Bariz - Problem

Source: 2022 Turkey JBMO TST P7 + 2023 Dutch BxMO TST, Problem 4

Tags: geometry, circumcircle

BarisKoyuncu

15.03.2022 23:41

In a triangle $\triangle ABC$ with $\angle ABC < \angle BCA$, we define $K$ as the excenter with respect to $A$. The lines $AK$ and $BC$ intersect in a point $D$. Let $E$ be the circumcenter of $\triangle BKC$. Prove that \[\frac{1}{|KA|} = \frac{1}{|KD|} + \frac{1}{|KE|}.\]

We need to prove that $\frac{|KA|}{|KD|}+\frac{|KA|}{|KE|}=1$. Simple Angle-Chasing gives us $\angle ADC=\frac{C-B}2$. Hence, $$\frac{|KA|}{|KD|}=\frac{|AC|}{|CD|}=\frac{\sin{ADC}}{\sin{CAD}}=\frac{\sin{(\frac{C-B}2)}}{\cos{(\frac A2)}}$$Also, $$\frac{|KA|}{|KE|}=\frac{2|KA|}{2R_{BKC}}=\frac{2|KA|\cdot\sin{(\frac B2)}}{|KC|}=\frac{2\cos{(\frac C2)\cdot \sin{(\frac B2)}}}{\cos{(\frac A2)}}$$Therefore, it suffices to show that $$1=\frac{\sin{(\frac{C-B}2)}}{\cos{(\frac A2)}}+\frac{2\cos{(\frac C2)\cdot \sin{(\frac B2)}}}{\cos{(\frac A2)}}=\frac{\sin{(\frac{C+B}2)}}{\cos{(\frac A2)}}\Leftrightarrow \sin{\left(\frac{C+B}2\right)}=\cos{\left(\frac A2\right)}$$, which is correct. Done.