1952 AMC 12 Problem 35

Question

With a rational denominator, the expression

\dfrac{\sqrt2}{\sqrt2+\sqrt3-\sqrt5}

is equivalent to:

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

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

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

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

None of these

First treat the denominator as

(\sqrt2+\sqrt3)-\sqrt5

and multiply by its conjugate

After the first rationalization, a denominator containing

\sqrt6

remains; use another conjugate

Multiply first by the conjugate

\sqrt2+\sqrt3+\sqrt5.

This gives

\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

\sqrt6

yields

\begin{gathered} \frac{2\sqrt6+6+\sqrt{60}}{12} \\ =\frac{3+\sqrt6+\sqrt{15}}6. \end{gathered}

Thus, the correct answer is A.

Problem 35 in other years