Question

Each of the two triples of numbers log a, log b, log c and log $a-log2b,log2b-log3c,log3c-loga$ are in A.P. Can the numbers a, b, c be the lengths of the sides of a triangle? If they can, what kind of triangle is it? Find the angles of the triangle provided that it exists.

Answer 1

We have $2logb=loga+logc$
or $b^2=ac$ .$\left(1\right)$
and $2[log2b-log3c]=log3c-log2b$
or $3(log2b-log3c)=0$
$\therefore 2b=3c$ .$\left(2\right)$
Solving $\left(1\right)$ and $\left(2\right)$ for a and b, we get $a=9c/4$, $b=3c/2$.
Thus the triple of numbers that satisfies the given conditions is $9c/4,3c/2,c(c\ne 0)$. Now $a=9c/4,b=3c/2$ and c will form a triangle if (i) $a+b>c$ (ii)$b+c>a$ and (iii) $c+a>b$.
But since $a+b=15c/4>c$,
$b+c=5c/2>9c/4=a$,
and $a+c=13c/4>3c/2=b(c>0):$ a triangle with sides a, b and c exists and since $a^2>b^2+c^2$, it is obtuse. To find the angles, we use the cosine formula.
Thus, $\cos A=\frac{b^2+c^2-a^2}{2bc}=\frac{29}{48}$,
$\cos B=\frac{a^2+c^2-b^2}{2ac}=\frac{61}{72}$
and $\cos C=\frac{a^2+b^2-c^2}{2ab}=\frac{101}{108}$.
Thus, bearing in mind that A, B, C are the angles of a triangle, we get
$A={\cos }^{-1}(-\frac{29}{48})=\pi -{\cos }^{-1}\frac{29}{48}$
$B={\cos }^{-1}\left(\frac{61}{72}\right)$ and $C={\cos }^{-1}\left(\frac{101}{108}\right)$.