Yeah Geometry Rocks

Geometry Level 3

A point inside an equilateral triangle with side 1 is at distance $a, b, c$ from the vertices.

Construct a triangle ABC with side lengths $BC = a, CA = b, AB = c$ .

Let D be a point in ABC such that $\angle ADB = \angle BDC = \angle CDA =120^\circ$ .

Find $DA+DB+DC$ .

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The answer is 1.

2 solutions

A Former Brilliant Member
Dec 21, 2014

Let D be the point inside ABC, so that ∠ADB = ∠BDC = 120°.

The key is to start from ABC and to rotate the triangle BDC through 60° away from the triangle ADB. After that everything is routine.

Suppose D goes to D' and C to C'. Then BD = BD' and ∠DBD' = 60°, so BDD' is equilateral. Hence ∠D'DB = 60°. ∠BDA = 120°, so ADD' is a straight line. Also ∠DD'B = 60° and ∠C'D'B = 120°, so DD'C' is a straight line. Thus AC' has length DA + DB + DC.

Note that BC = BC' and ∠CBC' = 60°, so CBC' is equilateral. Hence ∠CC'B = 60°.

Now take Y such that AC'Y is equilateral, Y is on the opposite side of AC' to C. Then ∠BC'Y = 60° - ∠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.

Punit Kalariya
Jan 4, 2015

The question do not tells about any type of relation between a,b and c. So we can assume a=b=c to deal question easily.hence D point will be at center of eq.triangle ABC . AD,BD,CD=1/3

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