Question

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 the numbers be $a$ and $b$.

Geometric mean of two numbers $a$ and $b$ is $\sqrt{ab}$.

According to the question,

$a+b=6\sqrt{ab}$

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

Apply componendo and dividendo.

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

$\frac{(\sqrt{a}{)}^2+(\sqrt{b}{)}^2+2(\sqrt{a}\times \sqrt{b})}{(\sqrt{a}{)}^2+(\sqrt{b}{)}^2-2(\sqrt{a}\times \sqrt{b})}=\frac{4}{2}$

${\left(\frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}}\right)}^2=\frac{2}{1}$

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

Apply componendo and dividendo again.

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

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

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

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

Square both the sides.

${\left(\sqrt{\frac{a}{b}}\right)}^2={\left(\frac{\sqrt{2}+1}{\sqrt{2}-1}\right)}^2$

$\frac{a}{b}=\left(\frac{\sqrt{2}+1}{\sqrt{2}-1}\right)$

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

Thus, $a$ and $b$ are in the ratio of $3+2\sqrt{2}:3-2\sqrt{2}$.