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\hfill \thepage} %} \input{tcilatex} \begin{document} \section{Eighth International Olympiad, 1966} \subsection{1966/1.} In a mathematical contest, three problems, $A,B,C$ were posed. Among the participants there were $25$ students who solved at least one problem each. Of all the contestants who did not solve problem $A,$ the number who solved $% B$ was twice the number who solved $C.$ The number of students who solved only problem $A$ was one more than the number of students who solved $A$ and at least one other problem. Of all students who solved just one problem, half did not solve problem $A.$ How many students solved only problem $B$? \subsection{1966/2.} Let $a,b,c$ be the lengths of the sides of a triangle, and $\alpha ,\beta ,\gamma$, respectively, the angles opposite these sides. Prove that if $a+b=\tan \frac{\gamma }{2}(a\tan \alpha +b\tan \beta ),$ the triangle is isosceles. \subsection{1966/3.} Prove: The sum of the distances of the vertices of a regular tetrahedron from the center of its circumscribed sphere is less than the sum of the distances of these vertices from any other point in space. \subsection{1966/4.} Prove that for every natural number $n,$ and for every real number $x\neq k\pi /2^{t}(t=0,1,...,n;k$ any integer) $\frac{1}{\sin 2x}+\frac{1}{\sin 4x}+\cdots +\frac{1}{\sin 2^{n}x}=\cot x-\cot 2^{n}x.$ \subsection{1966/5.} Solve the system of equations $\begin{array}{lllll} & \left| a_{1}-a_{2}\right| x_{2} & +\left| a_{1}-a_{3}\right| x_{3} & +\left| a_{1}-a_{4}\right| x_{4} & =1 \\ \left| a_{2}-a_{1}\right| x_{1} & & +\left| a_{2}-a_{3}\right| x_{3} & +\left| a_{2}-a_{3}\right| x_{3} & =1 \\ \left| a_{3}-a_{1}\right| x_{1} & +\left| a_{3}-a_{2}\right| x_{2} & & & =1 \\ \left| a_{4}-a_{1}\right| x_{1} & +\left| a_{4}-a_{2}\right| x_{2} & +\left| a_{4}-a_{3}\right| x_{3} & & =1 \end{array}$ where $a_{1},a_{2},a_{3},a_{4}$ are four different real numbers. \subsection{1966/6.} In the interior of sides $BC,CA,AB$ of triangle $ABC,$ any points $K,L,M,$ respectively, are selected. Prove that the area of at least one of the triangles $AML,BKM,CLK$ is less than or equal to one quarter of the area of triangle $ABC.$ \end{document}