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

To find the ratio \(a : b\) given that the arithmetic mean of two distinct positive real numbers \(a\) and \(b\) is twice their geometric mean, let's first define the arithmetic and geometric means:

The arithmetic mean (AM) of \(a\) and \(b\) is given by:

\(\text{AM} = \frac{a + b}{2}\)

The geometric mean (GM) of \(a\) and \(b\) is given by:

\(\text{GM} = \sqrt{ab}\)

According to the problem, the arithmetic mean is twice the geometric mean:

\(\frac{a + b}{2} = 2\sqrt{ab}\)

Multiply both sides by 2 to eliminate the fraction:

\(a + b = 4\sqrt{ab}\)

Square both sides to eliminate the square root:

\((a + b)^2 = 16ab\)

Expand the left-hand side:

\(a^2 + 2ab + b^2 = 16ab\)

Rearrange the equation:

\(a^2 + b^2 = 14ab\)

Now, we need to find the ratio \(\frac{a}{b}\). Let \(\frac{a}{b} = k\), which implies \(a = kb\). Substitute \(a = kb\) into the equation:

\((kb)^2 + b^2 = 14(kb)(b)\)

Simplify:

\(k^2b^2 + b^2 = 14kb^2\)

Factor out \(b^2\):

\((k^2 + 1)b^2 = 14kb^2\)

Divide both sides by \(b^2\) (since \(b \neq 0\)):

\(k^2 + 1 = 14k\)

Rearrange the quadratic equation:

\(k^2 - 14k + 1 = 0\)

Use the quadratic formula \(k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = 1\), \(b = -14\), and \(c = 1\):

\(k = \frac{14 \pm \sqrt{(-14)^2 - 4(1)(1)}}{2}\)

\(k = \frac{14 \pm \sqrt{196 - 4}}{2}\)

\(k = \frac{14 \pm \sqrt{192}}{2}\)

\(k = \frac{14 \pm 8\sqrt{3}}{2}\)

\(k = 7 \pm 4\sqrt{3}\)

Only positive ratio makes sense with \(a > b\), so we select \(k = 7 + 4\sqrt{3}\).

Thus, the ratio \(a : b = k : 1 = 7 + 4\sqrt{3} : 1\).

However, note that this ratio \(7 + 4\sqrt{3}\) must be expressed by simplifying and checking which one it matches with options. Rationalize by dividing the ratio of calculated roots, \(a:b\) at various trials:

Given \(a = 2 + \sqrt{3}\) and \(b = 2 - \sqrt{3}\), calculate:

Calculate \(k=\frac{2+\sqrt{3}}{2-\sqrt{3}}\), multiply numerator and denominator by \(2 + \sqrt{3}\):

\(k = \frac{(2+\sqrt{3})^{2}}{(2)^{2}-(\sqrt{3})^{2}}\)

\(k = \frac{4+4\sqrt{3}+3}{4-3}=7+4\sqrt{3}\)

Thus, the correct answer is \(2+\sqrt{3} : 2-\sqrt{3}\).

Answer 2

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).