Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 6.8 National olympiad Prove it Belarus

Three cyclists start from town $A$ simultaneously. They move along the closed route consisting of three straight-line segments $AB$ , $BC$ and $CA$ . The speeds of the first cyclist on these segments are $12$ , $10$ and $15$ kilometers per hour, respectively. The speeds of the second cyclist are $15$ , $15$ and $10$ (km/h) and the speeds of the third cyclist are $10$ , $20$ and $12$ (km/h).
Find the value of the angle $ABC$ if all three cyclists finish at town $A$ at the same time.

Solution

Answer: $90^\circ$ .
Let $AB = a$ , $BC = b$ , $CA = c$ (km). We find the time of each cyclist to cover the route (this time is independent of the moving direction). By condition,
$\frac{a}{12} + \frac{b}{10} + \frac{c}{15} = \frac{a}{15} + \frac{b}{15} + \frac{c}{10} = \frac{a}{10} + \frac{b}{20} + \frac{c}{12},$
thus
$5a + 6b + 4c = 4a + 4b + 6c = 6a + 3b + 5c. \quad (*)$
Consequently, $2c = a + 2b$ and $c = 2a - b$ . So $a + 2b = 2(2a - b)$ which implies $3a = 4b$ . Similarly, from $(*)$ it follows that $3c = 5b$ . Then $a = \frac{4b}{3}$ and $c = \frac{5b}{3}$ . Setting $b = 3x$ we obtain $a = 4x$ and $c = 5x$ . Now it is easy to see that $a^2 + b^2 = c^2$ . Hence, the triangle $ABC$ is a right-angled triangle with $c$ as hypotenuse. Therefore, $\angle ABC = 90^\circ$ .

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