In the triangle ABC, the midpoints of AC and AB are M and N respectively. BM and CN meet at P. Show that if it is possible to inscribe a circle in the quadrilateral AMPN (touching every side), then ABC is isosceles.
Solution
If the quadrilateral has an inscribed circle then AM + PN = AN + PM (consider the tangents to the circle from A, M, P, N). But if AB > AC, then BM > CN (see below). We have AN = AB/2, PM = BM/3, AM = AC/2, PN = CN/3, so it follows that AM + PN < AN + PM. Similarly, AB < AC implies AM + PN > AN + PM, so the triangle must be isosceles.
To prove the result about the medians, note that BM2 = BC2 + CM2 - 2 BC.CM cos C = (BC - CM cos C)2 + (CM sin C)2. Similarly, CN2 = (BC - BN cos B)2 + (BN sin B)2. But MN is parallel to BC, so CM sin C = BN sin B. But AB > AC, so BN > CM and B < C, so cos B > cos C, hence BN cos B > CM cos C and BC - CM cos C > BC - BN cos B. So BM > CN.
© John Scholes
jscholes@kalva.demon.co.uk
1 July 2002
Last corrected/updated 22 Oct 2002