The side lengths of a triangle are consecutive integers.
The largest interior angle is twice the smallest.

What are the side lengths of the triangle?

By law of sines I=II=III

sin(a)/x = sin(180-3a)/(x+1) = sin(2a)/(x+2)

By I and III
(x+2)/x = sin(2a)/sin(a) = 2sin(a)cos(a)/sin(a)
cos(a)= (x+2)/(2x)   ..........................................IV

By I and II
(x+1)/x = sin(180-3a)/sin(a)
= sin(3a)/sin(a)
= [sin(a)cos(2a) + 2sin(a)cos(a)^2]/sin(a)
= cos(2a) + 2cos(a)^2
= 2cos(a)^2 + 1 + 2cos(a)^2
(x+1)/x= 4cos(a)^2 + 1 ............................................V

Substituting IV into V

(x+1)/x = 4[(x+2)/(2x)]^2 - 1
(x+1)/x = (x^2 + 4x + 4)/x^2 - 1
x^2 + x = x^2 + 4x + 4 - x^2
x^2 - 3x - 4 = 0
(x-4)(x+1) = 0
x = 4 or x = -1
Only x=4 make sense so we have  4,5,6 as the side lengths

Edited on March 5, 2009, 2:13 pm

Posted by Jer on 2009-03-05 14:02:01