6. A circle passes through the vertices $K$ and $P$ of triangle $K P M$ and intersects its sides $K M$ and $P M$ at points $F$ and $B$ respectively, such that $K F: F M=3: 1, P B: B M=6: 5$. Find $K P$, if $B F=\sqrt{15}$.
Answer: $2 \sqrt{33}$.

Solution. Let $F M=x, B M=5 y$; then $K F=3 x, B P=6 y$. By the theorem of two secants $M F \cdot M K=M B \cdot M P$, from which $x \cdot 4 x=5 y \cdot 11 y, y=\frac{2 x}{\sqrt{55}}$. Triangles $K P M$ and $B F M$ are similar by two sides and the angle between them ( $K M: B M=P M: F M, \angle M-$ is common), and the similarity coefficient is $\frac{P M}{F M}=\frac{11 y}{x}=11 \sqrt{\frac{4}{55}}=\frac{2 \sqrt{11}}{\sqrt{5}}$. Therefore, $K P=B F \cdot \frac{2 \sqrt{11}}{\sqrt{5}}=2 \sqrt{33}$.

"Phystech-2016", mathematics, instructions for checking, tickets 13-16

The problem is considered fully solved (and the maximum possible number of points is awarded for it) only if all necessary transformations are provided in the solution text and all logical steps are fully explained, with the answers obtained reduced to a simplified form.

The presence of a correct answer does not guarantee a positive score for the problem.
1.(5) The inequality $y>0$ or $y \geq 0$ is solved ..... 2 points,
- the statement that it is necessary to find the number of natural numbers divisible by three from the obtained interval is formulated
2.(4) The equation is reduced to the form $\left(x-x_{1}\right)\left(x^{2}+a x+b\right)=0$ ..... 1 point,
- all roots of the equation are found and there are no further advances ..... 2 points for the entire problem,
- the theorem of Vieta is used for the equation $x^{2}+a x+b=0$ and it is not proven that its roots are the largest and smallest roots of the original equation ..... deduct 1 point.
3.(6) All possible options for the last digit of the number are indicated ..... 1 point,
- points are not added for formulating divisibility rules for 6 (for 15) ..... points are not added,
- if the solution by enumeration results in an incorrect answer ..... no more than 1 point for the problem,
- the answer is written in the form $6^{k} \cdot 8$ and so on, points are not deducted.
4.(6) The set of points satisfying the first inequality is constructed ..... 4 points,
- the set of points satisfying the second inequality is constructed ..... 1 point,
- the area is found ..... 1 point.
5.(6) The first equation is factored ..... 2 points,
- 2 points for each of the two resulting cases.
6.(5) The similarity of triangles with a common angle is proven ..... 2 points,
- the segment is calculated ..... 3 points.