# 1979 USA Mathematical Olympiad, Problem 4

Problem 4 of USA Mathematical Olympiad 1979

Show how to construct a chord FPE of a given angle A through a fixed point P within the angle A such that (1/FP) + (1/PE) is a maximum.

Solution by Steve Dinh, a.k.a. Vo Duc Dien Let EP = p, FP = q, AF = a, AP = l, AE = b, ∠EAP = α, ∠FAP = ε, ∠EPA = η, ∠EFA = γ. Extend FE and from A draw a perpendicular line to intercept this extension at G. Let AG = h.

1/EP + 1/FP = 1/p + 1/q = (p + q)/ (pq)

Applying the law of the sine function, we obtain (p + q)/sin(α + ε) = b /sinγ and b/p = sinη /sinα or

(p + q)/ (pq) = bsin(α + ε) / (pqsinγ) = sin(α + ε)sinη /(q sinγ sinα)

but α and ε are constants, so 1/p + 1/q is maximum when sinη /(q sinγ) is maximum.

We also have sinη = h/l and sinγ = h/a and now sinη/(q sinγ) = ha/(qlh) = a/(ql) but l is constant, so a/(ql) is maximum is when ratio a/q is maximum.

a/q = sin(180° - η)/sinε = sinη/sinε and with angle ε fixed, a/q is maximum when sinη is maximum or is equal to 1 when η = 90° as line MN represents.