[ Perpendicular Planes]

The base of the pyramid THRSK is a convex quadrilateral THRS, which is divided into two equal-area triangles by the diagonal HS. The length of the edge $TH$ is $4, \operatorname{ctg} \angle HCP = \sqrt{2}$. The sum of the lengths of the edges $TK$ and $CK$ is 4.

The volume of the pyramid is $5 \frac{1}{3}$. Find the radius of the sphere with the largest volume among the spheres that can fit inside the pyramid THRSK.
See the given problem.

## Answer

$$
2-\sqrt{2}
$$