If the length of the three sides of a triangle satisfies geometric sequence, determine the range of their common ratio.
(0,\dfrac{1+\sqrt{5} }{2} )
(\dfrac{1+\sqrt{5} }{2} ,\infty )
( \dfrac{-1+\sqrt{5} }{2} ,\dfrac{1+\sqrt{5} }{2} )
(1,\dfrac{1+\sqrt{5} }{2} )
Let q be the common ratio of the three sides of the triangle, which are
a, aq, aq^2
where a is one of the sides.
Using the theorem regarding the sides of a triangle, the sum of the length of any two sides is larger than the length of the third side, and the difference of the length of the two sides is less than the third side. We get the following inequalities.
a+aq>aq^2
a-aq < aq^2
\because a>0 , the above inequalities are simplified as, as
q^2-q-1<0
q^2+q-1>0
Solve the inequalities, the range of q is determined as
( \dfrac{-1+\sqrt{5} }{2} ,\dfrac{1+\sqrt{5} }{2} )