Question

The sum of two numbers is 6 times their geometric mean. Show that numbers are in the ratio $(3+2\sqrt{2}):(3-2\sqrt{2})$

Answer 1

Let 'a' and 'b' be two numbers.

Then $a+b=6\sqrt{ab}\Rightarrow \frac{a+b}{2\sqrt{ab}}=\frac{1}{3}$

Applying componendo and dividendo, we get

$\frac{a+b+2\sqrt{ab}}{a+b-2\sqrt{ab}}=\frac{3+1}{3-1}$

$\Rightarrow \frac{[\sqrt{a}+\sqrt{b}{]}^2}{[\sqrt{a}-\sqrt{b}{]}^2}=\frac{4}{2}$

$\Rightarrow \frac{[\sqrt{a}+\sqrt{b}{]}^2}{[\sqrt{a}-\sqrt{b}{]}^2}=\frac{2}{1}$

$\Rightarrow \frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}}=\frac{\sqrt{2}}{1}$

Again applying componendo and dividendo we have :

$\frac{\sqrt{a}+\sqrt{b}+\sqrt{a}-\sqrt{b}}{\sqrt{a}+\sqrt{b}-\sqrt{a}+\sqrt{b}}=\frac{\sqrt{2}+1}{\sqrt{2}-1}$

$\Rightarrow \frac{\sqrt{a}}{\sqrt{b}}=\frac{\sqrt{2}+1}{\sqrt{2}-1}$

Squaring both sides, we get

$\frac{a}{b}=\frac{2+1+2\sqrt{2}}{2+1-2\sqrt{2}}\Rightarrow \frac{a}{b}=\frac{3+2\sqrt{2}}{3-2\sqrt{2}}$

Thus, the numbers are in the ratio

$(3+2\sqrt{2}):(3-2\sqrt{2}).$