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2017 LMT Team Round - Radical center lies on OI - Lexington Math Tournament

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January 16, 2022
radical axisRadical centergeometryLMT

Problem Statement

Let PP be a point and ω\omega be a circle with center OO and radius rr . We define the power of the point PP with respect to the circle ω\omega to be OP2−r2OP^2 - r^2 , and we denote this by pow (P,ω)(P, \omega). We define the radical axis of two circles ω1\omega_1 and ω2\omega_2 to be the locus of all points P such that pow (P,ω1)=(P,\omega_1) = pow (P,ω2)(P,\omega_2). It turns out that the pairwise radical axes of three circles are either concurrent or pairwise parallel. The concurrence point is referred to as the radical center of the three circles.
In △ABC\vartriangle ABC, let II be the incenter, Γ\Gamma be the circumcircle, and OO be the circumcenter. Let A1,B1,C1A_1,B_1,C_1 be the point of tangency of the incircle of △ABC\vartriangle ABC with side BC,CA,ABBC,CA, AB, respectively. Let X1,X2∈ΓX_1,X_2 \in \Gamma such that X1,B1,C1,X2X_1,B_1,C_1,X_2 are collinear in this order. Let MAM_A be the midpoint of BCBC, and define ωA\omega_A as the circumcircle of △X1X2MA\vartriangle X_1X_2M_A. Define ωB\omega_B ,ωC\omega_C analogously. The goal of this problem is to show that the radical center of ωA\omega_A, ωB\omega_B, ωC\omega_C lies on line OIOI.
(a) LetA1′ A'_1 denote the intersection of B1C1B_1C_1 and BCBC. Show that A1BA1C=A1′BA1′C\frac{A_1B}{A_1C}=\frac{A'_1B}{A'_1C}. (b) Prove that A1A_1 lies on ωA\omega_A. (c) Prove that A1A_1 lies on the radical axis of ωB\omega_B and ωC\omega_C . (d) Prove that the radical axis of ωB\omega_B and ωC\omega_C is perpendicular to B1C1B_1C_1. (e) Prove that the radical center of ωA\omega_A, ωB\omega_B, ωC\omega_C is the orthocenter of △A1B1C1\vartriangle A_1B_1C_1. (f ) Conclude that the radical center of ωA\omega_A, ωB\omega_B, ωC\omega_C , OO, and II are collinear.
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