Difference between revisions of "2016 AMC 12A Problems/Problem 9"
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==Solution== | ==Solution== | ||
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+ | Let <math>s</math> be the side length of the small squares. | ||
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+ | 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>. | ||
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+ | 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> | ||
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+ | ==Solution 2== | ||
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+ | The diagonal of the small square can be written in two ways: <math>s \sqrt(2)</math> and <math>2*(1-2s).</math> Equating and simplifying gives <math>s = \frac{4 - \sqrt{2}}{7}</math>. Hence our answer is <math>4 + 7 \Rightarrow \boxed{\textbf{(E) } 11}.</math> | ||
+ | |||
+ | ==See Also== | ||
+ | {{AMC12 box|year=2016|ab=A|num-b=8|num-a=10}} | ||
+ | {{MAA Notice}} |
Latest revision as of 02:02, 22 January 2019
Contents
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 , where and are positive integers. What is ?
Solution
Let be the side length of the small squares.
The diagonal of the big square can be written in two ways: and .
Solving for , we get , so our answer is
Solution 2
The diagonal of the small square can be written in two ways: and Equating and simplifying gives . Hence our answer is
See Also
2016 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 8 |
Followed by Problem 10 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
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