The isosceles right triangle ABC has its right angle at B and has area 1. The rays trisecting ABC intersect AC at E and F, where E is closer to A than it is to C. The area of BEF can be written in the form a - b\(\sqrt{c}\) for positive integers a, b, and c such that c is square free. Determine a + b + c.

Guest Dec 23, 2018

#1**+1 **

BC = √2

Angle BFC = 105°

Angle BCF = 45°

Using the Law of Sines

sin BFC / √2 = sin BCF / BF

sin 105 / √2 = 1 / [ √2 BF ]

sin 105 = 1 /BF

sin[ 60 + 45] = 1/BF

sin 60cos45 + sin45cos60 = 1/BF

√3/2 *√2/ 2 + √2/2 *1/2 = 1/BF

[ √6 + √2 ] / 4 = 1/BF

4 / [ √6 + √2 ] = BF

4 [ √6 - √2] / 4 = BF

√6 - √2 = BF

And because by ASA triangle BAE is congruent to Triangle BCF....then BE = BF

So.....area of Triangle BEF =

(1/2) BE * BF sin (30) =

(1/2) [ √6 - √2 ] ^2 (1/2)

[ 6 - 2√12 + 2 ] / 4 =

[8 - 4√3 ] / 4 =

2 - √3 =

2 - 1 √3

So

a + b + c = 2 + 1 + 3 = 6

CPhill Dec 23, 2018