Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 8.2 Shortlist Prove it Turkey

In a triangle $ABC$ with $\hat{B} < \hat{C}$ , let $K$ be the center of the excircle that is tangent to the side $[AC]$ . The lines $AK$ and $BC$ intersect at $D$ , and $E$ is the center of the circumcircle of $BKC$ . Prove that
$\frac{1}{|KA|} = \frac{1}{|KD|} + \frac{1}{|KE|}.$

Solution

We will show that $\frac{|KA|}{|KD|} + \frac{|KA|}{|KE|} = 1$ . Let $\angle A = 2\alpha$ , $\angle B = 2\beta$ and $\angle C = 2\theta$ . Recall that $\alpha + \beta + \theta = 90^\circ$ . By angle chasing we obtain that $\angle ADC = \theta - \beta$ , $\angle ABK = \angle KBD = \beta$ , $\angle BAK = 180^\circ - \theta - \beta$ and hence we get
$\frac{|KA|}{|KD|} = \frac{|AB|}{|BD|} = \frac{\sin(\theta - \beta)}{\sin(180^\circ - \theta - \beta)} = \frac{\sin\theta \cos\beta - \cos\theta \sin\beta}{\cos\alpha} \quad (1)$
We also have $\angle CAK = \beta + \theta$ and $\angle ACK = \alpha + \beta$ . Therefore,
$\frac{|KA|}{|KE|} = \frac{|KA|}{|KC|} \cdot \frac{|KC|}{|KE|} = \frac{\sin(\alpha + \beta)}{\sin(\beta + \theta)} \cdot 2\sin\beta = \frac{2\sin\beta\cos\theta}{\cos\alpha} \quad (2)$
Adding up (1) and (2) yields
$\frac{|KA|}{|KD|} + \frac{|KA|}{|KE|} = \frac{\sin\theta \cos\beta + \cos\theta \sin\beta}{\cos\alpha} = \frac{\sin(\theta + \beta)}{\cos\alpha} = \frac{\cos\alpha}{\cos\alpha} = 1$
and the result follows.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.