6. An isosceles triangle \(ABC\) with base \(AC\) is inscribed in a circle \(\Omega\). Chords \(DN\) and \(LT\), parallel to the line \(AC\), intersect the side \(BC\) at points \(F\) and \(H\) respectively, and \(BF = FH = HC\). Find the radius of the circle \(\Omega\) and the area of triangle \(ABC\), if \(DN = 2\sqrt{30}\), \(LT = 2\sqrt{42}\), and the center \(O\) of the circle \(\Omega\) is located between the lines \(LT\) and \(AC\).
Answer: $R=6.5, S=54$.

Solution. The line $B O$ is perpendicular to the chords $A C, D N, L T$ and bisects each of them. Let points $P$, $Q$, and $E$ be the midpoints of $D N, L T$, and $A C$. Denote the radius of the circle $\Omega$ as $R; B P=P Q=Q E=x$.

Then $O Q=O B-B Q=R-2 x, O P=O B-B P=R-x$ and by the Pythagorean theorem for triangles $O Q T$ and $O P N$ we get $R^{2}=42+(R-2 x)^{2}, R^{2}=30+(R-x)^{2}$, from which after expanding the brackets and combining like terms, we obtain

$$
\left\{\begin{array}{l}
2 x^{2}-2 R x+21=0 \\
x^{2}-2 R x+30=0
\end{array}\right.
$$

Subtracting the second equation from the first, we find that $x^{2}-9=0, x=3$. Then $R=\frac{2 x^{2}+21}{2 x}=\frac{13}{2}$; $O E=B E-B O=3 x-R=\frac{5}{2}, C E^{2}=O C^{2}-O E^{2}=\left(\frac{13}{2}\right)^{2}-\left(\frac{5}{2}\right)^{2}=36$. Therefore, $S_{A B C}=B E \cdot C E=9 \cdot 6=54$.

"Phystech-2016", mathematics, instructions for checking, tickets 29-32

The 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, and the obtained answers are reduced to a simplified form.

The presence of a correct answer does not guarantee a positive score for the problem.
1.(5) Incorrect trigonometric formulas were used 0 points for all subsequent actions,

- $\operatorname{tg} 2 \gamma$ (tickets 29,31) or $\operatorname{tg} 2 \alpha$ (tickets 30,32) was found ..... 1 point,
- a fractional-rational equation in terms of $\operatorname{tg} \alpha$ (tickets 29,31) or $\operatorname{tg} \beta$ (tickets 30,32) was obtained ..... 1 point,
- this equation was reduced to a quadratic equation ..... 1 point.
2.(5) The modulus was correctly expanded ..... 1 point.
When solving using substitution:
- the substitution $a x-\sqrt{x}=t$ was made and the inequality was reduced to an inequality in terms of $t$ ..... 1 point,
- a quadratic inequality in terms of $t$ was solved ..... 1 point,
- for each of the two considered intervals for $t$ ..... 1 point each.
When using another method of solving:
- The inequality was reduced to the form $(\sqrt{x}+a)(\sqrt{x}+b)(\sqrt{x}+c)(\sqrt{x}+d)>0$ ..... 2 points,
- Negative values of $x$ were included in the answer ..... no more than 3 points for the problem.
3.(6) All possible variants for the last digit of the number were indicated ..... 1 point,
- points are not added for stating the divisibility rules for 12 (for 75) ..... points are not added,
- 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 $6^{k} .8$ and so on, points are not deducted.
4.(4) The set was constructed ..... 3 points,
- its area was found ..... 1 point.
5.(6) A linear relationship between the variables was obtained (in ticket 29: $y=x+3$, in ticket 30: $x+y=3$, in ticket 31: $y=x+2$, in ticket 32: $-x+y=5$) ..... 3 points,
- a cubic equation was solved ..... 2 points,
- the system was solved ..... 1 point.
6.(9) An answer was obtained to only one of the two questions ..... 6 points.

## CRITERIA

for determining the winners and prizewinners of the "Phystech" school olympiad in mathematics

$$
9 \text {th grade }
$$

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_d3dd8bb2c54e292d29beg-30.jpg?height=208&width=441&top_left_y=1723&top_left_x=969)

S.E. Gorodetsky

[^0]:    ' If the set of points satisfying the first equation is correctly constructed, the score for the problem cannot be less than 2 points.