# Problem 1. (Folklore)

There are balls in three boxes. In the first - red ones, in the second - white ones, in the third - both red and white balls. Each box has a label: "red," "white," "mixed," but it is known that none of the labels correspond to the actual contents. Seventh-grader Sergey wants to find out which balls are in which box. To do this, he can open exactly one box and take out exactly one ball. Can he achieve his goal?
Answer: Yes. Solution:

Take from the box "mixed":

Red $\Rightarrow$ in the box with mixed balls red $\Rightarrow$ in the box with white balls not white and not red (then mixed) $\Rightarrow$ in the box with red balls white

Criteria: Correct solution: 15 points (+)

Non-obvious and unclear consequence of the form “in mixed red $\Rightarrow$ in red white”: 11 points ( $\pm$ )

## Problem 2

(Stern A.) On the board, a positive number is written, with which the following operations are allowed:

1) multiply by two
2) add one.

Each of the three students applied the first operation once and the second operation twice in some order. As a result, all three numbers turned out to be different, and the number obtained by the first student exceeds the number obtained by the second student by more than $60 \%$. Prove that the number obtained by the third student exceeds the number obtained by the second student by more than $30 \%$.

Solution: From $x$ you can get $2 x+2,2 x+3,2 x+4$

For $x>0: 2 x+4$ cannot be more than $60 \%$ greater than $2 x+3$ (since 1 is more than $60 \%$ of $2 x+3$)

$(2 x+2) * 160 \%<2 x+4 \Rightarrow 1.2 x<0.8 \Rightarrow x<\frac{2}{3}$

Then $(2 x+3) * 130 \%<2 x+4$

Criteria: Correct solution: 15 points (+)

Not considered the case with a different correspondence of numbers and students: 11 points ( $\pm$ )

Obtained the estimate $x<\frac{2}{3}$ : no less than 8 points $(+/ 2)$

Only matched numbers and students: 6 points (†)

## Problem 3

(Stern A.) In a right triangle $K L M$, the bisectors $K E$ and $L F$ are drawn, intersecting at point $O$. The line that bisects angle $E O L$ cuts off an isosceles triangle from the original triangle. Find the acute angles of triangle $K L M$.

Answer: $\left(45^{\circ}, 45^{\circ}\right),\left(18^{\circ}, 72^{\circ}\right),\left(\frac{270^{\circ}}{7}, \frac{360^{\circ}}{7}\right)$

Solution:

Case 1: the bisectors of two acute angles are drawn.

Let the bisector of angle $E O L$ intersect the leg $K M$ at point $T$, the leg $L M$ at point $S$, and $\angle E K M=\alpha$.

Then $\angle K O F=45^{\circ} \Rightarrow \angle S T M=\alpha+22.5^{\circ}, \angle T S M=67.5^{\circ}-\alpha$.

The triangle $S T M$ is cut off, in which there is a right angle $\Rightarrow$ if it is isosceles,

$$
\angle S T M=\angle T S M \Rightarrow \alpha+22.5^{\circ}=67.5^{\circ}-\alpha \Rightarrow \alpha=22.5^{\circ} \Rightarrow \text { the sought angles }\left(45^{\circ}, 45^{\circ}\right)
$$

Case 2: the bisector of the acute angle $L$ and the right angle $K$ is drawn.

Let the bisector of angle $E O L$ intersect the leg $K M$ at point $S$, the leg $L M$ at point $T$, and $\angle K L O=\alpha$.

Then $\angle E O L=45^{\circ}+\alpha, \angle E T O=22.5^{\circ}+\frac{3 \alpha}{2} \Rightarrow \angle T S M=67.5^{\circ}+\frac{\alpha}{2}$

Thus, the angles of the cut-off triangle $STM$ are $22.5^{\circ}+\frac{3 \alpha}{2}, 67.5^{\circ}+\frac{\alpha}{2}$, and $90^{\circ}-2 \alpha$.

Among them, there are two equal ones.

The angles $22.5^{\circ}+\frac{3 \alpha}{2}$ and $67.5^{\circ}+\frac{\alpha}{2}$ cannot be equal.

The other two equalities lead to $\alpha=9^{\circ}$ and $\alpha=\frac{135^{\circ}}{7}$, that is, to the answers $\left(18^{\circ}, 72^{\circ}\right),\left(\frac{270^{\circ}}{7}, \frac{360^{\circ}}{7}\right)$.

## Criteria:

To solve the problem, you need to consider 4 sub-cases (one in the first case and 3 in the second).

If correctly considered

one: 6 points ( $($ )

two: 8 points $(+1 / 2)$

three: 11 points $( \pm)$

all four: : 15 points (+)

## Problem 4

(Based on materials from American mathematical competitions) A proper divisor of a number is any divisor other than 1 and the number itself. Find the number of ways to color the numbers $2,3,4,5,6,7,8,9$ in three colors such that the color of each number is different from the color of any of its proper divisors. Don't forget to explain your proposed method of counting.

Answer: 432 Solution:

Color 5 and 7 (9 options)

Color $2,4,8(3 * 2 * 1=6$ options $)$

Left to color 3,6 and 9

Color 3. If 3 and 2 are the same color - 2 ways for 6 and 9 each

If different - uniquely 6 and 2 ways for 9

In total $2 * 2+2=8$ ways for $3,6,9$ and $9 * 6 * 8$ ways in total

Criteria: Correct solution: 15 points (+)

Correctly calculated the number of ways to color $2,3,6: 8$ points ( + ( $/ 2$ )

Incorrectly calculated only the number of ways to color 2,3,6 (for example, always color 2 and 3 different colors) : no more than 6 points ( $(+)$