Difference between revisions of "2006 Cyprus MO/Lyceum/Problem 15"
(New page: ==Problem== {{empty}} ==Solution== {{solution}} ==See also== {{CYMO box|year=2006|l=Lyceum|num-b=14|num-a=16}}) |
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==Problem== | ==Problem== | ||
− | {{ | + | The expression <math>\frac{1}{2+\sqrt7} + \frac{1}{\sqrt7+\sqrt{10}}+ \frac{1}{\sqrt{10}+\sqrt{13}} + \frac{1}{\sqrt{13}+4}</math> equals |
+ | |||
+ | <math>\mathrm{(A)}\ \frac{3}{4}\qquad\mathrm{(B)}\ \frac{3}{2}\qquad\mathrm{(C)}\ \frac{2}{5}\qquad\mathrm{(D)}\ \frac{1}{2}\qquad\mathrm{(E)}\ \frac{2}{3}</math> | ||
==Solution== | ==Solution== | ||
− | {{ | + | Multiply all of the terms by their [[complex conjugate]]s to simplify: |
+ | |||
+ | <cmath>\frac{1}{\sqrt{7} + \sqrt{4}} \cdot \left(\frac{\sqrt{7}-\sqrt{4}}{\sqrt{7}-\sqrt{4}}\right) + \ldots + \frac{1}{\sqrt{16} + \sqrt{13}} \cdot \left(\frac{\sqrt{16}-\sqrt{13}}{\sqrt{16}-\sqrt{13}}\right)</cmath> | ||
+ | <cmath>= \frac{\sqrt{7} - \sqrt{4}}{3} + \frac{\sqrt{10} - \sqrt{7}}{3} + \frac{\sqrt{13} - \sqrt{10}}{3} + \frac{\sqrt{16} - \sqrt{13}}{3}</cmath> | ||
+ | |||
+ | This [[telescope]]s to <math>\frac{\sqrt{16} - \sqrt{4}}{3} = \frac{2}{3} \Longrightarrow \mathrm{(E)}</math>. | ||
==See also== | ==See also== | ||
{{CYMO box|year=2006|l=Lyceum|num-b=14|num-a=16}} | {{CYMO box|year=2006|l=Lyceum|num-b=14|num-a=16}} | ||
+ | |||
+ | [[Category:Introductory Algebra Problems]] |
Latest revision as of 09:29, 27 April 2008
Problem
The expression equals
Solution
Multiply all of the terms by their complex conjugates to simplify:
This telescopes to .
See also
2006 Cyprus MO, Lyceum (Problems) | ||
Preceded by Problem 14 |
Followed by Problem 16 | |
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