Difference between revisions of "2014 AMC 12B Problems/Problem 21"
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<math> \textbf{(A) }\frac{1}{2}(\sqrt{6}-2)\qquad\textbf{(B) }\frac{1}{4}\qquad\textbf{(C) }2-\sqrt{3}\qquad\textbf{(D) }\frac{\sqrt{3}}{6}\qquad\textbf{(E) } 1-\frac{\sqrt{2}}{2}</math> | <math> \textbf{(A) }\frac{1}{2}(\sqrt{6}-2)\qquad\textbf{(B) }\frac{1}{4}\qquad\textbf{(C) }2-\sqrt{3}\qquad\textbf{(D) }\frac{\sqrt{3}}{6}\qquad\textbf{(E) } 1-\frac{\sqrt{2}}{2}</math> | ||
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==Solution 1== | ==Solution 1== |
Revision as of 10:55, 13 August 2014
Problem 21
In the figure, is a square of side length . The rectangles and are congruent. What is ?
Solution 1
Draw the attitude from to and call the foot . Then . Consider . It is the hypotenuse of both right triangles and , and we know and , so we must have .
Notice that all four triangles in this picture are similar and thus we have . This means is the midpoint of . So , along with all other similar triangles in the picture, is a 30-60-90 triangle, and we have and subsequently . This means , which gives , so the answer is .
Solution 2
Let . Let . Because and , are all similar. Using proportions and the pythagorean theorem, we find Because we know that , we can set up a systems of equations Solving for in the second equation, we get Plugging this into the first equation, we get Plugging into the previous equation with , we get
Solution 3
Let , , and . Then and because and , . Furthermore, the area of the four triangles and the two rectangles sums to 1:
By the Pythagorean theorem:
Then by the rational root theorem, this has roots , , and . The first and last roots are extraneous because they imply and , respectively, thus .
See also
2014 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 20 |
Followed by Problem 22 |
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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