Question

If the harmonic mean between two positive numbers is to their geometric mean as $12 : 13$, then the numbers could be in the ratio:

• A. $12 : 13$
• B. $1/12 : 1/13$
• C. $4 : 9$
• D. $2 : 3$

Hint:

Always express HM and GM in terms of $x$ and $y$, then use a substitution like $t = \sqrt{x/y}$ to simplify the ratio conditions into a quadratic equation.

Answer 1

The Correct Option is C

Let the two positive numbers be $x$ and $y$.
Harmonic Mean (HM) = $\frac{2xy}{x + y}$.
Geometric Mean (GM) = $\sqrt{xy}$.
Given: HM : GM = $12 : 13$.
That is: \[ \frac{\frac{2xy}{x + y}}{\sqrt{xy}} = \frac{12}{13}. \] Simplify the left side: \[ \frac{2xy}{x + y} \times \frac{1}{\sqrt{xy}} = \frac{2\sqrt{xy}}{x + y}. \] So: \[ \frac{2\sqrt{xy}}{x + y} = \frac{12}{13}. \] Cross-multiply: \[ 26\sqrt{xy} = 12(x + y). \] Divide through by 2: \[ 13\sqrt{xy} = 6(x + y). \] Let $\sqrt{x/y} = t \Rightarrow x = t^2 y$.
Then $\sqrt{xy} = \sqrt{t^2 y^2} = ty$, and $x + y = t^2 y + y = y(t^2 + 1)$.
Substitute into $13\sqrt{xy} = 6(x + y)$: $13(ty) = 6[y(t^2 + 1)]$.
Cancel $y>0$: $13t = 6(t^2 + 1)$.
Rearrange: $6t^2 - 13t + 6 = 0$.
Multiply through by 1: $6t^2 - 13t + 6 = 0$.
Solve: $t = \frac{13 \pm \sqrt{169 - 144}}{12} = \frac{13 \pm 5}{12}$.
So $t = \frac{18}{12} = \frac{3}{2}$ or $t = \frac{8}{12} = \frac{2}{3}$.
If $t = 3/2$, then $x / y = t^2 = (9/4) \Rightarrow x : y = 9 : 4$.
If $t = 2/3$, then $x : y = (4/9) \Rightarrow 4 : 9$.
Since ratio order is not fixed, both 9:4 and 4:9 are equivalent forms. Given option (C) matches $4:9$, answer is $\boxed{4 : 9}$.