2016 AMC 12A Problem 9

Question

The five small shaded squares inside this unit square are congruent and have disjoint interiors. The midpoint of each side of the middle square coincides with one of the vertices of the other four small squares as shown. The common side length is

\dfrac{a-\sqrt{2}}{b},

where

a

and

b

are positive integers. What is

a+b?

7

8

9

10

11

Answer: E

Concepts:diagonal rationalizing denominator

Difficulty rating: 1510

Small hint:

Draw the diagonal of the unit square, which has length

\sqrt{2}

Big hint:

That diagonal equals two small-square diagonals plus one small-square side:

2x\sqrt2+x=\sqrt2

Solution:

Let

x

be the common side length. The diagonal of the unit square has length

\sqrt{2}

and consists of two small-square diagonals (each

x\sqrt2

) plus one small-square side length

x,

so

2x\sqrt2+x=\sqrt2.

Solving,

\begin{gathered} x=\dfrac{\sqrt2}{2\sqrt2+1}\\ =\dfrac{\sqrt2\,(2\sqrt2-1)}{(2\sqrt2+1)(2\sqrt2-1)}\\ =\dfrac{4-\sqrt2}{7}. \end{gathered}

Thus

a=4,

b=7,

and

a+b=11.

Thus, the correct answer is E.