+0  
 
0
88
1
avatar

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.

 Dec 23, 2018
 #1
avatar+98130 
+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

 

 

cool cool cool

 Dec 23, 2018
edited by CPhill  Dec 24, 2018
edited by CPhill  Dec 24, 2018

5 Online Users

avatar