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Difference between revisions of "1952 AHSME Problems/Problem 35"

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== Solution ==
 
== Solution ==
  
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Let <math>k=\sqrt{2}+\sqrt{3}</math>
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First let <math>k=\sqrt2+\sqrt3</math>.
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Then <math>\frac{\sqrt{2}}{k-\sqrt{5}}\implies \frac{\sqrt{2}(k+\sqrt{5})}{k^2-\sqrt{5}^2}\implies\frac{\sqrt{2}k+\sqrt{10}}{k^2-5}\implies \frac{\sqrt{2}(\sqrt{2}+\sqrt{3})+\sqrt{10}}{(\sqrt{2}+\sqrt{3})^2-5}\implies \frac{2+\sqrt{6}+\sqrt{10}}{2\sqrt{6}}\implies\frac{2\sqrt{6}+6+\sqrt{60}}{2}\implies \frac{\sqrt{6}+3+\sqrt{15}}{6}\fbox{A}</math>
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Then our original expression becomes <math>\frac{\sqrt2}{k-\sqrt5}</math>. Rationalizing the denominator, we have:
 +
\begin{align*}
 +
\frac{\sqrt2}{k-\sqrt5}&=\frac{\sqrt2(k+\sqrt5)}{(k-\sqrt5)(k+\sqrt5)}=\frac{\sqrt{2}k+\sqrt{10}}{k^2-(\sqrt5)^2}=\frac{\sqrt{2}k+\sqrt{10}}{k^2-5} \\
 +
&=\frac{\sqrt{2}(\sqrt{2}+\sqrt{3})+\sqrt{10}}{(\sqrt{2}+\sqrt{3})^2-5}= \frac{2+\sqrt{6}+\sqrt{10}}{2\sqrt{6}+2+3-5}=\frac{2+\sqrt6+\sqrt{10}}{2\sqrt{6}} \\
 +
&=\frac{2\sqrt{6}+6+\sqrt{60}}{12}=\boxed{\textbf{(A) }\frac{3+\sqrt{6}+\sqrt{15}}{6}}.
 +
\end{align*}
 +
 
 +
(reworked by Technodoggo)
  
 
== See also ==
 
== See also ==

Latest revision as of 23:50, 28 August 2025

Problem

With a rational denominator, the expression $\frac {\sqrt {2}}{\sqrt {2} + \sqrt {3} - \sqrt {5}}$ is equivalent to:

$\textbf{(A)}\ \frac {3 + \sqrt {6} + \sqrt {15}}{6} \qquad \textbf{(B)}\ \frac {\sqrt {6} - 2 + \sqrt {10}}{6} \qquad \textbf{(C)}\ \frac{2+\sqrt{6}+\sqrt{10}}{10} \qquad\\ \textbf{(D)}\ \frac {2 + \sqrt {6} - \sqrt {10}}{6} \qquad \textbf{(E)}\ \text{none of these}$

Solution

First let $k=\sqrt2+\sqrt3$. Then our original expression becomes $\frac{\sqrt2}{k-\sqrt5}$. Rationalizing the denominator, we have: \begin{align*} \frac{\sqrt2}{k-\sqrt5}&=\frac{\sqrt2(k+\sqrt5)}{(k-\sqrt5)(k+\sqrt5)}=\frac{\sqrt{2}k+\sqrt{10}}{k^2-(\sqrt5)^2}=\frac{\sqrt{2}k+\sqrt{10}}{k^2-5} \\ &=\frac{\sqrt{2}(\sqrt{2}+\sqrt{3})+\sqrt{10}}{(\sqrt{2}+\sqrt{3})^2-5}= \frac{2+\sqrt{6}+\sqrt{10}}{2\sqrt{6}+2+3-5}=\frac{2+\sqrt6+\sqrt{10}}{2\sqrt{6}} \\ &=\frac{2\sqrt{6}+6+\sqrt{60}}{12}=\boxed{\textbf{(A) }\frac{3+\sqrt{6}+\sqrt{15}}{6}}. \end{align*}

(reworked by Technodoggo)

See also

1952 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 34 		Followed by
Problem 36
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All AHSME Problems and Solutions

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