L, M, N are three lines through a point such that the angle between any pair is 60o. Show that the set of points P in the plane of ABC whose distances from the lines L, M, N are less than a, b, c respectively is the interior of hexagon iff there is a triangle with sides a, b, c. Find the perimeter of this hexagon.
Solution
P must lie in the strip a distance a either side of L and in the strip a distance b either side of M. So it must lie in the parallelogram PQRS. It must also lie in the strip a distance c either side of N. So we evidently require that AB lies between P and N. Put k = 2/√3. Then QR = 2ka (since the angles are all 60o). Similarly, PQ = 2kb. Also AQ + QC = 2kc, so PA + CR = 2k(a+b-c). But PA = CR by symmetry and PAB is equilateral, so AB = k(a+b-c). We get a hexagon iff PA = k(a+b-c) > 0 and similar conditions apply if we consider the strip about M relative to the parallelogram given by the strips about L and N, or the strip L about relative to the third parallelogram. Thus we require c < a+b, b < c+a, a < b+c, or a, b, c to form a triangle.
We have also shown that the perimeter of the hexagon is 2k(a+b-c) + 2k(b+c-a) + 2k(c+a-b) = 2k(a+b+c).
© John Scholes
jscholes@kalva.demon.co.uk
1 Nov 2003
Last corrected/updated 1 Nov 03