AB<BC$), $M$ is a midpoint of $AC$, $N$ is a midpoint of circumcircle's arc $ABC$. Prove that $\angle IMA=\angle INB$. ([i]A. Badzyan[/i])

Triangle $ \triangle ABC$ is given. Points $ D$ i $ E$ are on line $ AB$ such that $ D \minus{} A \minus{} B \minus{} E, AD \equal{} AC$ and $ BE \equal{} BC$. Bisector of internal angles at $ A$ and $ B$ intersect $ BC,AC$ at $ P$ and $ Q$, and circumcircle of $ ABC$ at $ M$ and $ N$. Line which connects $ A$ with center of circumcircle of $ BME$ and line which connects $ B$ and center of circumcircle of $ AND$ intersect at $ X$. Prove that $ CX \perp PQ$.

Two circles $K_{1}$ and $K_{2}$, centered at $O_{1}$ and $O_{2}$ with radii $1$ and $\sqrt{2}$ respectively, intersect at $A$ and $B$. Let $C$ be a point on $K_{2}$ such that the midpoint of $AC$ lies on $K_{1}$. Find the length of the segment $AC$ if $O_{1}O_{2}=2$

Show how to cut an isosceles right triangle into a number of triangles similar to it in such a way that every two of these triangles is of different size. (AV Savkin)