Given a triangle $ \triangle{ABC} $ and a point $ O $. $ X $ is a point on the ray $ \overrightarrow{AC} $. Let $ X' $ be a point on the ray $ \overrightarrow{BA} $ so that $ \overline{AX} = \overline{AX_{1}} $ and $ A $ lies in the segment $ \overline{BX_{1}} $. Then, on the ray $ \overrightarrow{BC} $, choose $ X_{2} $ with $ \overline{X_{1}X_{2}} \parallel \overline{OC} $. Prove that when $ X $ moves on the ray $ \overrightarrow{AC} $, the locus of circumcenter of $ \triangle{BX_{1}X_{2}} $ is a part of a line.
Problem
Source: 2018 Taiwan TST Round 1
Tags: geometry, circumcircle
AlastorMoody
29.11.2019 18:58
Taiwan TST 2018 Round 1 Quiz 2 P1 wrote:
Given a triangle $ \triangle{ABC} $ and a point $ O $. $ X $ is a point on the ray $ \overrightarrow{AC} $. Let $ X' $ be a point on the ray $ \overrightarrow{BA} $ so that $ \overline{AX} = \overline{AX_{1}} $ and $ A $ lies in the segment $ \overline{BX_{1}} $. Then, on the ray $ \overrightarrow{BC} $, choose $ X_{2} $ with $ \overline{X_{1}X_{2}} \parallel \overline{OC} $.
Prove that when $ X $ moves on the ray $ \overrightarrow{AC} $, the locus of circumcenter of $ \triangle{BX_{1}X_{2}} $ is a part of a line.
Solution: Let $OC \cap AB=M$ and Let $D$ be center of $\odot (BX_1X_2)$. So, as $X$ varies on $ \overrightarrow{AC} $, $X_1,X_2$ also vary linearly and $\angle BMC=\angle BX_1X_2$ $=\frac{1}{2}$ $\angle BDX_2$ is fixed or $\Delta BMC \sim \Delta BX_1X_2$. Hence, by homothety at $B$, $D$ varies on a line passing through $B$ $\qquad \blacksquare$