Difference between revisions of "2006 Cyprus MO/Lyceum/Problem 6"

Latest revision as of 10:43, 27 April 2008

The value of the expression $K=\sqrt{19+8\sqrt{3}}-\sqrt{7+4\sqrt{3}}$ is

$\mathrm{(A)}\ 4\qquad\mathrm{(B)}\ 4\sqrt{3}\qquad\mathrm{(C)}\ 12+4\sqrt{3}\qquad\mathrm{(D)}\ -2\qquad\mathrm{(E)}\ 2$

Suppose that $19 + 8\sqrt{3}$ can be written in the form of $(a+b\sqrt{3})^2$ , in order to eliminate the square root.

Then $19 = a^2 + 3b^2$ and $2ab\sqrt{3} = 8\sqrt{3} \Longrightarrow ab = 4$ , and we quickly find that $19 + 8\sqrt{3} = (4+\sqrt{3})^2$ .

Doing the same on the second radical gets us $(2 + \sqrt{3})^2$ .

Thus the expression evaluates to $\sqrt{(4+ \sqrt{3})^2} - \sqrt{(2 + \sqrt{3})^2} = 4 + \sqrt{3} - 2 - \sqrt{3} = 2\ \mathrm{(E)}$ .

@@ Line 1: / Line 1: @@
 ==Problem==
-{{empty}}
+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==
-{{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==
 {{CYMO box|year=2006|l=Lyceum|num-b=5|num-a=7}}
+[[Category:Introductory Algebra Problems]]