AoPSWiki
Want to learn how to tackle those tough MATHCOUNTS and AMC counting and probability problems? Check out Art of Problem Solving's Introduction to Counting & Probability by David Patrick.

2006 Cyprus MO/Lyceum/Problem 6

From AoPSWiki

Problem

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

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)}.

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
Retrieved from "http://www.artofproblemsolving.com/Wiki/index.php/2006_Cyprus_MO/Lyceum/Problem_6"
Looking for a challenging geometry text? Preparing for MATHCOUNTS or the AMC exams? Check out Art of Problem Solving's Introduction to Geometry by Richard Rusczyk.
© Copyright 2008 AoPS Incorporated. All Rights Reserved. • Foundation • Privacy • Contact Us