Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 4.7 AIME Find the answer

13. In $\triangle A B C$ , $\angle A, \angle B, \angle C$ are opposite to sides $a, b, c$ respectively. Let
$\begin{array}{l} f(x)=\boldsymbol{m} \cdot \boldsymbol{n}, \boldsymbol{m}=(2 \cos x, 1), \\ \boldsymbol{n}=(\cos x, \sqrt{3} \sin 2 x), \\ f(A)=2, b=1, S_{\triangle A B C}=\frac{\sqrt{3}}{2} . \\ \text { Then } \frac{b+c}{\sin B+\sin C}= \end{array}$

Your answer

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

Solution

$=13.2$ .
It is easy to know, $f(x)=1+2 \sin \left(2 x+\frac{\pi}{6}\right)$ .
Combining the conditions, we get $\angle A=\frac{\pi}{3}, c=2, a=\sqrt{3}$ . Therefore, $\frac{a}{\sin A}=\frac{b+c}{\sin B+\sin C}=2$ .

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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.