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Belarus — Selection and Training Session

Geometry Difficulty 4.7 AIME Prove it Belarus

Given a triangle $ABC$ . Let $S$ be the circle passing through $C$ centered at $A$ . Let $X$ be a variable point on $S$ , and let $K$ be the midpoint of the segment $CX$ .
Find the locus of the midpoints of $BK$ , when $X$ moves along $S$ .

Solution

Let $R$ be the radius of the given circle, $M$ the midpoint of $AC$ , $N$ the midpoint of $BM$ , $T$ the midpoint of $BK$ . Let $a, b, c, m, n, t, k$ , $x$ be the (complex) coordinates of the points $A, B, C, M, N, T, X$ , respectively. Then $m = (a+c)/2$ , $n = (b+m)/2$ , $k = (c+x)/2$ , $t = (k+b)/2$ . So we can easily find $x = 4t - 2b - c$ , $n = (2b + c + a)/2$ . The complex equation of the given circle is $|x - a| = R$ , i.e. $|4t - 2b - c - a| = R$ , or $|t - (2b + c + a)/4| = R/4$ , or $|t - n| = R/4$ . That is the locus of points $T$ is the circle of radius $R/4$ with the center at $N$ .

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