Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.0 AIME, harder Find the answer

Question 4 In $\triangle A B C$ , the sides opposite to angles $A, B, C$ are $a, b, c$ respectively, and $a^{n}+b^{n}=c^{n}(n \geqslant 3)$ . Then the shape of $\triangle A B C$ is $(\quad)$ .

Pick one

Solution

Solution: Clearly, $c$ is the largest side. Since $\left(\frac{a}{c}\right)^{n}+\left(\frac{b}{c}\right)^{n}=1$ ( $n \geqslant 3$ ), we have
$\left(\frac{a}{c}\right)^{2}+\left(\frac{b}{c}\right)^{2}>\left(\frac{a}{c}\right)^{n}+\left(\frac{b}{c}\right)^{n}=1$

Thus, $a^{2}+b^{2}>c^{2}$ . By the cosine rule, the largest angle $C$ of the triangle is acute, hence $\triangle A B C$ is an acute triangle. The correct choice is B.

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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.