Problem: In the right triangle shown the sum of the distances B M B M M A M A B C B C C A C A M B = x , C B = h M B = x , C B = h C A = d C A = d x x
Answer Choices:
A. h d 2 h + d 2 h + d h d ​
B. d − h d − h
C. 1 2 d 2 1 ​ d
D. h + d − 2 d h + d − 2 d ​
E. h 2 + d 2 − h h 2 + d 2 ​ − h Solution:
x + d 2 + ( h + x ) 2 = h + d , d 2 + ( h + x ) 2 = h + d − x x + d 2 + ( h + x ) 2 ​ = h + d , d 2 + ( h + x ) 2 ​ = h + d − x d 2 + h 2 + 2 h x + x 2 = h 2 + d 2 + x 2 + 2 h d − 2 h x − 2 d x , 2 h x + d x = h d ; d 2 + h 2 + 2 h x + x 2 = h 2 + d 2 + x 2 + 2 h d − 2 h x − 2 d x , 2 h x + d x = h d ; ∴ x = h d / ( 2 h + d ) ∴ x = h d / ( 2 h + d )
