6. A circle passes through the vertices $P$ and $T$ of triangle $MPT$ and intersects its sides $MP$ and $MT$ at points $D$ and $E$, respectively, different from the vertices of the triangle. The ratio of the area of triangle $MDE$ to the area of triangle $MPT$ is $\frac{1}{4}$.

a) Find the ratio $DE: TP$.

b) Suppose it is additionally known that the ratio of the areas of triangles $MDT$ and $EMP$ is $\frac{4}{9}$. Find the ratio $TE: PD$.
Answer: a) $D E: T P=1: 2$, b) $T E: P D=1: 4$.

Solution. a) By the theorem of two secants $M T \cdot M E=M P \cdot M D$. Therefore, triangles $M P T$ and $M E D$ are similar by two sides and the angle between them ($M P: M E=M T: M D, \angle M$ - common). The areas of similar figures are proportional to the square of the similarity coefficient, so the similarity coefficient is 2. Therefore, $D E: T P=1: 2$.

b) From the similarity proven in the first part, we also get that $M T=2 M D, M P=2 M E$. By the theorem of the ratio of areas of triangles with a common angle $\frac{S_{M D T}}{S_{E M P}}=\frac{M D}{M P} \cdot \frac{M T}{M E}$, hence $\frac{4}{9}=\frac{M D}{2 M E} \cdot \frac{2 M D}{M E}$. Therefore, $\frac{M D}{M E}=\frac{2}{3}$. Let $M D=2 x$, then $M E=3 x$, $M P=2 M E=6 x$, $M T=2 M D=4 x$, $E T=M T-M E=x$, $D P=M P-M D=4 x$, so $T E: P D=1: 4$.

“Phystech-2016”, mathematics, instructions for checking, tickets $25-28$

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

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

1.(5) The inequality $y>0$ (or $y \geq 0$) is solved ..... 2 points,
- the statement that the number of natural numbers divisible by three from the obtained interval must be found is formulated ..... 2 points.
2.(5) The modulus is expanded justifiably ..... 1 point.
When solving using substitution:
- the substitution $a x-\sqrt{x}=t$ is made and the inequality is reduced to an inequality in terms of $t$ ..... 1 point,
- the quadratic inequality in terms of $t$ is solved ..... 1 point,
- for each of the two intervals for $t$ ..... 1 point each.
When solving in another way:
- The inequality is reduced to the form $(\sqrt{x}+a)(\sqrt{x}+b)(\sqrt{x}+c)(\sqrt{x}+d)>0$ ..... 2 points.
3.(6) All possible options for the last digit of the number are indicated ..... 1 point,
- points are not added for stating the divisibility rules for 12 (for 75),
- if the solution by enumeration results in an incorrect answer ..... no more than 1 point for the problem,
- if the answer is written in the form $6^{k} \cdot 8$ and so on, points are not deducted.
4.(4) The set is constructed ..... 3 points,
- its area is found ..... 1 point.
5.(6) A linear relationship between the variables is obtained (in ticket $25-y=x+3$, in ticket $26-$
$x+y=3$, in ticket $27-y=x+2$, in ticket $28-x+y=5$) ..... 3 points,
- the cubic equation is solved ..... 2 points,
- the solution to the system is obtained. ..... 1 point.
6.(6) The similarity of triangles with a known area ratio is proven ..... 2 points,
- the ratio of part a) is found ..... 1 point,
- the ratio of part b) is found ..... 3 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_97b37135d89d4349f43ag-28.jpg?height=208&width=441&top_left_y=1723&top_left_x=969)

S.E. Gorodetsky