Difference between revisions of "1952 AHSME Problems/Problem 35"
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== Solution == | == Solution == | ||
| − | + | First let <math>k=\sqrt2+\sqrt3</math>. | |
| − | Then <math></math> | + | 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 
is equivalent to:
Solution
First let 
.
Then our original expression becomes 
. 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 | |
| 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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
