20th USAMO 1991 Problem 1 An obtuse angled triangle has integral sides and one acute angle is twice the other. Find the smallest possible perimeter. Solution
: 77 = 16 + 28 + 33. Let the triangle be ABC with angle A obtuse and angle B = 2 angle C. Let the sides be a, b, c as usual. Note that a > b > c. We have b sin C = c sin 2C and c 2 = a 2 + b 2 - 2ab sin C. Hence, b/2c = sin 2C/sin C = cos C = (a 2 + b 2 - c 2 )/2ab. So ab 2 = a 2 c + b 2 c - c 3 . Hence b 2 (a - c) = c(a 2 - c 2 ). Dividing by a - c we get b 2 = c(a + c). Now the triangle with smallest perimeter will have a, b, c relatively prime (otherwise we could divide by the common factor). Hence c must be a square. For if c and a+c have a common factor, then so do a and c and hence a, b and c, which means they cannot be the minimal set. Clearly c is not 1 (or the triangle would have nil area). c = 4 gives a = 5, which is too short, or a ≥ 21, which is too long. c = 9 gives a = 7 (too short), 18 (3 divides a, b, c), or ≥ 40 (too long). c = 16 gives a = 20 (too short) or a = 33 which works. Larger c gives a larger perimeter. Eg c = 25 gives a = 56, b = 45 (perimeter 126). 20th USAMO 1991 © John Scholes jscholes@kalva.demon.co.uk 11 May 2002