Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 4.7 AIME Find the answer

4. Given the three sides of $\triangle A B C$ are $a, b, c$ , and
$\frac{a}{b}+\frac{a}{c}=\frac{b+c}{b+c-a} \text {. }$

Then the triangle that satisfies this condition is

Pick one

Solution

4. B.

Given $\frac{a(b+c)}{b}=\frac{b+c}{b+c-a}$ , eliminating the denominators and factoring yields $(b+c)(b-a)(c-a)=0$ . Since $a$ , $b$ , and $c$ are the lengths of the sides of a triangle, $b+c>0$ . Therefore, $b-a=0$ or $c-a=0$ , i.e., $b=a$ or $c=a$ . Thus, the triangle that satisfies this condition is an isosceles triangle with $a$ as the base.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.