1976 AMC 12 Problem 27

Attempt Problem 27 of the 1976 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1976 AMC 12 solutions, or check the answer key.

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27.

If N=5+2+5−25+1−3−22, \begin{aligned} N&=\frac{\sqrt{\sqrt5+2}+\sqrt{\sqrt5-2}} {\sqrt{\sqrt5+1}}\\ &\quad-\sqrt{3-2\sqrt2}, \end{aligned} then NN equals
11
22−12\sqrt2-1
52\frac{\sqrt5}{2}
52\sqrt{\frac52}
none of these
Answer: A
Concepts:radicalrationalizing denominatoralgebraic manipulation
Small hint:
Square the numerator of the large fraction before simplifying it
Big hint:
Recognize 3−223-2\sqrt2 as (2−1)2(\sqrt2-1)^2
Solution:
Let TT be the first fraction. Its numerator squared is (5+2)+(5−2)+2(5+2)(5−2)=25+2. \begin{aligned} &(\sqrt5+2)+(\sqrt5-2)\\ &\quad+2\sqrt{(\sqrt5+2)(\sqrt5-2)}\\ &=2\sqrt5+2. \end{aligned} Thus T2=25+25+1=2,T^2=\frac{2\sqrt5+2}{\sqrt5+1}=2, and T=2.T=\sqrt2. Also 3−22=2−1.\sqrt{3-2\sqrt2}=\sqrt2-1. Hence N=2−(2−1)=1.N=\sqrt2-(\sqrt2-1)=1.
Therefore, the correct answer is A.
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