TST for BMO

Geometry Difficulty 8.3 Shortlist Prove it Bulgaria

Given a scalene triangle $ABC$ . On the rays $AC \rightarrow$ and $BC \rightarrow$ , the points $C_a$ and $C_b$ are chosen, respectively, such that $AC_a = BC_b = AB$ . We denote by $O_c$ the center of the circumcircle about $\triangle CC_a C_b$ . Analogously, we define the points $O_a$ and $O_b$ . Prove that the lines $AO_a$ , $BO_b$ and $CO_c$ intersect at a point that lies on the circumcircle of the triangle $ABC$ .

(Alexander Ivanov)

Solution

Let $O$ and $I$ be the centers of the circumscribed and inscribed circle of $ABC$ , and $A_1$ , $B_1$ , $C_1$ be the centers of the arcs $\widehat{AB}$ , $\widehat{AC}$ , $\widehat{BC}$ (not containing the third vertices) of the circumscribed circle. By symmetry with respect to $AA_1$ we have $A_1C_a = A_1B = A_1C$ and together with $O_cC = O_cC_a$ it follows that $A_1O_c$ is the segment bisector of $CC_a$ . Analogously, $B_1O_c$ is the segment bisector of $CC_b$ . Thus, the quadrilateral $OA_1O_cB_1$ is a parallelogram with $OA_1 = OB_1$ , i.e. rhombus. In other words, $O$ and $O_c$ are symmetric about $A_1B_1$ , and this is also true for $C$ and $I$ , i.e. $CO_c$ is symmetric to $OI$ with respect to $A_1B_1$ . Analogously, $AO_a$ and $BO_b$ are symmetric on $OI$ with respect to $B_1C_1$ and $A_1C_1$ , respectively. Since $I$ is the orthocenter of $A_1B_1C_1$ , finally the lines $AO_a$ , $BO_b$ , and $CO_c$ intersect at the Anti-Steiner point for the line $OI$ and triangle $A_1B_1C_1$ (circumscribed about $ABC$ ). $\square$

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