7. Given a right triangular prism $A B C A_{1} B_{1} C_{1}$. A sphere with diameter $A_{1} B_{1}$ intersects the edges $A_{1} C_{1}$ and $B_{1} C_{1}$ at points $T_{1}$ and $L_{1}$, respectively, different from the vertices of the prism. Segments $B T_{1}$ and $A L_{1}$ intersect at point $S$, and $A L_{1}=7, S T_{1}=2$.

a) Find the angle $S T_{1} A_{1}$.

b) Find the ratio $A_{1} T_{1}: T_{1} C_{1}$.

c) Suppose it is additionally known that $A C=5$. Find the volume of the prism.
Answer: a) $90^{\circ}$, b) $3: 2$, c) $V=35 \sqrt{3}$.

Solution. a) Points $T_{1}$ and $L_{1}$ lie on a circle with diameter $A_{1} B_{1}$; hence, $\angle A_{1} L_{1} B_{1}=90^{\circ}, \angle A_{1} T_{1} B_{1}=90^{\circ}$ (i.e., $A_{1} L_{1}$ and $B_{1} T_{1}$ are altitudes of triangle $A_{1} B_{1} C_{1}$). The line $B_{1} T_{1}$ is the projection of the line $B T_{1}$ onto the base plane, and since $B_{1} T_{1} \perp A_{1} C_{1}$, by the theorem of three perpendiculars, $B T_{1} \perp A_{1} C_{1}$, i.e., $\angle S T_{1} A_{1}=90^{\circ}$.

b) Since the lines $B T_{1}$ and $A L_{1}$ intersect, all four points $T_{1}, L_{1}, A$, and $B$ lie in the same plane (let's call it $\alpha$). Therefore, the lines $A B$ and $T_{1} L_{1}$ lie in the same plane $\alpha$, and since they do not intersect (as they lie in parallel bases of the prism), $A B \| T_{1} L_{1}$. Thus, $T_{1} L_{1} \| A_{1} B_{1}$. The trapezoid $A_{1} T_{1} L_{1} B_{1}$ is inscribed in a circle, so it is isosceles, and the angles at its base $A_{1} B_{1}$ are equal, making triangle $A_{1} B_{1} C_{1}$ isosceles ($A_{1} C_{1}=B_{1} C_{1}$).

Triangles $T_{1} L_{1} S$ and $A B S$ are similar by two angles. From the equality of triangles $B B_{1} T_{1}$ and $A A_{1} L_{1}$, it follows that $A L_{1}=B T_{1}$, so both triangles $T_{1} L_{1} S$ and $A B S$ are isosceles with bases $T_{1} L_{1}$ and $A B$ respectively. Therefore, $T_{1} L_{1}: A_{1} B_{1}=T_{1} L_{1}: A B=T_{1} S: S B=T_{1} S:\left(T_{1} B-T_{1} S\right)=T_{1} S:\left(A L_{1}-T_{1} S\right)=2: 5$, hence $\frac{A_{1} T_{1}}{T_{1} C_{1}}=\frac{C_{1} A_{1}-T_{1} C_{1}}{T_{1} C_{1}}=\frac{C_{1} A_{1}}{T_{1} C_{1}}-1=\frac{A_{1} B_{1}}{T_{1} L_{1}}-1=\frac{5}{2}-1=\frac{3}{2}$.

c) If $A C=5$, then $C_{1} T_{1}=2, A_{1} T_{1}=3, B_{1} C_{1}=5 ; B_{1} T_{1}=\sqrt{B_{1} C_{1}^{2}-C_{1} T_{1}^{2}}=\sqrt{21} ; B B_{1}=\sqrt{B T_{1}^{2}-B_{1} T_{1}^{2}}=2 \sqrt{7}$. Thus, the area of the base of the prism is $S_{A_{1} B_{1} C_{1}}=\frac{1}{2} \cdot 5 \cdot \sqrt{21}$, and the volume of the prism is $V=S_{A_{1} B_{1} C_{1}} \cdot B B_{1}=35 \sqrt{3}$.

"Phystech-2016", mathematics, instructions for checking, tickets $21-24$

A problem is considered fully solved (and the maximum possible number of points is awarded) only if all necessary transformations are provided in the solution text and all logical steps are fully explained, with the answers given in simplified form.

## The presence of a correct answer does not guarantee a positive score for the problem.

1.(6) The equation in terms of $\sin x(\cos x)$ has been obtained .................................................................... 2 points,

- for each case of modulus expansion ....................................................................................................
- not done (or incorrectly done) selection, arising from the consideration of cases when expanding the modulus, deduct 2 points,
- not considered the domain (zeros of the denominator) deduct 1 point.

2.(6) The first equation has been factored .................................................................................................... points,

- for each of the two resulting cases 2 points.

3.(6) Inequivalent transformation of the inequality .................. 0 points for all subsequent actions,

- the inequality has been reduced to a rational inequality 2 points,
- the inequality has been transformed to the form $\frac{P(x) Q(x)}{R(x)} \geq 0$, where $P(x), Q(x), R(x)$ are linear functions of $x$ (or to the form $\frac{G(x)}{R(x)} \geq 0$, where $R(x)$ is a linear function, $G(x)$ is a quadratic function, and the roots of $G(x)$ have been found) 2 points,
- the boundaries of the intervals are given in the answer in an unsimplified form deduct 1 point,
- the points are placed on the number line in the wrong order deduct 1 point,
- an integer value of $x$ that satisfies the inequality has been guessed ... points are not added.

4.(8) Part a) has been solved 2 points,

- for each of the questions in part b) (bisector, area) .................................................................................. 3 points.

5.(6) All possible variants for the last digit of the number have been indicated ................................................................ 1 point,

- for the formulation of the divisibility criteria for 6 (for 15) ................................................................ points are not added,
- an incorrect answer has been obtained by enumeration .................................... no more than 1 point for the problem,
- the answer is written in the form $6^{k} \cdot 8$ and so on ................................................................................................

6.(8) a) (4) The set has been constructed .......................................................................................................... 3 points,

- its area has been found ...................................................................................................................... 1 point.

b) (4) It has been established that all lines defined by the second equation of the system pass through one point 2 points,

- for each found parameter value 1 point;

for another solution method:

- necessary conditions for a line and a segment to have exactly one common point (tangency or the line passing through the "vertex" of the segment) have been indicated

1 point,

- it has been shown that there are no tangents ............................................................................................... 1 point,
- the parameter values for which the line passes through the "vertices" of the segment have been found ....... 1 point,
- it has been verified that these lines do not have common points with the circle 1 point.

7.(8) It has been shown that the segment connecting the points of intersection of the sphere with the edges is parallel to the given diameter of the sphere 2 points,

- it has been proven that the base of the prism is an isosceles triangle (or an equivalent statement) ............................................................................................................................ 1 point,
- the angle of part a) has been found ........................................................................................................................ 1 point,
- the ratio of part b) has been found ................................................................................................. 2 points,
- the volume of the prism has been found ............................................................................................................ 2 points.