Given any triangle ABC, construct external triangles ABR, BCP, CAQ on the sides, so that ∠PBC = 45o, ∠PCB = 30o, ∠QAC = 45o, ∠QCA = 30o, ∠RAB = 15o, ∠RBA = 15o. Prove that ∠QRP = 90o and QR = RP.
Solution
Trigonometry provides a routine solution. Let BC = a, CA = b, AB = c. Then, by the sine rule applied to AQC, AQ = b/(2 sin 105o) = b/(2 cos 15o). Similarly, PB = a/(2 cos 15). Also AR = RB = c/(2 cos 15o). So by the cosine rule RP2 = (a2 + c2 - 2ac cos(B+60o))/(4 cos215o), and RQ2 = (b2 + c2 - 2bc cos(A+60o))/(4 cos215o). So RP = RQ is equivalent to: a2 - 2ac cos(60o+B) = b2 - 2bc cos(60o+A) and hence to a2 - ac cos B + √3 ac sin B = b2 - bc cos A + √3 bc sin A. By the sine rule, the sine terms cancel. Also b - b cos A = a cos C, and a - c cos B = b cos C, so the last equality is true and hence RP = RQ. We get an exactly similar expression for PQ2 and show that it equals 2 RP2 in the same way.
A more elegant solution is to construct S on the outside of AB so that ABS is equilateral. Then we find that CAS and QAR are similar and that CBS and PBR are similar. So QR/CS = PR/CS. The ratio of the sides is the same in each case (CA/QA = CB/PB since CQA and CPB are similar), so QR = PR. Also there is a 45o rotation between QAR and CAS and another 45o rotation between CBS and PBR, hence QR and PR are at 90o.
Solutions are also available in: Samuel L Greitzer, International Mathematical Olympiads 1959-1977, MAA 1978, and in István Reiman, International Mathematical Olympiad 1959-1999, ISBN 189-8855-48-X.
© John Scholes
jscholes@kalva.demon.co.uk
11 Oct 1998