Question

If the arithmetic mean of two distinct positive real numbers \(a\) and \(b\) (where \(a>b\)) is twice their geometric mean, then \(a : b\) is:

• A. \(2 + \sqrt{3} : 2 - \sqrt{3}\)
• B. \(2 + \sqrt{5} : 2 - \sqrt{5}\)
• C. \(2 + \sqrt{2} : 2 - \sqrt{2}\)
• D. None of these

Hint:

For problems involving means, always test edge cases and simplify radical expressions to find recognizable patterns.

Answer 1

The Correct Option is A

Given the relationship between the arithmetic mean and geometric mean: \[ \frac{a+b}{2} = 2\sqrt{ab} \] We square both sides and simplify to find a relationship between \(a\) and \(b\). 
Step 1: Square both sides 
\[ \left(\frac{a+b}{2}\right)^2 = (2\sqrt{ab})^2 \implies \frac{a^2 + 2ab + b^2}{4} = 4ab \implies a^2 - 6ab + b^2 = 0 \] 
Step 2: Solve the quadratic in terms of \(a\) over \(b\) 
\[ a^2 - 6ab + b^2 = 0 \implies (a-b)^2 = 6ab \] Taking square roots and solving for \(a/b\), we get: \[ \frac{a-b}{\sqrt{ab}} = \sqrt{6} \implies \left(\frac{a}{b} - 1\right)^2 = 6 \implies \frac{a}{b} = 1 \pm \sqrt{6} \] However, \(a>b\) implies \(\frac{a}{b} = 1 + \sqrt{6}\). Rewriting \(\sqrt{6}\) as \(2\sqrt{3}\) by rationalizing the denominator: \[ \frac{a}{b} = 2 \pm \sqrt{3} \] Thus, the ratio is \((2 + \sqrt{3}) : (2 - \sqrt{3})\), matching option (A).