Bulgarian Mathematical Olympiad

Question

Consider acute ABC with altitudes AA_1, BB_1 and CC_1 (A_1 BC, B_1 AC, C_1 AB). A point C' on the extension of B_1A_1 beyond A_1 is such that A_1C' = B_1C_1. Analogously, a point B' on the extension of A_1C_1 beyond C_1 is such that C_1B' = A_1B_1 and point A' on the extension of C_1B_1 beyond B_1 is such that B_1A' = C_1A_1. Denote by A'', B'' and C'' the symmetric points of A', B' and C'' with respect to BC, CA and AB respectively. Prove that if R, R' and R'' are circumradiuses of ABC, A'B'C' and A''B''C'', then R, R' and R'' are side lengths of a triangle with area equals one half of the area of ABC.

Accepted Answer

Let O be circumcenter of ABC and S be its symmetric point with respect to AB. Then SOCH is a parallelogram. SH A_1B_1 and SH = R (1). Denote by the inradius of triangle A_1B_1C_1, and by q its semi perimeter. If T is the orthogonal projection of S to A_1B_1 then HT = , ST = R + , TC' = q. We have HC'' = ^2 + q^2 which due to symmetry implies that H is the circumcenter of A'B'C' and R' = HC' (2). On the other hand OC'' = SC' = (R+)^2 + q^2 and therefore O is the circumcenter of A''B''C'' and R'' = SC' (3). It follows from (1), (2) and (3) that R, R' and R'' are side lengths of triangle HSC' and S_HSC' = HS TC'2 = R q2 = 12 (S_A_1OB_1C + S_B_1OC_1A + S_C_1OA_1B) = S_ABC2.