7. The height of a regular triangular prism $A B C A_{1} B_{1} C_{1}$ is 5. A sphere $\Omega$ with radius $r=\frac{\sqrt{33}}{4}$ touches all the lateral faces of the prism. Points $F$ and $L$ are chosen on segments $C C_{1}$ and $B B_{1}$, respectively, such that $F L \| B C$, and the planes $L A C$ and $F A_{1} B_{1}$ touch the sphere $\Omega$. Find the volume of the prism and the length of segment $C F$.
Answer: $V=\frac{495 \sqrt{3}}{16}, C F=2$ or $C F=3$.

Solution. Since the sphere $\Omega$ of radius $r$ touches all the lateral faces of the prism, circles of the same radius $r$ can be inscribed in the bases of the prism. Therefore, the side of the base is $2 r \sqrt{3}=\frac{3 \sqrt{11}}{2}$, the area of the base $S=\frac{\sqrt{3}}{4}\left(\frac{3 \sqrt{11}}{2}\right)^{2}=\frac{99 \sqrt{3}}{16}$, and the volume of the prism is $S h=\frac{99 \sqrt{3}}{16} \cdot 5=\frac{495 \sqrt{3}}{16}$.

From the fact that the prism is regular and the sphere touches all its faces, it follows that the plane $L A_{1} C_{1}$ also touches the sphere; the points of tangency of the sphere with the planes $L A C$ and $L A_{1} C_{1}$ lie on the segments $L Q$ and $L Q_{1}$, where points $Q$ and $Q_{1}$ are the midpoints of the edges $A C$ and $A_{1} C_{1}$.

Consider the plane of the rectangle $B B_{1} Q_{1} Q$. Let the center of the circle $\omega$, obtained by the intersection of the sphere with this plane, be denoted by $O$, and the points of tangency of the segments $L Q_{1}, Q Q_{1}$, and $L Q$ with the circle be denoted by $G, J$, and $P$ respectively. The height $L H$ of the triangle $L Q Q_{1}$ is equal to the height of the triangle lying in the base of the prism, that is, $L H=3 r$.

Write the area $S_{0}$ of the triangle $L Q Q_{1}$ in two ways: $S_{0}=p r$ and $S_{0}=\frac{1}{2} L H \cdot Q Q_{1}=\frac{1}{2} \cdot 3 r \cdot 5=\frac{15 r}{2}$, where $p$ is the semiperimeter of the triangle $L Q Q_{1}$. Therefore, $p=\frac{15}{2}$. Let $Q J=x$; then $Q_{1} G=Q_{1} J=5-x$, $P Q=x, L G=L P=\frac{1}{2}(15-2 x-2(5-x))=\frac{5}{2}$.

Thus, $Q Q_{1}=5, Q L=\frac{5}{2}+x, Q_{1} L=\frac{15}{2}-x$. Then Heron's formula gives $S_{0}=\sqrt{\frac{15}{2} \cdot \frac{5}{2} \cdot x \cdot(5-x)}$, from which $\sqrt{\frac{15}{2} \cdot \frac{5}{2} \cdot x \cdot(5-x)}=\frac{15 r}{2}, x(5-x)=3 r^{2}$. Substituting the value of the radius from the condition, we get the equation $x^{2}-5 x+\frac{99}{16}=0$, hence $x=\frac{11}{4}$ or $x=\frac{9}{4}$.

Then $Q L=\frac{21}{4}$ or $Q L=\frac{19}{4}$, and since $B L=\sqrt{Q L^{2}-B Q^{2}}=\sqrt{Q L^{2}-9 r^{2}}$, we have $B L=3$ or $B L=2$. It remains to note that $C F=B L=Q H$.

“Phystech-2016”, mathematics, solutions for ticket 36

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 the correct answer does not guarantee a positive score for the problem.

1.(6) It is not specified that the base of the power is positive
no points should be deducted,

- non-equivalent transformation of the inequality
0 points for all subsequent actions,
- for considering each of the cases (base of the power $>1$, base of the power $<1$, base of the power equals 1)
2 points,
for another method of solving:
- the inequality is reduced to an algebraic one ..... 2 points,
- a factorization is obtained ..... 1 point,
- an isolated point is lost in the method of intervals ..... deduct 2 points.
2.(6) The equation is reduced to the form $\frac{\cos 3 x \sin x}{\cos x}=a|\sin 2 x|$ (ticket 33), $\frac{\sin 3 x \cos x}{\sin x}=a|\sin 2 x|$ (ticket ..... $34)$
$\cos 3 x=a|\sin 2 x|$ (ticket 35 ) , $\sin 3 x=a|\sin 2 x|$ (ticket 36 ) 1 point,
- in each of the two cases of opening the modulus, an equation is obtained in terms of $\sin x(\cos x)$ ..... 2 points (1 point per case),
- one of these equations (or both) is solved ..... 1 point,
- the selection arising from the sign of the modulus is made 2 points (1 point per case),
- in versions 33-35, the series $x=k \pi$ is not discarded, in version 36 the series $x=k \pi$ is lost ..... deduct 1 point,
- non-equivalent transformation with the modulus (for example, $|\sin 2 x|=2 \sin x|\cos x|$ )
no more than 2 points for the problem.
3.(6) A linear relationship between the variables is obtained (in ticket $33-y=x+3$, in ticket $34-$
$x+y=3$, in ticket $35-y=x+2$, in ticket $36-x+y=5$ ) ..... 3 points,
- the cubic equation is solved ..... 2 points,
- the system solution is obtained. ..... 1 point.
4.(8) An answer to one of the questions of the problem is given ..... 2 points,
- an answer to two questions of the problem is given ..... 5 points.
5.(6) All possible options for the last digit of the number are indicated ..... 1 point,
- points are not added for formulating the divisibility criteria by 12 (by 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 $6^{k} \cdot 8$ and so on, points are not deducted.
6.(8) a) (4) The set is constructed ..... 3 points,
- its area is found ..... 1 point.
b) (4) it is indicated that exactly three solutions are possible if and only if the second equation defines a circle circumscribed around the triangle defined by the first equation ..... 2 points,
- for each found parameter value ..... 1 point each.
7.(8) - The volume of the prism is found ..... 1 point,
- the segment is found ..... 7 points.


## CRITERIA

for determining the winners and prizewinners of the "Phystech" mathematics olympiad for schoolchildren

$$
9 \text {th grade }
$$

Maximum number of points - 32 .

- Winner - from 30 to 32 points inclusive,
- 2nd place prizewinner - from 26 to 29 points inclusive,
- 3rd place prizewinner - from 20 to 25 points inclusive.

10th grade

Maximum number of points - 35 .

- Winner - from 30 to 35 points inclusive,
- 2nd place prizewinner - from 24 to 29 points inclusive,
- 3rd place prizewinner - from 18 to 23 points inclusive.

11th grade

Maximum number of points - 48 .

- Winner - from 40 to 48 points inclusive,
- 2nd place prizewinner - from 31 to 39 points inclusive,
- 3rd place prizewinner - from 24 to 30 points inclusive.

Chairman of the jury of the olympiad

![](https://cdn.mathpix.com/cropped/2024_05_06_6b29ef4d534e57085f3bg-40.jpg?height=208&width=441&top_left_y=1723&top_left_x=969)

S.E. Gorodetsky


[^0]:    ${ }^{1}$ For this, we consider 6 possible cases: $x=2 k \pi, x=2 k \pi+\frac{\pi}{3}, x=2 k \pi+\frac{2 \pi}{3}$, etc. Each of these values of $x$ is substituted into the first equation, the period $2 k \pi$ can be dropped.

[^1]:    ${ }^{2}$ For this, we consider 5 possible cases: $x=\frac{\pi}{10}+k \pi, x=k \pi+\frac{3 \pi}{10}, x=k \pi+\frac{5 \pi}{10}$, etc. Each of these values of $x$ is substituted into the first equation, the period $k \pi$ can be dropped.

[^2]:    ${ }^{3}$ For this, we consider 6 possible cases: $x=2 k \pi, x=2 k \pi+\frac{\pi}{3}, x=2 k \pi+\frac{2 \pi}{3}$, etc. Each of these values of $x$ is substituted into the first equation, the period $2 k \pi$ can be dropped.

[^3]:    ${ }^{4}$ For this, we consider 5 possible cases