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1952 AMC 12 Problem 35

Attempt Problem 35 of the 1952 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 1952 AMC 12 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

35.

With a rational denominator, the expression 22+3−5\dfrac{\sqrt2}{\sqrt2+\sqrt3-\sqrt5} is equivalent to:

3+6+156\dfrac{3+\sqrt6+\sqrt{15}}6

6−2+106\dfrac{\sqrt6-2+\sqrt{10}}6

2+6+1010\dfrac{2+\sqrt6+\sqrt{10}}{10}

2+6−106\dfrac{2+\sqrt6-\sqrt{10}}6

None of these

Answer: A
Concepts:radicalrationalizing denominatoralgebraic manipulation
Difficulty rating: 2210
Small Hint:

First treat the denominator as (2+3)−5(\sqrt2+\sqrt3)-\sqrt5 and multiply by its conjugate

Big Hint:

After the first rationalization, a denominator containing 6\sqrt6 remains; use another conjugate

Solution:

Multiply first by the conjugate 2+3+5.\sqrt2+\sqrt3+\sqrt5. This gives 2(2+3+5)(2+3)2−5=2+6+1026. \begin{aligned} &\frac{\sqrt2(\sqrt2+\sqrt3+\sqrt5)} {(\sqrt2+\sqrt3)^2-5}\\ &\qquad=\frac{2+\sqrt6+\sqrt{10}}{2\sqrt6}. \end{aligned} Multiplying numerator and denominator by 6\sqrt6 yields 26+6+6012=3+6+156. \begin{gathered} \frac{2\sqrt6+6+\sqrt{60}}{12} \\ =\frac{3+\sqrt6+\sqrt{15}}6. \end{gathered}

Thus, the correct answer is A.

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