Difference between revisions of "2006 Cyprus MO/Lyceum/Problem 6"
(typo?? w/e, I'll just pretend that the 7 = 8) |
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==Problem== | ==Problem== | ||
| − | The value of the expression <math>K=\sqrt{19+ | + | The value of the expression <math>K=\sqrt{19+8\sqrt{3}}-\sqrt{7+4\sqrt{3}}</math> is |
A. <math>4</math> | A. <math>4</math> | ||
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B. <math>4\sqrt{3}</math> | B. <math>4\sqrt{3}</math> | ||
| − | C. | + | C. <math>12+4\sqrt{3}</math> |
D. <math>-2</math> | D. <math>-2</math> | ||
| Line 13: | Line 13: | ||
==Solution== | ==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>. |
==See also== | ==See also== | ||
{{CYMO box|year=2006|l=Lyceum|num-b=5|num-a=7}} | {{CYMO box|year=2006|l=Lyceum|num-b=5|num-a=7}} | ||
| + | |||
| + | [[Category:Introductory Algebra Problems]] | ||
Revision as of 17:53, 19 October 2007
Problem
The value of the expression 
is
A. 
B. 
C. 
D. 
E. 
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 | ||
