11.3. On the edge $A A_{1}$ of the cube $A B C D A_{1} B_{1} C_{1} D_{1}$ with edge length $3 \sqrt{2}$, a point K is marked. In space, a point $T$ is marked such that $T B=7$ and $T C=\sqrt{67}$. What are the smallest and largest values that the length of the segment TK can have? Justify your answer.

![](https://cdn.mathpix.com/cropped/2024_05_06_de0eff2f5c97fa664b16g-06.jpg?height=663&width=671&top_left_y=157&top_left_x=133)

K to the solution of problem 11.3
Solution: Consider the triangle $TBC$. Since $TC^2 = 67 = 49 + 18 = TB^2 + BC^2$, this triangle is a right triangle with the right angle at $TBC$. Therefore, the line $BC$ is perpendicular to the plane $TBA$ by the criterion of perpendicularity of a line and a plane. This means that point $T$ lies in the plane $ABA_1$. Since the distance $TB$ is fixed, point $T$ lies on a circle with center at point $B$ and radius 7 (see figure).

Since $TK \leq TB + BK = 7 + BK$, and equality is achieved when point $B$ lies on the segment $TK$, the maximum value of $TK$ is achieved when $BK$ is maximized, i.e., when $K = A_1$. In this case, $BK = 6$, and $TK = 13$. On the other hand, $TB > BA_1 \geq BK$, so $TK \geq TB - BK = 7 - BK \geq 7 - 6 = 1$. Equality is achieved again when $K = A_1$, and point $T$ is chosen on the extension of the line $BA_1$ beyond point $A_1$.

Answer: the maximum value is 13; the minimum value is 1.

Recommendations for checking:

| is in the work | points |
| :--- | :--- |
| Correct answers with the positions of the points indicated, where they are achieved | 7 points |
| Correct answers without the positions of the points indicated, where they are achieved | 6 points |
| Only one of the required values is correctly indicated with the position of the point, where it is achieved | 4 points |
| Only one of the required values is correctly indicated without the position of the point, where it is achieved | 3 points |
| Answers not justified (or justified incorrectly) | 0 points |