Each of the two triplets of numbers $\log a$, $\log b$, $\log c$ $\;$ and $\;$ $\log a - \log 2b$, $\log 2b - \log 3c$, $\log 3c - \log a$ is an arithmetic progression (A.P). Can the numbers $a$, $b$ and $c$ be the lengths of the sides of a triangle? If they can, then what triangle is it? Find the angles of the triangle provided that it exists.
Given: $\;$ $\log a$, $\log b$, $\log c$ are in A.P
$\implies$ $2 \log b = \log a + \log c$
i.e. $\log b^2 = \log ac$
i.e. $b^2 = ac$ $\;\;\; \cdots \; (1)$
And: $\;$ $\log a - \log 2b$, $\log 2b - \log 3c$, $\log 3c - \log a$ are in A.P
$\implies$ $2 \left(\log 2b - \log 3c\right) = \log a - \log 2b + \log 3c - \log a$
i.e. $\;$ $2 \times \log \left(\dfrac{2b}{3c}\right) = \log \left(\dfrac{3c}{2b}\right)$
i.e. $\;$ $\log \left(\dfrac{2b}{3c}\right)^2 = \log \left(\dfrac{3c}{2b}\right)$
i.e. $\;$ $\left(\dfrac{2b}{3c}\right)^2 = \dfrac{3c}{2b}$
i.e. $\;$ $\left(2b\right)^3 = \left(3c\right)^3$
i.e. $\;$ $2b = 3c$ $\implies$ $b = \dfrac{3c}{2} = 1.5 c$ $\;\;\; \cdots \; (2)$
$\therefore \;$ In view of equation $(2)$, we have from equation $(1)$
$\dfrac{9c^2}{4} = ac$ $\implies$ $a = \dfrac{9c}{4} = 2.25c$ $\;\;\; \cdots \; (3)$
$\therefore \;$ The three numbers that satisfy the given conditions are
$a = \dfrac{9c}{4} = 2.25c$, $\;$ $b = \dfrac{3c}{2} = 1.5c$, $\;$ $c \; \left(c > 0\right)$
Check the existance of a triangle with sides $a$, $b$ and $c$:
For $a$, $b$ and $c$ to form the sides of a triangle, the condition to be satisfied is
$a + b >c$ $\;\;\; \cdots \; (4a)$, $\;$ $b + c > a$ $\;\;\; \cdots \; (4b)$, $\;$ $c + a > b$ $\;\;\; \cdots \; (4c)$
In view of equations $(2)$ and $(3)$,
equation $(4a)$ $\implies$ $2.25 c + 1.5c = 3.75 c > c$ $\;$ which is true
equation $(4b)$ $\implies$ $1.5 c + c = 2.5 c > 2.25c$ $\;$ which is true
and equation $(4c)$ $\implies$ $c + 2.25 c = 3.25 c > 1.5c$ $\;$ which is true
$\implies$ A triangle can be formed with sides $a$, $b$ and $c$.
Type of triangle:
From equation $(2)$, $\;$ $b^2 = \dfrac{9c^2}{4} = 2.25 c^2$ $\;\;\; \cdots \; (5a)$
From equation $(3)$, $\;$ $a^2 = \dfrac{81 c^2}{16} = 5.0625 c^2$ $\;\;\; \cdots \; (5b)$
$\therefore \;$ We have from equations $(5a)$ and $(5b)$
$a^2 = 5.0625 c^2 > b^2 + c^2 = 2.25 c^2 + c^2 = 3.25 c^2$
$\implies$ The triangle with sides $a$, $b$ and $c$ is an obtuse triangle.
Angles of the triangle:
By cosine rule,
$\cos A = \dfrac{b^2 + c^2 - a^2}{2bc} = \dfrac{\dfrac{9c^2}{4} + c^2 - \dfrac{81 c^2}{16}}{2 \times \dfrac{3c}{2} \times c} = \dfrac{-29}{48}$
$\implies$ $A = \cos^{-1} \left(\dfrac{-29}{48}\right) = \pi - \cos^{-1} \left(\dfrac{29}{48}\right)$
$\cos B = \dfrac{c^2 + a^2 - b^2}{2ca} = \dfrac{c^2 + \dfrac{81 c^2}{16} - \dfrac{9c^2}{4}}{2 \times c \times \dfrac{9c}{4}} = \dfrac{61}{72}$
$\implies$ $B = \cos^{-1} \left(\dfrac{61}{72}\right)$
$\cos C = \dfrac{a^2 + b^2 - c^2}{2ab} = \dfrac{\dfrac{81 c^2}{16} + \dfrac{9c^2}{4} - c^2}{2 \times \dfrac{9c}{4} \times \dfrac{3c}{2}} = \dfrac{101}{108}$
$\implies$ $C = \cos^{-1} \left(\dfrac{101}{108}\right)$
$\therefore \;$ The angles of the triangle are
$\angle A = \pi - \cos^{-1} \left(\dfrac{29}{48}\right)$, $\;$ $\angle B = \cos^{-1} \left(\dfrac{61}{72}\right)$, $\;$ $\angle C = \cos^{-1} \left(\dfrac{101}{108}\right)$