A point inside an equilateral triangle with side 1 is a distance a, b, c from the vertices. The triangle ABC has BC = a, CA = b, AB = c. The sides subtend equal angles at a point inside it. Show that sum of the distances of the point from the vertices is 1.
Solution
Let D be the point inside ABC, so that ∠ADB = ∠BDC = 120o. The key is to start from ABC and to rotate the triangle BDC through 60o away from the triangle ADB. After that everything is routine.
Suppose D goes to D' and C to C'. Then BD = BD' and ∠DBD' = 60o, so BDD' is equilateral. Hence ∠D'DB = 60o. ∠BDA = 120o, so ADD' is a straight line. Also ∠DD'B = 60o and ∠C'D'B = 120o, so DD'C' is a straight line. Thus AC' has length DA + DB + DC.
Note that BC = BC' and ∠CBC' = 60o, so CBC' is equilateral. Hence ∠CC'B = 60o. Now take Y such that AC'Y is equilateral, Y is on the opposite side of AC' to C. Then ∠BC'Y = 60o - ∠AC'B = ∠CC'A. Also BC' = CC' and YC' = AC', so triangles BC'Y and CC'A are congruent. Hence BY = CA = b. Also BC' = BC = a and BA = c. Thus B is a point inside an equilateral triangle and distances a, b, c from the vertices. Hence the triangle must have side 1. So DA + DB + DC = AC' = 1.
© John Scholes
jscholes@kalva.demon.co.uk
19 Aug 2002