Trigonometry! #63

Given condition is $2b = a + c$ .

Using sine formula, $2\sin B = \sin C + \sin A$ , or , $4\sin \frac{B}{2} \cos \frac{B}{2} = 2 \sin (\frac{C+A}{2}) \cos (\frac{C-A}{2})$ , or $2\sin \frac{B}{2} \cos \frac{B}{2} = \sin (\frac{\pi - B}{2}) \cos (\frac{C-A}{2})$ , or $2\sin \frac{B}{2} \cos \frac{B}{2} = \sin (\frac{\pi }{2} - \frac{B}{2}) \cos (\frac{C-A}{2})$ , or $2\sin \frac{B}{2} \cos \frac{B}{2} = \cos \frac{B}{2} \cos (\frac{C-A}{2})$ , or $2\sin \frac{B}{2} = \cos (\frac{C-A}{2})$ .

This implies that $\sin \frac{B}{2} \leq \frac{1}{2}$ . Only one value from the options satisfies.

As the side lengths are in A.P. let them be $a = b - k, b$ and $c = b + k$ for some real number $k.$ Then with $\angle ABC$ being opposite side $b$ , we have by the Cosine rule that

$b^{2} = (b - k)^{2} + (b + k)^{2} - 2(b - k)(b + k)\cos(\angle ABC)$

$\Longrightarrow b^{2} = 2b^{2} + 2k^{2} - 2(b^{2} - k^{2})\cos(\angle ABC)$

$\Longrightarrow \cos(\angle ABC) = \dfrac{b^{2} + 2k^{2}}{2(b^{2} - k^{2})}.$

Now as $k^{2} \ge 0$ this expression will achieve a minimum value at $k = 0$ , i.e., when $\cos(\angle ABC) = \frac{1}{2}$ . Thus we are looking for angles $\angle ABC$ such that $\cos(\angle ABC) \ge \frac{1}{2}$ on the interval $[0^{\circ}, 90^{\circ}]$ . The only such angles are those between $0^{\circ}$ and $60^{\circ}$ inclusive, so the only possible value for $\angle ABC$ from the given options is $\boxed{45^{\circ}}$ .

Comments: In order to have a non-degenerate triangle we will require that $|k| \lt b.$ So assuming, without loss of generality, that $k \gt 0$ , and letting $k = rb$ for some positive real number $r \lt 1$ , we see that

$\cos(\angle ABC) = \dfrac{1 + 2r^{2}}{2(1 - r^{2})}$ .

Now since $\frac{1}{2} \le \cos(\angle ABC) \le 1$ , we must have $0 \le r \le \frac{1}{2}$ .