IMO 1981 SL 11
On a semicircle with unit radius four consecutive chords AB, BC,
IMO 1981 SL 11
Origin: NET
Problem
On a semicircle with unit radius four consecutive chords AB, BC, CD, DE with lengths a, b, c, d, respectively, are given. Prove that a2 + b2 + c2 + d2 + abc + bcd < 4.
Solution
Let us denote the center of the semicircle by O, and \angleAOB = 2\alpha, \angleBOC = 2\beta, AC = m, CE = n. We claim that a2 +b2+n2+abn = 4. Indeed, since a = 2 sin \alpha, b = 2 sin \beta, n = 2 cos(\alpha + \beta), we have a2 + b2 + n2 + abn = 4(sin2 \alpha + sin2 \beta + cos2(\alpha + \beta) + 2 sin \alpha sin \beta cos(\alpha + \beta)) = 4 + 4 −cos 2\alpha −cos 2\beta
- cos(\alpha + \beta) cos(\alpha −\beta) = 4 + 4 (cos(\alpha + \beta) cos(\alpha −\beta) −cos(\alpha + \beta) cos(\alpha −\beta)) = 4. Analogously, c2 + d2 + m2 + cdm = 4. By adding both equalities and subtracting m2 + n2 = 4 we obtain a2 + b2 + c2 + d2 + abn + cdm = 4. Since n > c and m > b, the desired inequality follows.