Problem 4. Three frogs were sitting on a straight line. At first, one of them jumped, then another, and finally the third. Each frog landed exactly in the middle of the segment between the other two. It turned out that the lengths of two of these three jumps were 60 cm.

a) What could the length of the remaining jump be?

[2 points]

b) What could the distance between the two outermost frogs have been initially?

[4 points]

(A. V. Shapovalov)

Compare with Problem 2 for 6th grade (p. 3).
Answer. a) 30 cm or 120 cm; b) 100 cm or 160 cm.

Solution. Regardless of the initial positions of the frogs, after the first jump, the first frog (A) will land exactly in the middle of the segment between the other two.

![](https://cdn.mathpix.com/cropped/2024_05_06_c2c81ae80e218af9a7a0g-3.jpg?height=60&width=525&top_left_y=1290&top_left_x=466)

Next, one of the frogs sitting at the ends of this segment (let's say frog B) will jump. It will cover half the length of the segment and an additional quarter, i.e., a total of $3 / 4$ of its length. Now the frogs will be positioned as follows:

![](https://cdn.mathpix.com/cropped/2024_05_06_c2c81ae80e218af9a7a0g-3.jpg?height=87&width=523&top_left_y=1567&top_left_x=463)

Note that the distance between the outermost frogs was equal to the length of segment BC, and now it is half of that (AB is half the length of the initial segment BC).

The third jump will have the same property: its length will be $3 / 4$ of the current segment (AB), and the distance between the outermost frogs will again be halved.

From this, it follows that the length of the third jump is definitely less than the length of the second. Therefore, the lengths that can match are either the first and second jumps or the first and third jumps. Let's consider both cases.

1) The lengths of the first and second jumps are 60 cm. Therefore, the distance between the outermost frogs before the second jump was $60: 3 \cdot 4 = 80$ cm, and after this jump, it will become $80: 2 = 40$ cm. Then the length of the third jump will be $3 / 4$ of 40 cm, i.e., 30 cm, and the initial distance between the outermost frogs will be $60 + 40 = 100$ cm (frog A was initially 60 cm from the midpoint of the 80 cm segment).

![](https://cdn.mathpix.com/cropped/2024_05_06_c2c81ae80e218af9a7a0g-4.jpg?height=117&width=644&top_left_y=872&top_left_x=405)

2) The lengths of the first and third jumps are 60 cm. Therefore, the distance between the two outermost frogs before the third jump was $60: 3 \cdot 4 = 80$ cm. This means that before the second jump, the distance between the outermost frogs was twice as much - 160 cm. And since the first frog only jumped 60 cm, it was somewhere inside this segment.

![](https://cdn.mathpix.com/cropped/2024_05_06_c2c81ae80e218af9a7a0g-4.jpg?height=126&width=1023&top_left_y=1352&top_left_x=219)

Therefore, the initial distance between the outermost frogs (before the first jump) was also 160 cm.

Thus, both cases are possible. In the first case, the initial distance between the outermost frogs is 100 cm, and the lengths of the jumps are 60 cm, 60 cm, 30 cm. In the second case, the initial distance between the outermost frogs is 160 cm, and the lengths of the jumps are 60 cm, 120 cm, 60 cm.