# Difference between revisions of "1960 AHSME Problems/Problem 39"

## Problem

To satisfy the equation $\frac{a+b}{a}=\frac{b}{a+b}$, $a$ and $b$ must be:

$\textbf{(A)}\ \text{both rational}\qquad\textbf{(B)}\ \text{both real but not rational}\qquad\textbf{(C)}\ \text{both not real}\qquad$ $\textbf{(D)}\ \text{one real, one not real}\qquad\textbf{(E)}\ \text{one real, one not real or both not real}$

## Solution

First, note that $a \neq 0$ and $a \neq -b$. Cross multiply both sides to get $$a^2 + 2ab + b^2 = ab$$ Subtract both sides by $ab$ to get $$a^2 + ab + b^2 = 0$$ From the quadratic formula, $$a = \frac{-b \pm \sqrt{b^2 - 4b^2}}{2}$$ $$a = \frac{-b \pm \sqrt{-3b^2}}{2}$$ If $b$ is real, then $\sqrt{-3b^2}$ is imaginary because $-3b^2$ is negative. If $b$ is not real, where $b = m+ni$ and $n \neq 0$, then $\sqrt{-3b^2}$ evaluates to $\sqrt{-3m^2 - 6mni + 3n^2}$. As long as $m \neq 0$, the expression can also be imaginary because a real number squared will be a real number. From these two points, the answer is $\boxed{\textbf{(E)}}$.

## See Also

 1960 AHSC (Problems • Answer Key • Resources) Preceded byProblem 38 Followed byProblem 40 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 All AHSME Problems and Solutions
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