Difference between revisions of "2006 Cyprus MO/Lyceum/Problem 6"
(typo?? w/e, I'll just pretend that the 7 = 8) |
I like pie (talk | contribs) (Standardized answer choices; minor edits) |
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The value of the expression <math>K=\sqrt{19+8\sqrt{3}}-\sqrt{7+4\sqrt{3}}</math> is | The value of the expression <math>K=\sqrt{19+8\sqrt{3}}-\sqrt{7+4\sqrt{3}}</math> is | ||
− | + | <math>\mathrm{(A)}\ 4\qquad\mathrm{(B)}\ 4\sqrt{3}\qquad\mathrm{(C)}\ 12+4\sqrt{3}\qquad\mathrm{(D)}\ -2\qquad\mathrm{(E)}\ 2</math> | |
− | + | ==Solution== | |
+ | Suppose that <math>19 + 8\sqrt{3}</math> can be written in the form of <math>(a+b\sqrt{3})^2</math>, in order to eliminate the [[square root]]. | ||
− | + | Then <math>19 = a^2 + 3b^2</math> and <math>2ab\sqrt{3} = 8\sqrt{3} \Longrightarrow ab = 4</math>, and we quickly find that <math>19 + 8\sqrt{3} = (4+\sqrt{3})^2</math>. | |
− | + | Doing the same on the second radical gets us <math>(2 + \sqrt{3})^2</math>. | |
− | + | Thus the expression evaluates to <math>\sqrt{(4+ \sqrt{3})^2} - \sqrt{(2 + \sqrt{3})^2} = 4 + \sqrt{3} - 2 - \sqrt{3} = 2\ \mathrm{(E)}</math>. | |
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==See also== | ==See also== |
Latest revision as of 09:43, 27 April 2008
Problem
The value of the expression is
Solution
Suppose that can be written in the form of , in order to eliminate the square root.
Then and , and we quickly find that .
Doing the same on the second radical gets us .
Thus the expression evaluates to .
See also
2006 Cyprus MO, Lyceum (Problems) | ||
Preceded by Problem 5 |
Followed by Problem 7 | |
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 |