5. Prove that the ratio of the area of the triangle formed by the points of contact of the inscribed circle of a triangle $ABC$ to the area of $ABC$ is half the ratio of the radii of the inscribed and circumscribed circles of the triangle $ABC$.
Solution:
$\;\;\;$ Let $L, M, N$ be the points of contact. Then if $\Delta'$ be the area of $LMN,$ we have
$\displaystyle \Delta' = \frac12r^2 \cdot \Sigma \sin A = 2r^2 \cdot \Pi \cos \frac{A}2,\;\;$ and $\;\;\Delta = 2R^2 \cdot \Pi \sin A.$
Also $\;\;\displaystyle r = 4R \cdot \Pi \sin \frac{A}2,\;\;$ whence $\;\;\displaystyle \frac{\Delta'}{\Delta} = \frac{r}{2R}.$
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