The triangle ABC satisfies the relation cot2A/2 + 4 cot2B/2 + 9 cot2C/2 = 9(a+b+c)2/(49r2), where r is the radius of the incircle (and a = |BC| etc, as usual). Show that ABC is similar to a triangle whose sides are integers and find the smallest set of such integers.
Solution
Answer: a =13, b = 40, c = 45.
Let the incenter be I. Consider the triangle IBC. It has angle IBC = B/2, angle ICB = C/2 and height r. Hence a = r cot B/2 + r cot C/2. With the two similar relations for the other sides, that gives 2r cot A/2 = (b + c - a), 2r cot B/2 = (c + a - b), 2r cot C/2 = (a + b - c). So the given relation becomes: 49( (b + c - a)2 + 4(c + a - b)2 + 9(a + b - c)2) = 36(a + b + c)2.
Multiplying out is a mistake. It leads nowhere. It is more helpful to change variable to d = b + c - a, e = c + a - b, f = a + b - c giving 49(d2 + 4e2 + 9f2) = 36(d + e + f)2, or 13d2 + 160e2 + 405f2 - 72(de + ef + fd) = 0. We would like to express this as (hd + ke)2 + (h'e + k'f)2 + (h''f + k''d)2 = 0. Presumably 13 = 32 + 22. Then 72 = 2·3·something and 2·2·something, giving 12 and 18. Squares 144, 324. Fortunately, we see that 160 = 122+ 42, 405 = 182 + 92 and 2·4·9 = 72. So putting that together we get: (2d - 18f)2 + (3d - 12e)2 + (4e - 9f)2 = 0.
So we conclude that b + c - a = 9(a + b - c) = 4(c + a - b), or 5a + 4b = 5c, 5a + 3c = 5b, or a = 13k, b = 40k, c = 45k. We get the smallest triangle with integer sides by taking k = 1.
© John Scholes
jscholes@kalva.demon.co.uk
11 May 2002