Problem 5.5. The houses of Andrey, Boris, Vova, and Gleb are located in some order on a straight street. The distance between Andrey's and Boris's houses, as well as the distance between Vova's and Gleb's houses, is 600 m. What can the distance in meters between Andrey's and Gleb's houses be, if it is known that it is 3 times greater than the distance between Boris's and Vova's houses? List all possible options.
Answer: $900, 1800$.

Solution. For brevity, we will denote the houses of the residents by the first letter of their names. We will denote the distance between the houses of the corresponding people by two capital letters in a row.

Without loss of generality, A is to the left of B (otherwise, we will look at everything from the other side of the street). First, let's assume that V is to the right of G. Since A is 600 meters to the left of B, and G is 600 meters to the left of V, then AG equals BV (if we mentally move the segment connecting B and V 600 meters to the left, we get a segment connecting A and G). But according to the condition, they differ by 3 times, which is a contradiction.

![](https://cdn.mathpix.com/cropped/2024_05_06_6da73bfd3e09e8b55e3fg-10.jpg?height=85&width=505&top_left_y=1241&top_left_x=478)

Therefore, A is to the left of B, and V is to the left of G. Since AB = VG, V and G cannot simultaneously be between A and B, and A and B cannot simultaneously be between V and G. Then there are 4 possible cases for the arrangement of the houses.

Case 1. The order of the houses is: A, B, V, G. Since AB = 600, BV = x, VG = 600, then according to the condition, 600 + x + 600 = 3x, from which x = 600, then AG = 1800.

![](https://cdn.mathpix.com/cropped/2024_05_06_6da73bfd3e09e8b55e3fg-10.jpg?height=101&width=507&top_left_y=1599&top_left_x=477)

Case 2. The order of the houses is: A, V, B, G. Let BV = x. Since AB = VG = 600, then AB = BG = 600 - x. Then according to the condition, (600 - x) + x + (600 - x) = 3x, from which x = 300, then AG = 900.

![](https://cdn.mathpix.com/cropped/2024_05_06_6da73bfd3e09e8b55e3fg-11.jpg?height=101&width=509&top_left_y=95&top_left_x=476)

Case 3. The order of the houses is: V, A, G, B. It is easy to see that AG cannot be greater than BV, which is a contradiction.

![](https://cdn.mathpix.com/cropped/2024_05_06_6da73bfd3e09e8b55e3fg-11.jpg?height=107&width=517&top_left_y=322&top_left_x=468)

Case 4. The order of the houses is: V, G, A, B. It is easy to see that AG cannot be greater than BV, which is a contradiction.

![](https://cdn.mathpix.com/cropped/2024_05_06_6da73bfd3e09e8b55e3fg-11.jpg?height=111&width=507&top_left_y=550&top_left_x=477)

Therefore, only two variants are possible: 900 and 1800.