3rd Eötvös 1896 Problem 3 Given three points P, Q, R in the plane, find points A, B, C such that P is the foot of the perpendicular from A to BC, Q is the foot of the perpendicular from B to CA, and R is the foot of the perpendicular from C to AB. Find the lengths AB, BC, CA in terms of PQ, QR and RP.
Take I to be the incenter of PQR and A, B, C to be the three excenters. Then AP is the internal bisector of ∠QPR and is perpendicular to BC, the external bisector. Similarly for the other points. Of course, we could also take I, B, C or A, I, C or A, B, I as the three points. We derive some bookwork formulae which may not be familiar. For a triangle ABC with sides a,b,c as usual, put s = (a+b+c)/2 and take the inradius to be r. Suppose the vertex A is a distance x from the two points of contact of the incircle with the sides AB, AC. Then B and C must be c-x and b-x from their points of contact, so a = (c-x)+(b-x). Hence x = s-a. So tan A/2 = r/(s-a). But the area of the triangle is √(s(s-a)(s-b)(s-c)) by Heron or rs by considering it as made up of ABI, BCI, CAI. Hence r = √((s-a)(s-b)(s- c)/s) and tan A/2 = √((s-b)(s-c)/((s-a)s)). Now using sec2A/2 = 1 + tan2A/2 and noting that s(s-a) + (s-b)(s- c) = bc, we find sec 2 A/2 = bc/(s(s-a)). Hence sin A/2 = tan A/2 cos A/2 = √( (s-b)(s-c)/bc) (*). ARPC is cyclic (because ∠APC = ∠ARC = 90 o ), so x = ∠APR = ∠ACR = 90 o - ∠A. Similarly, y = 90 o - ∠B, z = 90 o - ∠C. Hence ABC is similar to AQR. So QR/BC = AQ/AB = cos A = sin x. Hence BC = QR/sin x. We are now home, because we can use the formulae (*) to get sin x, sin y, sin z. 3rd Eötvös 1896 © John Scholes jscholes@kalva.demon.co.uk 29 Oct 2003 Last corrected/updated 29 Oct 03