Authors: Vysokiy I. Raskina I.V.

In the aquarium, there are three types of fish: gold, silver, and red. If the cat eats all the gold fish, the number of fish will be one less than $2 / 3$ of the original number. If the cat eats all the red fish, the number of fish will be four more than $2 / 3$ of the original number. Which fish, gold or silver, are there more of, and by how many?
From the first condition, it is clear that the number of goldfish is 1 more than a third. From the second condition, it follows that the number of red fish is 4 less than a third. Therefore, the number of silver fish is 3 more than a third.

## Answer

The number of silver fish is 2 more

Authors: Baksueva E.V., Khachatryan A.V.

The Smeshariki live on the shores of a pond in the shape of an equilateral triangle with a side length of 600 meters. Krysh and Barash live on the same shore, 300 meters apart. In the summer, it is 900 meters for Losyash to walk to Krysh, and 900 meters for Barash to walk to Nyusha. Prove that in the winter, when the pond freezes and it is possible to walk directly on the ice, the distance for Losyash to walk to Krysh will again be the same number of meters as for Barash to walk to Nyusha.

## Solution

Let Krysh live at a distance of $x$ from the nearest corner of the pond: $A K=x$. Then the distance $B V$ from Barash to his corner of the pond is $300-x$.

According to the condition, $V L=900-B K=300+x$ (since 900 is exactly half the perimeter of the pond, it does not matter which of the two paths Losyash takes to get to Krysh),

$A H=900-A B=600-x$. It remains to note that triangles AHB and VKL are equal by angle and two sides adjacent to it $(A B=300+x=B L$,

$A H=B K)$. Therefore, their corresponding sides LK and BN are also equal.

![](https://cdn.mathpix.com/cropped/2024_05_06_c9b930b00ea53112180dg-20.jpg?height=698&width=744&top_left_y=860&top_left_x=657)

Send a comment