1990 APMO Problems/Problem 1

Let .I be the intersection of" AG and EF. Let 6 = AI.IG - FI.IE. Then AI:ADlz, IG:ADl6, FI:BCl4:IE. (1) Fur"ther, applying the cosine rule to triangles ABD, ACD we get i AB2 : BC2I4 + AD2 _ AD.BC.cos LBDA, .. ACz : aC2l+ + AD2 + AD.BC.cos IBDA, f ,.*j;-\, so ADz : (Or, + AC2 - BCr 12) 12. -H"rr"" 6 - (or, + ACz - zBC2) lz4

(+ea.eC.cos LBAC - AB2 - AC') .

d.] Now AEFG is a cyclic quadrilateral if and only if 6 : 0, i.e. if and only if cos LBAC _ (Or, + ABr) I $.AB.AC)

(AB IAC + AC IAB) 14.

EI EI -sB EI Now AB/A C + AC IAB > 2. Hence cos LBAC > Llz and so LBAC S 60'. For LBAC > 60o there is no triangle with the required property. For LBAC:60o there exists, within similarity, precisely one triangle (which is equilateral) having the required property. For LBAC < 60o there exists, within similaritS again precisely one triangle having the required property (even though for fixed median AE there arq two, but one arises from the other by interchanging point B with point C, thus proving them to be similar)