Found problems: 15
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 right in $C$ and the points $D, E$ on the sides $BC$ and $CA$ respectively, such that $\frac{BD}{AC} =\frac{AE}{CD} = k$. Lines $BE$ and $AD$ intersect at $O$. Show that the angle $\angle BOD = 60^o$ if and only if $k =\sqrt3$.
Two segments $A_1B_1$ and $A_2B_2$ are given on the plane, with $\frac{A_2B_2}{A_1B_1} = k < 1$. On segment $A_1A_2$, point $A_3$ is taken, and on the extension of this segment beyond point $A_2$, point $A_4$ is taken, so $\frac{A_3A_2}{A_3A_1} =\frac{A_4A_2}{A_4A_1}= k$. Similarly, point $B_3$ is taken on segment $B_1B_2$ , and on the extension of this the segment beyond point $B_2$ is point $B_4$, so $\frac{B_3B_2}{B_3B_1} =\frac{B_4B_2}{B_4B_1}= k$. Find the angle between lines $A_3B_3$ and $A_4B_4$.
(Netherlands)
$ABCD$ is a convex quadrilateral. $X$ lies on the segment $AB$ with $\frac{AX}{XB} = \frac{m}{n}$. $Y$ lies on the segment $CD$ with $\frac{CY}{YD} = \frac{m}{n}$. $AY$ and $DX$ intersect at $P$, and $BY$ and $CX$ intersect at $Q$. Show that $\frac{S_{XQYP}}{S_{ABCD}} < \frac{mn}{m^2 + mn + n^2}$.
On the sides $ BC $, $ CA $, $ AB $ of the triangle $ ABC $, the points $ M $, $ N $, $ P $ are taken, respectively, in such a way that $$\frac{BM}{MC} = \frac{CN}{NA} = \frac{AP}{PB} = k, $$
where $ k $ means a given number greater than $ 1 $, then the segments $ AM $, $ BN $, $ CP $ were drawn . Given the area $ S $ of the triangle $ ABC $, calculate the area of the triangle bounded by the lines $ AM $, $ BN $ and $ CP $.
$ABC$ is a triangle. $A' , B' , C'$ are points on the segments $BC, CA, AB$ respectively. $\angle B' A' C' = \angle A$ , $\frac{AC'}{C'B} = \frac{BA' }{A' C} = \frac{CB'}{B'A}$. Show that $ABC$ and $A'B'C'$ are similar.
Let $ABC$ be a triangle such that $AB = AC$ and let $M$ be a point interior to the triangle. If $BM$ meets $AC$ at $D$. show that $\frac{DM}{DA}=\frac{AM}{AB}$ if and only if $\angle AMB = 2\angle ABC$.
Michel Bataille
Let $ABC$ be an acute triangle with orthocenter $H$, and $P$ a point interior to the triangle. Points $D,E,F$ are the reflections of $P$ about $BC,CA,AB$. If $Q$ is the intersection of $HD$ and $EF$, prove that the ratio $HQ/HD$ is independent of the choice of $P$.
Luis González
Given a triangle $ABC$, a line $\delta$ and a constant $k$, distinct from $0$ and $1,M$ a variable point on the line $\delta$. Points $E, F$ are on $MB,MC$ respectively such that $\frac{\overline{ME}}{\overline{MB}} = \frac{\overline{MF}}{\overline{MC}} = k$. Points $P,Q$ are on $AB,AC$ such that $PE, QF$ are perpendicular to $\delta$. Prove that the line through $M$ perpendicular to $PQ$ has a fixed point.
Nguyễn Minh Hà
Given $\vartriangle ABC$, points $C_1, A_1, B_1$ on sides $AB, BC, CA$, respectively, such that $\frac{AC_1}{C_1B}= \frac{BA_1}{A_1C}= \frac{CB_1}{B_1A}=\frac{1}{n}$ and points $C_2, A_2, B_2$ on sides $A_1B_1, B_1C_1, C_1A_1$ of $\vartriangle A_1B_1C_1$, respectively, such that $\frac{A_1C_2}{C_2B_1}= \frac{B_1A_2}{A_2C_1}= \frac{C_1B_2}{B_2A_1}= n$. Prove that $A_2C_2 //AC, C_2B_2 // CB, B_2A_2 // BA$.
Triangles $ABC$ and $A_1B_1C_1$ are similar and differently oriented. On the segment $AA_1$, a point $A'$ is taken such that $AA' / A_1A'= BC / B_1C_1$. We similarly construct $B'$ and $C'$. Prove that $A', B',C'$ lie on one straight line.
Prove that if $ M $, $ N $, $ P $ are the feet of the altitudes of acute-angled triangle $ ABC $, then the ratio of the perimeter of triangle $ MNP $ to the perimeter of triangle $ ABC $ is equal to the ratio of the radius of the circle inscribed in triangle $ ABC $ to the radius of the circle circumscribed about triangle $ ABC $.
A non-equilateral triangle $ABC$ is inscribed in a circle $\Gamma$ with centre $O$, radius $R$ and its incircle has centre $I$ and touches $BC,CA,AB$ at $D,E, F$, respectively. A circle with centre $I$ and radius $r$ intersects the rays $[ID), [IE), [IF)$ at $A',B',C'$. Show that the orthocentre $K$ of $\vartriangle A'B'C'$ is on the line $OI$ and that $\frac{IK}{IO}=\frac{r}{R}$
Michel Bataille
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Point $D$ lies on side $AB$ of triangle $ABC$, and $$\frac{AD}{DC} = \frac{AB}{BC}.$$
Prove that angle $C$ is obtuse.
(Sergey Berlov)
Let $ABC$ be an acute-angled triangle. Let $D$ and $E$ be the feet of the altitudes on the sides $BC$ or $AC$. Points $F$ and $G$ are located on the lines $AD$ and $BE$ in such a way that$ \frac{AF}{FD}=\frac{BG}{GE}$. The line passing through $C$ and $F$ intersects $BE$ at point $H$, and the line passing through $C$ and $G$ intersects $AD$ at point $I$. Prove that points $F, G, H$ and $I$ lie on a circle.
(Walther Janous)