Cosine Rule

In triangle $$ABC$$, $$BC = a = 7$$, $$AC = b = 8$$, $$AB = c = \alpha \in \mathbb{N}$$, and $$\cos A = 2/3$$.

Find $$\alpha$$ using the cosine rule.

$$a^2 = b^2 + c^2 - 2bc\cos A$$

$$49 = 64 + \alpha^2 - 2(8)(\alpha)(2/3) = 64 + \alpha^2 - \frac{32\alpha}{3}$$

$$\alpha^2 - \frac{32\alpha}{3} + 15 = 0$$

$$3\alpha^2 - 32\alpha + 45 = 0$$

$$\alpha = \frac{32 \pm \sqrt{1024 - 540}}{6} = \frac{32 \pm \sqrt{484}}{6} = \frac{32 \pm 22}{6}$$

$$\alpha = 9$$ or $$\alpha = 5/3$$ (not a natural number). So $$\alpha = 9$$.

Find $$\cos C$$ using the cosine rule.

$$c^2 = a^2 + b^2 - 2ab\cos C$$

$$81 = 49 + 64 - 2(7)(8)\cos C$$

$$81 = 113 - 112\cos C$$

$$\cos C = \frac{32}{112} = \frac{2}{7}$$

Compute $$\cos 3C$$.

$$\cos 3C = 4\cos^3 C - 3\cos C = 4\left(\frac{2}{7}\right)^3 - 3\left(\frac{2}{7}\right) = \frac{32}{343} - \frac{6}{7} = \frac{32 - 294}{343} = \frac{-262}{343}$$

Compute the expression.

$$49\cos 3C + 42 = 49 \cdot \frac{-262}{343} + 42 = \frac{-262}{7} + 42 = \frac{-262 + 294}{7} = \frac{32}{7}$$

So $$\frac{m}{n} = \frac{32}{7}$$ with $$\gcd(32, 7) = 1$$. Therefore $$m + n = 32 + 7 = 39$$.

The answer is 39.

We are given $$AB = 5$$, $$\angle B = \cos^{-1}\!\left(\frac{3}{5}\right)$$ (so $$\cos B = \frac{3}{5}$$ and $$\sin B = \frac{4}{5}$$), and circumradius $$R = 5$$.

By the law of sines, the side opposite to $$B$$ (i.e., $$AC$$) satisfies: $$\frac{AC}{\sin B} = 2R \implies AC = 2 \times 5 \times \frac{4}{5} = 8.$$

Now applying the law of cosines with $$BC = a$$: $$AC^2 = AB^2 + BC^2 - 2 \cdot AB \cdot BC \cdot \cos B$$ $$64 = 25 + a^2 - 2(5)(a)\!\left(\frac{3}{5}\right) = 25 + a^2 - 6a.$$

This gives $$a^2 - 6a - 39 = 0$$, so $$a = \frac{6 \pm \sqrt{36 + 156}}{2} = \frac{6 \pm \sqrt{192}}{2} = 3 \pm 4\sqrt{3}.$$

Since $$a = BC > 0$$, we have $$BC = 3 + 4\sqrt{3}$$.

The area of triangle $$ABC$$ is: $$\text{Area} = \frac{1}{2} \cdot AB \cdot BC \cdot \sin B = \frac{1}{2} \cdot 5 \cdot (3 + 4\sqrt{3}) \cdot \frac{4}{5} = 2(3 + 4\sqrt{3}) = 6 + 8\sqrt{3}.$$