Berkeley Math Circle: Monthly Contest 2

Solution:
We prove the following more general result, called Carnot's theorem: in a triangle $ABC$ we have
$$\cos A + \cos B + \cos C = 1 + \frac{r}{R}$$
where $r$ and $R$ are the inradius and circumradius of $\triangle ABC$.

The altitude from vertex $C$ divides side $AB$ into two segments (one of which may be negative), giving $b \cos A + a \cos B = c$. The other two altitudes give $b \cos C + c \cos A = b$ and $c \cos B + b \cos C = a$.

Adding all three equations to $a \cos A + b \cos B + c \cos C$ gives
$$(a + b + c)(\cos A + \cos B + \cos C) = (a + b + c) + a \cos A + b \cos B + c \cos C$$
So
$$\cos A + \cos B + \cos C = 1 + \frac{a \cos A + b \cos B + c \cos C}{a + b + c}.$$
The area of $ABC$ is $r(a + b + c)/2$ and from the 3 triangles into which circumradii divide the triangle $R(a \cos A + b \cos B + c \cos C)$. Hence $r/R = \frac{a \cos A + b \cos B + c \cos C}{a + b + c}$. Substituting this in, we have proved Carnot's theorem.

In the present problem, we have $\cos A = \frac{r}{R}$ since the distance from $O$ to $BC$ equals $r$. This implies the problem.