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Geometry Difficulty 4.7 AIME Prove it Belarus

Given a triangle ABCABC. Let SS be the circle passing through CC centered at AA. Let XX be a variable point on SS, and let KK be the midpoint of the segment CXCX.
Find the locus of the midpoints of BKBK, when XX moves along SS.

Solution

Let RR be the radius of the given circle, MM the midpoint of ACAC, NN the midpoint of BMBM, TT the midpoint of BKBK. Let a,b,c,m,n,t,ka, b, c, m, n, t, k, xx be the (complex) coordinates of the points A,B,C,M,N,T,XA, B, C, M, N, T, X, respectively. Then m=(a+c)/2m = (a+c)/2, n=(b+m)/2n = (b+m)/2, k=(c+x)/2k = (c+x)/2, t=(k+b)/2t = (k+b)/2. So we can easily find x=4t2bcx = 4t - 2b - c, n=(2b+c+a)/2n = (2b + c + a)/2. The complex equation of the given circle is xa=R|x - a| = R, i.e. 4t2bca=R|4t - 2b - c - a| = R, or t(2b+c+a)/4=R/4|t - (2b + c + a)/4| = R/4, or tn=R/4|t - n| = R/4. That is the locus of points TT is the circle of radius R/4R/4 with the center at NN.

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