Trigonometry! #67

For any triangle we have

$\cos A + \cos B + \cos C = 1 + \frac{r}{R}$

where $r$ and $R$ are the radius of the incircle and circumcircle respectively.

Since

$\cos A + \cos B + \cos C =\frac{3}{2}$

we have

$\frac{r}{R}=\frac{1}{2}$

However from $\textbf{Euler's Inequality}$ we also have $R \ge 2r$ with equality iff the triangle is equilateral. Therefore the triangle is equilateral.

All triangles obey this rule: Cos A + Cos B + Cos C = (r/R) + 1; where r and R represent the radii of the inscribed and circumscribed circles respectively. Now, Euler's Inequality Relation states that for R> or = 2, triangle is equilateral. Given that Cos A + Cos B + Cos C = 3/2; We have that 3/2 = (r/R) + 1;

》》r/R = 1/2.

》R = 2r.

This satisfies Euler's Inequality, hence triangle is equilateral. ..