Problem. (Proposed by Floor van Lamoen, Goes, Holland.) Let ABC be a triangle with orthocenter H, incenter I, and circumcenter O. Let Z[P,r] denote a circle Z with P as center and r as radius. Prove that the radical center of Z₁[A, CA+AB], Z₂[B, AB+BC], Z₃[C, BC+CA] is the point obtained by reflecting H through O and then reflecting the result through I.

    Proof.We shall project everything to the line BC. Let P' denote the projection of any point P on BC. Using the usual notation, then
BH' = c cos B = 2 R sin C cos B, where R is circumradius,
BI' = s - b = R (sin A + sin C - sin B), and BO' = a/2 = R sin A.
Let X be the reflection of H through O, then since O' is the midpoint of BC, it is the midpoint of H'X', so that BX' = CH' = 2 R sin B cos C.
Let Y be the reflection of X through I, then
BY' = 2 BI' - BX' = 2 R (sin A + sin C - sin B - sin B cos C)
= 2 R (sin C - sin B + sin C cos B) = c - b + c cos B.
    On the other hand, the circles Z₂, Z₃ must intersect at some common chord QQ", which is their radical axis. BC is the perpendicular bisector of QQ", cutting QQ" at Q'.
Then BQ'² - CQ'² = BQ² - CQ² = (a + c)² - (a + b)² = 2 a (c - b) + c² - b².
Divide by BQ' + CQ' = a, \ BQ' - CQ' = 2 (c - b) + (c² - b²)/a,
or BQ' = (½)((BQ' - CQ') + (BQ' + CQ')) = c - b + (a² + c² - b²)/ 2 a
= [by law of cosines] c - b + c cos B.     \ BQ' = BY'.
    This means the radical center Z₀ of Z₁, Z₂, Z₃ has the same projection on BC as Y. By symmetry, Z₀ and Y have the same projections on BC, CA, and AB. Hence Y = Z₀. QED