Question

Each of the two triplets of numbers $\log a,\log b,\log c$ and $\log a-\log 2b,\log 2b-\log 3c,\log 3c-\log a$ 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

By the given condition

$2(\log 2b-\log 3c)=\log a-\log 2b+\log 3c-\log a=\log 3c-\log 2b$

$\Rightarrow 3(\log 2b-\log 3c)=0\Rightarrow 2b=3c\Rightarrow b=\frac{3c}{2}$

Also, as $\log a,\log b,\log c$ are in $A.P.$

$2\log b=\log a+\log c\Rightarrow \log b^2=\log ac\Rightarrow b^2=ac$

$\therefore \frac{9c^2}{4}=ac\Rightarrow c=\frac{4a}{9}$

And also, $b=\frac{3c}{2}=\frac{3}{2}.\frac{4a}{9}=\frac{2a}{3}$

Since $a+b=a+\frac{2a}{3}=\frac{5a}{3}>c$

$b+c=\frac{2a}{3}+\frac{4a}{9}=\frac{10a}{9}>a$

$c+a=\frac{13a}{9}>b$

$\therefore a,b,c$ are the sides of a triangle.

Also as $a$ is the greatest side, let us find angle $A$ of $△ABC$

$\cos A=\frac{b^2+c^2-a^2}{2bc}=\frac{\frac{4a^2}{9}+\frac{16a^2}{81}-a^2}{2.\frac{2a}{3}.\frac{4a}{9}}=-\frac{29}{48}<0$

Hence, $△ABC$ is an obtuse-angled triangle.