Bariz - Problem

Source: 2023 Macedonian Team Selection Test P2

Tags: geometric inequality, inequalities, geometry, geometric transformation, reflection

steppewolf

21.05.2023 16:02

Let $ABC$ be an acute triangle such that $AB<AC$ and $AB<BC$. Let $P$ be a point on the segment $BC$ such that $\angle APB = \angle BAC$. The tangent to the circumcircle of triangle $ABC$ at $A$ meets the circumcircle of triangle $APB$ at $Q \neq A$. Let $Q'$ be the reflection of $Q$ with respect to the midpoint of $AB$. The line $PQ$ meets the segment $AQ'$ at $S$. Prove that $$\frac{1}{AB}+\frac{1}{AC} > \frac{1}{CS}.$$ Proposed by Nikola Velov

Let $AB =c, BC=a, CA=b$. We have $\angle BAC= \angle APB = \alpha$. Clearly, $\triangle BAC \sim \triangle BPA \Rightarrow BA^2=BP\cdot BC \Rightarrow BP = \frac{c^2}{a}$. Angle chasing gives: $\angle ACB = \angle QAB =\angle QPB= \angle BAP = \gamma$. Therefore, $QP\| AC$. Now $\angle BAQ^{'}= \angle QBA = \angle QPA= \angle PAC \Rightarrow \angle CAS = \gamma$. From $\angle CAS = \angle PCA= \gamma$ and because $SP\|AC$, we get that $ASPC$ is inscribed $\Rightarrow \angle ACS=\angle APQ$. Hence, $\triangle AQB \sim \triangle ASC \Rightarrow \frac{AB}{AC} = \frac{QB}{SC} \Rightarrow SC = \frac{b\cdot QB}{c}$. Since $\angle QAB = \angle BAP = \gamma$ and because $QABP$ is inscribed, we obtain that $BQ=BP =\frac{c^2}{a} $. So $SC = \frac{b\cdot BP}{c} = \frac{bc^2}{ac} = \frac{bc}{a}$. We want to show that $\frac{1}{b}+ \frac{1}{c} = \frac{a}{bc} \Leftrightarrow b+c>a$, which is clear.