6. Point $N$ lies on side $DE$ of triangle $CDE$ with an angle of $60^{\circ}$ at vertex $C$. Incircles are inscribed in triangles $CNE$ and $CDE$ with centers $K$ and $P$ respectively. Find the radius of the circumcircle of triangle $CKP$, if $KN=8, NP=7$.
Answer: $\sqrt{113}$.

Solution. The center of the circle inscribed in an angle lies on the bisector of this angle, so rays $N K$ and $N P$ are the bisectors of angles $E N C$ and $D N C$. Since the angle between the bisectors of adjacent angles is right, then $\angle K N P=90^{\circ}$, and then by the Pythagorean theorem we find that $K P=\sqrt{K N^{2}+P N^{2}}=\sqrt{113}$.

Since $C K$ and $C P$ are the bisectors of angles $E C N$ and $D C N$, then

$$
\angle K C P=\angle K C N+\angle P C N=\frac{1}{2} \angle E C N+\frac{1}{2} \angle D C N=\frac{1}{2} \angle E C D=30^{\circ}
$$

By the generalized sine theorem for triangle СКР, we find that the desired radius is $\frac{K P}{2 \sin \angle K C P}=\sqrt{113}$.

"Phystech-2016", mathematics, instructions for checking, tickets $1-4$
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 the number of even natural numbers from the obtained interval needs to be found is formulated
2 points.
2.(5) Inequivalent transformation of the inequality
0 points for all subsequent actions,
- the domain of definition (ODZ) is not taken into account ..... deduct 2 points.
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 12 (for 75)
- if the answer is obtained by enumeration and is incorrect ..... no more than 1 point for the problem,
- the answer is written in the form $6^{k} .8$ and so on ..... points are not deducted.
4.(5) The case when one of the branches of the graph $y=a\left|x-x_{0}\right|$ is parallel to the line is not taken into account ..... deduct 1 point,
- the case when one of the branches of the graph passes through the punctured point is not taken into account ..... deduct 1 point.
5.(6) A substitution of variables (as in the solution) is made ..... 1 point,
- a cubic equation in one of the new variables is obtained ..... 1 point,
- the cubic equation is solved ..... 2 points,
- extraneous solutions are obtained ..... deduct 1 point.
6.(5) It is proven that the segment connecting the centers of the circles is seen from the base of the bisector at a right angle ..... 1 point,
- the segment connecting the centers of the circles is found ..... 1 point,
- the angle of the desired triangle, lying opposite the segment connecting the centers of the circles, is found ..... 1 point.