Cities A, B and C, together with the straight roads connecting them, form a triangle. It is known that the direct

Cities A, B and C, together with the straight roads connecting them, form a triangle. It is known that the direct route from A to B is 300 kilometers shorter than the detour through C, and the direct route from A to C is 100 kilometers shorter than the detour through B. What is the smallest number of kilometers between cities B and C?

Solution. Let the length of the road AB = x km, the length of the road AC = y km, the length of the road BC = z km, since it is known from the condition of the problem that the cities A, B and C together with the straight roads connecting them form a triangle. Since the direct path from A to B is 300 kilometers shorter than the detour through C, we get the equality AC + BC = AB + 300 or y + z = x + 300. The direct path from A to C is 100 kilometers shorter than the detour through B, we get the equality AB + ВС = АС + 100 or х + z = у + 100. Let us add term-by-term equalities and find the smallest number of kilometers that can be between cities B and C, that is ВС:
(y + z) + (x + z) = (x + 300) + (y + 100); 2 ∙ z = 300 + 100; z = 200 (km) – the smallest number of kilometers that can be between cities B and C.
Answer: between cities B and C the least number of kilometers can be 200.