Show that a line passing through the feet of two altitudes of an acute-angled triangle cuts off a similar triangle.

Let $\triangle ABC$ be an acute triangle, and let $D$ be the projection of $A$ on $BC$. Let $M,N$ be the midpoints of $AB$ and $AC$ respectively. Let $\Gamma_1$ and $\Gamma_2$ be the circumcircles of $\triangle BDM$ and $\triangle CDN$ respectively, and let $K$ be the other intersection point of $\Gamma_1$ and $\Gamma_2$. Let $P$ be an arbitrary point on $BC$ and $E,F$ are on $AC$ and $AB$ respectively such that $PEAF$ is a parallelogram. Prove that if $MN$ is a common tangent line of $\Gamma_1$ and $\Gamma_2$, then $K,E,A,F$ are concyclic.

Let $O$ be the circumcenter of an acute triangle $ABC$. Line $OA$ intersects the altitudes of $ABC$ through $B$ and $C$ at $P$ and $Q$, respectively. The altitudes meet at $H$. Prove that the circumcenter of triangle $PQH$ lies on a median of triangle $ABC$.

Color all points on a plane in red or blue. Prove that there exists two similar triangles, their similarity ratio is $1995$, and apexes of both triangles are in the same color.

Let $ r$ and $ s$ two orthogonal lines that does not lay on the same plane. Let $ AB$ be their common perpendicular, where $ A\in{}r$ and $ B\in{}s$(*).Consider the sphere of diameter $ AB$. The points $ M\in{r}$ and $ N\in{s}$ varies with the condition that $ MN$ is tangent to the sphere on the point $ T$. Find the locus of $ T$. Note: The plane that contains $ B$ and $ r$ is perpendicular to $ s$.

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] $\textbf{(A)}\ 115^\circ \qquad \textbf{(B)}\ 120^\circ \qquad \textbf{(C)}\ 130^\circ \qquad \textbf{(D)}\ 135^\circ \qquad \textbf{(E)}\ \text{not uniquely determined}$

Let $ \overline{AB}$ be a diameter of a circle and $ C$ be a point on $ \overline{AB}$ with $ 2 \cdot AC \equal{} BC$. Let $ D$ and $ E$ be points on the circle such that $ \overline{DC} \perp \overline{AB}$ and $ \overline{DE}$ is a second diameter. What is the ratio of the area of $ \triangle DCE$ to the area of $ \triangle ABD$? [asy]unitsize(2.5cm); defaultpen(fontsize(10pt)+linewidth(.8pt)); dotfactor=3; pair O=(0,0), C=(-1/3.0), B=(1,0), A=(-1,0); pair D=dir(aCos(C.x)), E=(-D.x,-D.y); draw(A--B--D--cycle); draw(D--E--C); draw(unitcircle,white); drawline(D,C); dot(O); clip(unitcircle); draw(unitcircle); label("$E$",E,SSE); label("$B$",B,E); label("$A$",A,W); label("$D$",D,NNW); label("$C$",C,SW); draw(rightanglemark(D,C,B,2));[/asy]$ \textbf{(A)} \ \frac {1}{6} \qquad \textbf{(B)} \ \frac {1}{4} \qquad \textbf{(C)}\ \frac {1}{3} \qquad \textbf{(D)}\ \frac {1}{2} \qquad \textbf{(E)}\ \frac {2}{3}$

Let $\triangle{ABC}$ be a triangle with orthocentre $H$. The tangent lines from $A$ to the circle with diameter $BC$ touch this circle at $P$ and $Q$. Prove that $H,P$ and $Q$ are collinear.

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]

$\vartriangle ABC \sim \vartriangle A'B'C'$. $\vartriangle ABC$ has sides $a,b,c$ and $\vartriangle A'B'C'$ has sides $a',b',c'$. Two sides of $\vartriangle ABC$ are equal to sides of $\vartriangle A'B'C'$. Furthermore, $a < a'$, $a < b < c$, $a = 8$. Prove that there is exactly one pair of such triangles with all sides integers.

Given convex quadrilateral $ABCD$. We construct the similar triangles $APB, BQC, CRD, DSA$ outside $ABCD$ so that $\angle PAB = \angle QBC = \angle RCD = \angle SDA, \angle PBA = \angle QCB = \angle RDC = \angle SAD$. Prove that if $PQRS$ is a parallelogram, so is $ABCD$.

Given an isosceles triangle $ABC$ with base $BC$. On the sides $BC$, $AC$ and $AB$ points $X,Y$ and $Z$ are chosen respectively such that triangles $ABC$ and $YXZ$ are similar. Point $W$ is symmetric to point $X$ with respect to the midpoint of $BC$. Prove that points $X,Y,Z$ and $W$ lie on a circle.

Triangle $ ABC$ has $ AC \equal{} 450$ and $ BC \equal{} 300$. Points $ K$ and $ L$ are located on $ \overline{AC}$ and $ \overline{AB}$ respectively so that $ AK \equal{} CK$, and $ \overline{CL}$ is the angle bisector of angle $ C$. Let $ P$ be the point of intersection of $ \overline{BK}$ and $ \overline{CL}$, and let $ M$ be the point on line $ BK$ for which $ K$ is the midpoint of $ \overline{PM}$. If $ AM \equal{} 180$, find $ LP$.

Let \(\triangle ABC\) have side lengths \(AB=30\), \(BC=32\), and \(AC=34\). Point \(X\) lies in the interior of \(\overline{BC}\), and points \(I_1\) and \(I_2\) are the incenters of \(\triangle ABX\) and \(\triangle ACX\), respectively. Find the minimum possible area of \(\triangle AI_1I_2\) as \( X\) varies along \(\overline{BC}\).

Let $ABC$ be a triangle and $D$ a point in $AC$. If $\angle{CBD} - \angle{ABD} = 60^{\circ}, \hat{BDC} = 30^{\circ}$ and also $AB \cdot BC = BD^{2}$, determine the measure of all the angles of triangle $ABC$.

Let $ABCD$ be a quadrilateral with $AB \parallel CD$, $AB > CD$. Prove that the line passing through $AC \cap BD$ and $AD \cap BC$ passes through the midpoints of $AB$ and $CD$.

In rectangle $ABCD$, $AB=6$, $AD=30$, and $G$ is the midpoint of $\overline{AD}$. Segment $AB$ is extended $2$ units beyond $B$ to point $E$, and $F$ is the intersection of $\overline{ED}$ and $\overline{BC}$. What is the area of $BFDG$? $ \textbf{(A)}\ \frac{133}{2}\qquad\textbf{(B)}\ 67\qquad\textbf{(C)}\ \frac{135}{2}\qquad\textbf{(D)}\ 68\qquad\textbf{(E)}\ \frac{137}{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]

The difference in the areas of two similar triangles is $18$ square feet, and the ratio of the larger area to the smaller is the square of an integer. The area of the smaller triange, in square feet, is an integer, and one of its sides is $3$ feet. The corresponding side of the larger triangle, in feet, is: $\textbf{(A)}\ 12\quad \textbf{(B)}\ 9\qquad \textbf{(C)}\ 6\sqrt{2}\qquad \textbf{(D)}\ 6\qquad \textbf{(E)}\ 3\sqrt{2}$

A triangular piece of paper of area $1$ is folded along a line parallel to one of the sides and pressed flat. What is the minimum possible area of the resulting figure?

In the parallelogram $ABCD$, $AC \cap BD = { O }$, and $M$ is the midpoint of $AB$. Let $P \in (OC)$ and $MP \cap BC = { Q }$. We draw a line parallel to $MP$ from $O$, which intersects line $CD$ at point $N$. Show that $A,N,Q$ are collinear if and only if $P$ is the midpoint of $OC$.

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$

In the adjoining figure, the triangle $ABC$ is a right triangle with $\angle BCA=90^\circ$. Median $CM$ is perpendicular to median $BN$, and side $BC=s$. The length of $BN$ is [asy] size(200); defaultpen(linewidth(0.7)+fontsize(10));real r=54.72; pair B=origin, C=dir(r), A=intersectionpoint(B--(9,0), C--C+4*dir(r-90)), M=midpoint(B--A), N=midpoint(A--C), P=intersectionpoint(B--N, C--M); draw(M--C--A--B--C^^B--N); pair point=P; markscalefactor=0.005; draw(rightanglemark(C,P,B)); label("$A$", A, dir(point--A)); label("$B$", B, dir(point--B)); label("$C$", C, dir(point--C)); label("$M$", M, S); label("$N$", N, dir(C--A)*dir(90)); label("$s$", B--C, NW);[/asy] $\textbf {(A) } s\sqrt 2 \qquad \textbf {(B) } \frac 32s\sqrt2 \qquad \textbf {(C) } 2s\sqrt2 \qquad \textbf {(D) } \frac 12s\sqrt5 \qquad \textbf {(E) } \frac 12s\sqrt6$

Let $n$ be an integer number greater than or equal to $2$, and let $K$ be a closed convex set of area greater than or equal to $n$, contained in the open square $(0, n) \times (0, n)$. Prove that $K$ contains some point of the integral lattice $\mathbb{Z} \times \mathbb{Z}$. [i]Marius Cavachi[/i]

Let $h_a$, $h_b$ and $h_c$ be the altitudes of a triangle $\triangle ABC$ ejected from the vertices $A$,$B$ and $C$, respectively. Similarly, let $h_x$, $h_y$ and $h_z$ be the altitudes of an another triangle $\triangle XYZ$. Show that if $$h_a : h_b : h_c = h_x : h_y : h_z, $$ then the triangles $\triangle ABC$ and $\triangle XYZ$ are similar.