2. On the border of a circular glade, points $A, B, C, D$ are marked clockwise. At point $A$ is a squirrel named An, at point $B$ is a squirrel named Bim, at point $C$ stands a pine tree, and at point $D$ stands an oak tree. The squirrels started running simultaneously, An towards the pine tree, and Bim towards the oak tree. They collided at point $M$, which is closer to the pine tree than to the oak tree. Is it true that if An had run from point $A$ to the oak tree, and Bim from point $B$ to the pine tree, An would have arrived first? Each squirrel runs in a straight line and at a constant speed.
Answer: No.

Solution: Let the speed of Ana be $-v$, and the speed of Bim be $-u$. They both reached point $M$ at the same time, so $\frac{A M}{v}=\frac{B M}{u}$. Triangles $D A M$ and $C B M$ are similar (by three angles).

Therefore, $\frac{A D}{B C}=\frac{A M}{B M}$. But $\frac{A M}{B M}=\frac{v}{u}$, so $\frac{A D}{B C}=\frac{v}{u}$, which means $\frac{A D}{v}=\frac{B C}{u}$. Thus, the squirrels will arrive at the same time.

![](https://cdn.mathpix.com/cropped/2024_05_06_88c5abeee9973c9a71c8g-6.jpg?height=257&width=277&top_left_y=545&top_left_x=1643)

Comment: Correct solution - 20 points. Correct solution not fully completed - 15 points. Solution with inaccuracies - 15 points. Solution based on an example - 10 points. Some progress - 5 points. Incorrect or no solution - 0 points.

3. In how many ways can 30 apples be distributed into 3 baskets such that the first basket has fewer apples than the second, the second has fewer apples than the third, and no basket is empty?

## Answer: 61.

Solution: Consider the equation $a+b+c=30$. Arrange 30 units in a row, choose two gaps between the units, and place a dash in each. The number of units to the left of the first dash is $a$ (the number of apples in the first basket), and the number to the right of the second dash is $c$ (the number of apples in the third basket). The number of ways to choose two gaps is $\frac{29 \cdot 28}{2}=406$. We need to subtract the number of cases where among the numbers $a, b, c$ there are equal numbers. Let $a=b$. Then $2 a+c=30$, this equation has 14 non-zero solutions ( $c$ is even and can vary from 2 to 28). Similarly, there will be 14 cases when $a=c$ or $b=c$. Thus, $406-3 \cdot 14=364$. The case $a=b=c$ is counted once in the total number of ways and subtracted three times, but it should be excluded only once, so we need to add 2. The number $364+2=366$ is the number of ordered triples of distinct $a, b, c$. But we need the order $a<b<c$, so we divide by the number of permutations of three elements: $\frac{366}{6}=61$.

Comment: Correct solution - 20 points. Correct solution not fully completed - 15 points. Significant progress in the solution - 10 points. Only the correct answer without a solution - 0 points.