Problem 5. On the edge $A A^{\prime}$ of the cube $A B C D A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ with edge length 2, a point $K$ is marked. In space, a point $T$ is marked such that $T B=\sqrt{11}$ and $T C=\sqrt{15}$. Find the length of the height of the tetrahedron $T B C K$, dropped from vertex $C$.
Answer: 2.

![](https://cdn.mathpix.com/cropped/2024_05_06_bb219abe882ecbba3775g-4.jpg?height=460&width=505&top_left_y=1290&top_left_x=480)

Fig. 5: to problem 5

Solution. Notice that

$$
T B^{2}+B C^{2}=11+4=15=T C^{2}
$$

From this, by the converse of the Pythagorean theorem, it follows that angle $T B C$ is a right angle. Therefore, $T B \perp B C$, which means $T$ lies in the plane of the face $A A^{\prime} B^{\prime} B$. Thus, $B C$ is the height dropped from vertex $C$, and its length is 2.

Remark. There are two possible positions of point $T$, symmetric with respect to the plane $K B C$.

## Criteria

## 4 p. Correct solution.

0 p. Only the correct answer.