1969 Canadian MO Problems/Problem 5

Problem

Let $\displaystyle ABC$ be a triangle with sides of length $\displaystyle a$ , $\displaystyle b$ and $\displaystyle c$ . Let the bisector of the $\displaystyle \angle C$ cut $\displaystyle AB$ at $\displaystyle D$ . Prove that the length of $\displaystyle CD$ is $\displaystyle \frac{2ab\cos \frac{C}{2}}{a+b}.$

Solution

Let $\displaystyle CD=d.$ Note that $\displaystyle [\triangle ABC]=[\triangle ACD]+[\triangle BCD].$ This can be rewritten as $\displaystyle \frac12 a b \sin C=\frac12 a d \sin \frac C2 +\frac12 b d \sin \frac C2 .$

Because $\displaystyle \sin C=2\sin \frac C2 \cos \frac C2,$ the expression can be written as $\displaystyle2ab\cos \frac C2=d(a+b).$ Dividing by $\displaystyle a+b,$ $CD=\frac{2ab\cos \frac C2}{a+b},$ as desired.

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1969 Canadian MO Problems/Problem 5

Problem

Solution