Competition problem — Geometry — Olympiad Maths Prep

Library / /14 of 16

Geometry Difficulty 5.8 AIME, harder Prove it Ukraine

Three cyclists start off at the same time and ride along the sides of a triangle $ABC$ along the route $AB \to BC \to CA$ . Their speeds on each of the segments $AB$ , $BC$ , $CA$ are known: the first cyclist has speeds $12$ , $10$ and $20$ mph respectively on the three sides, the second one rides $15$ , $15$ and $10$ mph, the third one rides $10$ , $20$ and $12$ mph respectively. What can be the angle measure of $\angle ABC$ , if all three cyclists arrived back at the point $A$ simultaneously?

Solution

Denote the sides of the triangle by $AB = x$ , $BC = y$ , $CA = z$ . Then the following equality must hold:
$\frac{x}{12} + \frac{y}{10} + \frac{z}{20} = \frac{x}{15} + \frac{y}{15} + \frac{z}{10} = \frac{x}{10} + \frac{y}{20} + \frac{z}{12} \text{ or } 5x + 6y + 3z = 4x + 4y + 6z = 6x + 3y + 5z.$
Hence, $x + 2y - 3z = 0$ and $2x - y - z = 0$ , which implies $x = y$ and $z = y$ . Therefore, $\triangle ABC$ is equilateral and all its angles are equal to $60^\circ$ .

Looking for a route rather than an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.