Found problems: 4
1991 Tournament Of Towns, (313) 3
Point $D$ lies on side $AB$ of triangle $ABC$, and $$\frac{AD}{DC} = \frac{AB}{BC}.$$
Prove that angle $C$ is obtuse.
(Sergey Berlov)
1989 All Soviet Union Mathematical Olympiad, 497
$ABCD$ 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 $\frac{S_{XQYP}}{S_{ABCD}} < \frac{mn}{m^2 + mn + n^2}$.
1989 All Soviet Union Mathematical Olympiad, 492
$ABC$ 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.
1955 Moscow Mathematical Olympiad, 312
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 $A_2C_2 //AC, C_2B_2 // CB, B_2A_2 // BA$.