1958 AHSME Problems/Problem 5

Problem

The expression $2 + \sqrt{2} + \frac{1}{2 + \sqrt{2}} + \frac{1}{\sqrt{2} - 2}$ equals:

$\textbf{(A)}\ 2\qquad  \textbf{(B)}\ 2 - \sqrt{2}\qquad  \textbf{(C)}\ 2 + \sqrt{2}\qquad  \textbf{(D)}\ 2\sqrt{2}\qquad  \textbf{(E)}\ \frac{\sqrt{2}}{2}$

Solution

To make this problem easier to solve, lets get the radicals out of the denominator. For $\frac{1}{2 + \sqrt2}$, we will multiply the numerator and denominator by $2- \sqrt2$ so,

$\frac{1}{2 + \sqrt2} \cdot \frac{2 - \sqrt2}{2 - \sqrt2} \Rightarrow \frac{2 - \sqrt2}{2}$.

Now, the other fraction we need to get the radical out of the denominator is $\frac{1}{\sqrt2 - 2}$. Here, we will multiply by the conjugate again, $\sqrt2 + 2$. So that simplifies to

$\frac{1}{\sqrt2 - 2} \cdot \frac{\sqrt2 + 2}{\sqrt2 + 2} \Rightarrow \frac{\sqrt2 + 2}{-2}$.

So now our simplified equation is

$2 + \sqrt2 + \frac{2- \sqrt2}{2} + \frac{\sqrt2 + 2}{-2} \Rightarrow 2+ \sqrt2 + \frac{2 - \sqrt2}{2} - \frac{\sqrt2 +2}{2}$

Bringing everything to the same denominator and combining like terms, we get

$\frac{4 + 2\sqrt2 + 2 - \sqrt2 - \sqrt2 - 2}{2} \Rightarrow \frac{4}{2} \Rightarrow 2 \Rightarrow \boxed{A}$

See also

1958 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 4
Followed by
Problem 6
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 41 42 43 44 45 46 47 48 49 50
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS