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Part of 1992 IMO Longlists

Problems(1)

Geometric equations IMO LongList 1992 SPA2

Source:

9/2/2010
In a triangle, a symmedian is a line through a vertex that is symmetric to the median with the respect to the internal bisector (all relative to the same vertex). In the triangle ABCABC, the median mam_a meets BCBC at A′A' and the circumcircle again at A1A_1. The symmedian sas_a meets BCBC at MM and the circumcircle again at A2A_2. Given that the line A1A2A_1A_2 contains the circumcenter OO of the triangle, prove that:
(a) AA′AM=b2+c22bc;\frac{AA'}{AM} = \frac{b^2+c^2}{2bc} ;
(b) 1+4b2c2=a2(b2+c2)1+4b^2c^2 = a^2(b^2+c^2)
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