Question

If the AM and GM between two numbers are in the ratio $m:n$, then the numbers are in the ratio

• A. $m+\sqrt{m^2-n^2}:n+\sqrt{m^2-n^2}$
• B. $m+\sqrt{m^2-n^2}:m+\sqrt{m^2-n^2}$
• C. $m+\sqrt{m^2+n^2}:n+\sqrt{m^2+n^2}$
• D. $m+\sqrt{m^2+n^2}:m+\sqrt{m^2-n^2}$

Answer 1

The correct option is B $m+\sqrt{m^2-n^2}:m+\sqrt{m^2-n^2}$

$\frac{AM}{GM}=\frac{m}{n}$

$\frac{\frac{a+b}{2}}{\sqrt{ab}}=\frac{m}{n}$

By applying componendo and dividendo.

$\frac{a+b+2\sqrt{ab}}{a+b-2\sqrt{ab}}=\frac{m+n}{m-n}$

$\frac{{(\sqrt{a}+\sqrt{b})}^2}{{(\sqrt{a}-\sqrt{b})}^2}=\frac{m+n}{m-n}$

$\frac{(\sqrt{a}+\sqrt{b})}{(\sqrt{a}-\sqrt{b})}=\frac{\sqrt{m+n}}{\sqrt{m-n}}$

By applying componendo and dividend again

$\frac{(\sqrt{a}+\sqrt{b})+(\sqrt{a}-\sqrt{b})}{(\sqrt{a}+\sqrt{b})-(\sqrt{a}-\sqrt{b})}=\frac{\sqrt{m+n}+\sqrt{m-n}}{\sqrt{m+n}-\sqrt{m-n}}$

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

$\frac{a}{b}=\frac{2m+\sqrt{m^2-n^2}}{2m-2\sqrt{m^2-n^2}}$

$\frac{a}{b}=\frac{m+\sqrt{m^2-n^2}}{m-\sqrt{m^2-n^2}}$