2454.11 – How Mean?

You are given two numbers, $x$ and $y$ . If the arithmetic mean of these two numbers happens to be double their geometric mean, and $x > y > 0$ , then which of the following is a possible value for $\frac{x}{y}$ , rounded to the nearest whole number?

5
8
11
14
none of these

Solution

The arithmetic mean, $AM$ , of $n$ numbers (commonly called the $average$ of the numbers) is equal to

$AM = \frac{\sum_{i=1}^{n} a_{i}}{n},$

which is the mathematical way of saying that to find $AM$ you add up the numbers and then divide how many numbers there are.

The geometric mean, $GM$ , of $n$ numbers is equal to

$GM = \sqrt[n]{\prod_{i=1}^{n} a_{i}},$

which is the mathematical way of saying that to find $GM$ you multiply the numbers then take the $n^{th}$ root.

With just two numbers $x$ and $y$ , we have

$AM = \frac{x+y}{2}$

and

$GM = \sqrt{xy},$

and we are told that $AM = 2 GM$ , which means that $\frac{x+y}{2} = 2\sqrt{xy}$ . This can be simplified by squaring and gathering terms. Here are the details.

\begin{aligned} (\frac{x+y}{2})^2 & = & 4 xy\\ \frac{x^2 + 2 x y + y^2}{4} & = & 4 xy \\ x^2 + 2 x y + y^2 & = & 16 x y \\ x^2 - 14 x y + y^2 & = & 0 \end{aligned}

Because we are looking for $\frac{x}{y}$ , we divide everything by $y^2$ :

\begin{aligned} \frac{x^2}{y^2} - 14 \frac{xy}{y^2} + \frac{y^2}{y^2}& = & 0\\ (\frac{x}{y})^2 - 14 \frac{x}{y} + 1 & = & 0 \end{aligned}

We recognize this as a quadratic with $\frac{x}{y}$ as the variable. The best way to solve this is with the quadratic formula:

\begin{aligned} \frac{x}{y} & = & \frac{-b ± \sqrt{b^2 - 4 a c}}{2 a}\\ \frac{x}{y} & = & \frac{14 ± \sqrt{192}}{2}\\ & = & 7 ± \sqrt{48} \\ & \approx & 13.93 \hspace{.1in} \text{or} \hspace{.1in} 0.07 \end{aligned}

Rounding these to the nearest whole numbers, we get 14 or 0, and 14 is in the list of possible values for $\frac{x}{y}$ . The correct answer is d.