Baltic Way 1996 · Problem 5

Solution:

For a triangle $MNK$ with in-radius $r$ and circumradius $R$, the equality

$$\cos \angle M + \cos \angle N + \cos \angle K = 1 + \frac{r}{R}$$

holds; this follows from the cosine theorem and formulas for $r$ and $R$.

We have $\angle ACB = \angle ADB$, $\angle BDC = \angle BAC$, $\angle CAD = \angle CBD$ and $\angle DBA = \angle DCA$. Denoting these angles by $\alpha, \beta, \gamma$ and $\delta$, respectively, we get $r_{a} = (\cos \beta + \cos \gamma + \cos (\alpha + \delta) - 1) R$ and $r_{c} = (\cos \alpha + \cos \delta + \cos (\beta + \gamma) - 1) R$.

Since $\cos (\alpha + \delta) = -\cos (\beta + \gamma)$, we get

$$r_{a} + r_{c} = (\cos \alpha + \cos \beta + \cos \gamma + \cos \delta - 2) R.$$

Similarly,

$$r_{b} + r_{d} = (\cos \alpha + \cos \beta + \cos \gamma + \cos \delta - 2) R,$$

where $R$ is the circumradius of the quadrangle $ABCD$.