Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 3.4 AMC 10/12 Find the answer

Given triangle $ABC$ , let the sides opposite to angles $A$ , $B$ , and $C$ be $a$ , $b$ , and $c$ respectively, with $a=1$ and $a\cos C +\frac{1}{2}c = b$ . Find the range of the perimeter $l$ of triangle $ABC$ .

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

From $a\cos C +\frac{1}{2}c = b$ , we obtain $\sin A\cos C +\frac{1}{2}\sin C = \sin B$ .

Now, $\sin B = \sin(A+C) = \sin A\cos C + \cos A\sin C$ .

Thus, $\frac{1}{2}\sin C = \cos A\sin C$ . Since $\sin C \neq 0$ , we have $\cos A = \frac{1}{2}$ .

Given that $0 a = 1$ ,

We have $l = a + b + c > 2$ .

Thus, the range of the perimeter $l$ of triangle $ABC$ is $(\boxed{2, 3}]$ .

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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.