Assume $A,B,C$ are three collinear points that $B \in [AC]$. Suppose $AA'$ and $BB'$ are to parrallel lines that $A'$, $B'$ and $C$ are not collinear. Suppose $O_1$ is circumcenter of circle passing through $A$, $A'$ and $C$. Also $O_2$ is circumcenter of circle passing through $B$, $B'$ and $C$. If area of $A'CB'$ is equal to area of $O_1CO_2$, then find all possible values for $\angle CAA'$

The diagonals of a convex quadrilateral dissect it into four similar triangles. Prove that this quadrilateral can also be dissected into two congruent triangles.

The areas of two right-angled triangles have ratio equal to the squares of their hypotenuses. Show that these triangles are similar.

The straight line $ \overline{AB}$ is divided at $ C$ so that $ AC\equal{}3CB$. Circles are described on $ \overline{AC}$ and $ \overline{CB}$ as diameters and a common tangent meets $ AB$ produced at $ D$. Then $ BD$ equals: $ \textbf{(A)}\ \text{diameter of the smaller circle} \\ \textbf{(B)}\ \text{radius of the smaller circle} \\ \textbf{(C)}\ \text{radius of the larger circle} \\ \textbf{(D)}\ CB\sqrt{3}\\ \textbf{(E)}\ \text{the difference of the two radii}$

Given a positive integer $n > 1$ and an angle $\alpha < 90^o$, Jaime draws a spiral $OP_0P_1...P_n$ of the following form (the figure shows the first steps): $\bullet$ First draw a triangle $OP_0P_1$ with $OP_0 = 1$, $\angle P_1OP_0 = \alpha$ and $P_1P_0O = 90^o$ $\bullet$ then for every integer $1 \le i \le n$ draw the point $P_{i+1}$ so that $\angle P_{i+1}OP_i = \alpha$, $\angle P_{i+1}P_iO = 90^o$ and $P_{i-1}$ and $P_{i+1}$ are in different half-planes with respect to the line $OP_i$ [img]https://cdn.artofproblemsolving.com/attachments/f/2/aa3913989dac1cf04f2b42b5d630b2e096dcb6.png[/img] a) If $n = 6$ and $\alpha = 30^o$, find the length of $P_0P_n$. b) If $n = 2018$ and $\alpha= 45^o$, find the length of $P_0P_n$.

Prove that there are infinite pairs of non-congruent triangles that have the same angles and two of their equal sides. Develop an algorithm or rule to obtain these pairs of triangles and indicate at least one pair that satisfies the asserted.

The circles $k_1$ and $k_2$ are tangent at the point $P$. A line is drawn through $P$, cutting $k_1$ at $A_1$ and $k_2$ at $A_2$. A second line is drawn through $P$, cutting $k_1$ at $B_1$ and $k_2$ at $B_2$. Prove that the triangles $PA_1B_1$ and $PA_2B_2$ are similar.

Let $ABC$ be an acute triangle with circumcenter $O$, and let $D$, $E$, and $F$ be the feet of altitudes from $A$, $B$, and $C$ to sides $BC$, $CA$, and $AB$, respectively. Denote by $P$ the intersection of the tangents to the circumcircle of $ABC$ at $B$ and $C$. The line through $P$ perpendicular to $EF$ meets $AD$ at $Q$, and let $R$ be the foot of the perpendicular from $A$ to $EF$. Prove that $DR$ and $OQ$ are parallel.

The circle $\omega_1$ with diameter $[AB]$ and the circle $\omega_2$ with center $A$ intersects at points $C$ and $D$. Let $E$ be a point on the circle $\omega_2$, which is outside $\omega_1$ and at the same side as $C$ with respect to the line $AB$. Let the second point of intersection of the line $BE$ with $\omega_2$ be $F$. For a point $K$ on the circle $\omega_1$ which is on the same side as $A$ with respect to the diameter of $\omega_1$ passing through $C$ we have $2\cdot CK \cdot AC = CE \cdot AB$. Let the second point of intersection of the line $KF$ with $\omega_1$ be $L$. Show that the symmetric of the point $D$ with respect to the line $BE$ is on the circumcircle of the triangle $LFC$.

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$.

A non-right triangle $A_1A_2A_3$ is given. Circles $C_1$ and $C_2$ are tangent at $A_3, C_2$ and $C_3$ are tangent at $A_1$, and $C_3$ and $C_1$ are tangent at $A_2$. Points $O_1,O_2,O_3$ are the centers of $C_1, C_2, C_3$, respectively. Supposing that the triangles $A_1A_2A_3$ and $O_1O_2O_3$ are similar, determine their angles.

Given a rectangle $ABCD$ such that $AB = b > 2a = BC$, let $E$ be the midpoint of $AD$. On a line parallel to $AB$ through point $E$, a point $G$ is chosen such that the area of $GCE$ is $$(GCE)= \frac12 \left(\frac{a^3}{b}+ab\right)$$ Point $H$ is the foot of the perpendicular from $E$ to $GD$ and a point $I$ is taken on the diagonal $AC$ such that the triangles $ACE$ and $AEI$ are similar. The lines $BH$ and $IE$ intersect at $K$ and the lines $CA$ and $EH$ intersect at $J$. Prove that $KJ \perp AB$.

Let $ABCD$ be a square, and let $E$ and $F$ be points in $AB$ and $BC$ respectively such that $BE=BF$. In the triangle $EBC$, let N be the foot of the altitude relative to $EC$. Let $G$ be the intersection between $AD$ and the extension of the previously mentioned altitude. $FG$ and $EC$ intersect at point $P$, and the lines $NF$ and $DC$ intersect at point $T$. Prove that the line $DP$ is perpendicular to the line $BT$.

$A_0B_0C_0$ and $A_1B_1C_1$ are acute-angled triangles. Describe, and prove, how to construct the triangle $ABC$ with the largest possible area which is circumscribed about $A_0B_0C_0$ (so $BC$ contains $B_0, CA$ contains $B_0$, and $AB$ contains $C_0$) and similar to $A_1B_1C_1.$

The triangle was divided into five triangles similar to it. Is it true that the original triangle is right-angled? (S. Markelov)

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 $ABC$.

Consider the $ 12$-sided polygon $ ABCDEFGHIJKL$, as shown. Each of its sides has length $ 4$, and each two consecutive sides form a right angle. Suppose that $ \overline{AG}$ and $ \overline{CH}$ meet at $ M$. What is the area of quadrilateral $ ABCM$? [asy]unitsize(13mm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=4; pair A=(1,3), B=(2,3), C=(2,2), D=(3,2), Ep=(3,1), F=(2,1), G=(2,0), H=(1,0), I=(1,1), J=(0,1), K=(0,2), L=(1,2); pair M=intersectionpoints(A--G,H--C)[0]; draw(A--B--C--D--Ep--F--G--H--I--J--K--L--cycle); draw(A--G); draw(H--C); dot(M); label("$A$",A,NW); label("$B$",B,NE); label("$C$",C,NE); label("$D$",D,NE); label("$E$",Ep,SE); label("$F$",F,SE); label("$G$",G,SE); label("$H$",H,SW); label("$I$",I,SW); label("$J$",J,SW); label("$K$",K,NW); label("$L$",L,NW); label("$M$",M,W);[/asy]$ \textbf{(A)}\ \frac {44}{3}\qquad \textbf{(B)}\ 16 \qquad \textbf{(C)}\ \frac {88}{5}\qquad \textbf{(D)}\ 20 \qquad \textbf{(E)}\ \frac {62}{3}$

Pinocchio claims that he can take some non-right-angled triangles , all of which are similar to one another and some of which may be congruent to one another, and put them together to form a rectangle. Is Pinocchio lying? (A Fedotov)

Let $ ABC$ be a non-obtuse triangle with $ CH$ and $ CM$ are the altitude and median, respectively. The angle bisector of $ \angle BAC$ intersects $ CH$ and $ CM$ at $ P$ and $ Q$, respectively. Assume that \[ \angle ABP\equal{}\angle PBQ\equal{}\angle QBC,\] (a) prove that $ ABC$ is a right-angled triangle, and (b) calculate $ \dfrac{BP}{CH}$. [i]Soewono, Bandung[/i]

Let $ABC$ be a triangle with $\angle A=90^{\circ}$ and $AB=AC$. Let $D$ and $E$ be points on the segment $BC$ such that $BD:DE:EC = 1:2:\sqrt{3}$. Prove that $\angle DAE= 45^{\circ}$

Let $ABC$ be a right triangle with a right angle at $A$, such that $AB < AC$. Let $M$ be the midpoint of $BC$ and $D$ the intersection of $AC$ with the perpendicular on $BC$ passing through $M$. Let $E$ be the intersection of the parallel to $AC$ that passes through $M$, with the perpendicular on $BD$ passing through $B$. Show that the triangles $AEM$ and $MCA$ are similar if and only if $\angle ABC = 60^o$.

The diagonals of a convex quadrilateral divide it into four similar triangles. Prove that is possible to inscribe a circle into this quadrilateral

In triangle $ ABC$, internal bisector of angle $ A$ intersects with $ BC$ at $ D$. Let $ E$ be a point on $ \left[CB\right.$ such that $ \left|DE\right|\equal{}\left|DB\right|\plus{}\left|BE\right|$. The circle through $ A$, $ D$, $ E$ intersects $ AB$ at $ F$, again. If $ \left|BE\right|\equal{}\left|AC\right|\equal{}7$, $ \left|AD\right|\equal{}2\sqrt{7}$ and $ \left|AB\right|\equal{}5$, then $ \left|BF\right|$ is $\textbf{(A)}\ \frac {7\sqrt {5} }{5} \qquad\textbf{(B)}\ \sqrt {7} \qquad\textbf{(C)}\ 2\sqrt {2} \qquad\textbf{(D)}\ 3 \qquad\textbf{(E)}\ \sqrt {10}$

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.

Let $ABC$ be a triangle. $D$ is the midpoint of $AB$, $E$ is a point on the side $BC$ such that $BE = 2 EC$ and $\angle ADC = \angle BAE$. Find $\angle BAC$.