Difference between revisions of "2016 AMC 12A Problems/Problem 9"

(Solution)
(Solution)
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Let <math>s</math> be the side length of the small squares.
 
Let <math>s</math> be the side length of the small squares.
  
The diagonal of the big square can be written in two ways: <math>\sqrt{2} and </math>s \sqrt{2} + s + s \sqrt{2}<math>.
+
The diagonal of the big square can be written in two ways: <math>\sqrt{2}</math> and <math>s \sqrt{2} + s + s \sqrt{2}</math>.
  
Solving for </math>s<math>, we get </math>s = \frac{4 - \sqrt{2}}{7}<math>, so our answer is </math>4 + 7 \Rightarrow \boxed{\textbf{(E)} 11}$
+
Solving for <math>s</math>, we get <math>s = \frac{4 - \sqrt{2}}{7}</math>, so our answer is <math>4 + 7 \Rightarrow \boxed{\textbf{(E)} 11}</math>

Revision as of 15:20, 4 February 2016

Problem 9

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 $\tfrac{a-\sqrt{2}}{b}$, where $a$ and $b$ are positive integers. What is $a+b$ ?

[asy] real x=.369; draw((0,0)--(0,1)--(1,1)--(1,0)--cycle); filldraw((0,0)--(0,x)--(x,x)--(x,0)--cycle, gray); filldraw((0,1)--(0,1-x)--(x,1-x)--(x,1)--cycle, gray); filldraw((1,1)--(1,1-x)--(1-x,1-x)--(1-x,1)--cycle, gray); filldraw((1,0)--(1,x)--(1-x,x)--(1-x,0)--cycle, gray); filldraw((.5,.5-x*sqrt(2)/2)--(.5+x*sqrt(2)/2,.5)--(.5,.5+x*sqrt(2)/2)--(.5-x*sqrt(2)/2,.5)--cycle, gray); [/asy]

$\textbf{(A)}\ 7\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 11$

Solution

Let $s$ be the side length of the small squares.

The diagonal of the big square can be written in two ways: $\sqrt{2}$ and $s \sqrt{2} + s + s \sqrt{2}$.

Solving for $s$, we get $s = \frac{4 - \sqrt{2}}{7}$, so our answer is $4 + 7 \Rightarrow \boxed{\textbf{(E)} 11}$

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