# 1960 AHSME Problems/Problem 9

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## Problem

The fraction $\frac{a^2+b^2-c^2+2ab}{a^2+c^2-b^2+2ac}$ is (with suitable restrictions of the values of a, b, and c): $\text{(A) irreducible}\qquad$ $\text{(B) reducible to negative 1}\qquad$ $\text{(C) reducible to a polynomial of three terms}\qquad$ $\text{(D) reducible to} \frac{a-b+c}{a+b-c}\qquad$ $\text{(E) reducible to} \frac{a+b-c}{a-b+c}$

## Solution

Use the commutative property to get $$\frac{a^2+2ab+b^2-c^2}{a^2+2ac+c^2-b^2}$$ Factor perfect square trinomials to get $$\frac{(a+b)^2-c^2}{(a+c)^2-b^2}$$ Factor difference of squares to get $$\frac{(a+b+c)(a+b-c)}{(a+b+c)(a-b+c)}$$ Cancel out like terms (with suitable restrictions of a, b, c) to get $$\frac{a+b-c}{a-b+c}$$ The answer is $\boxed{\textbf{(E)}}$.

## See Also

 1960 AHSME (Problems • Answer Key • Resources) Preceded byProblem 8 Followed byProblem 10 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 All AHSME Problems and Solutions
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