Question

Suppose sides of a triangle makes a gemetric progression of common ratio r then r lies in the interval

Answer 1

Let sides of traingle are a/r , a , ar .
we know , according to properties of traingle : sum of two sides of traingle is always greater than rest one side.

so,
a/r + a > ar
1/r + 1 > r
1 + r > r²
r² - r - 1 < 0
[r - (1+√5)/2][r - (1-√5)/2]<0
(1-√5)/2 < r < (1+√5)/2 -------------(1)

or,
ar + a/r > a
1 + r² > r
r² - r + 1 > 0 this is correct for all real numbers .
hence, r∈ R -----------(2)

or,
a + ar > a/r
1 + r > 1/r
r + r² > 1
r² + r -1 > 0
[r -(-1+√5)/2][r-(-1-√5)/2]>0
r>(-1+√5)/2 or , r< (-1-√5)/2 --------------(3)


we know, length of side can't be negative so, a>0 and r > 0 .
but we obeserved that from equation (1), (2) and (3) .

hence, we get
r∈ [(-1+√5)/2,(1+√5)/2]