2006 Cyprus MO/Lyceum/Problem 6

Problem

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

A. $4$

B. $4\sqrt{3}$

C. $12+4\sqrt{3}$

D. $-2$

E. $2$

Solution

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 $\sqrt{(4+ \sqrt{3})^2} - \sqrt{(2 + \sqrt{3})^2} = 4 + \sqrt{3} - 2 - \sqrt{3} = 2\ \mathrm{(E)}$ .

2006 Cyprus MO, Lyceum (Problems)
Preceded by Problem 5	Followed by Problem 7
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2006 Cyprus MO/Lyceum/Problem 6

Problem

Solution

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