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> | ||
− | == | + | ==Solution1== |
+ | Draw the attitude from <math>H</math> to <math>AB</math> and call the foot <math>L</math>. Then <math>HL=1</math>. Consider <math>HJ</math>. It is the hypotenuse of both right triangles <math>\triangle HGJ</math> and <math>\triangle HLJ</math>, and we know <math>HG=x</math> and <math>JG=1</math>, so we must have <math>LJ=BE</math>. | ||
+ | |||
+ | Notice that all four triangles in this picture are similar and thus we have <math>LA=HD=EJ=1-BE</math>. This means <math>L</math> is the midpoint of <math>AB</math>. So <math>\triangle AJG</math>, along with all other similar triangles in the picture, is a 30-60-90 triangle, and we have <math>AG=KE=\sqrt{3}/2</math> and subsequently <math>GD=\frac{2-\sqrt{3}}{2}</math>. This means <math>EJ=\frac{2\sqrt{3}-3}{2}</math>, which gives <math>BE=\frac{1}{2}-EJ=\frac{4-2\sqrt{3}{2}=2-\sqrt{3}</math>, so the answer is <math>\textbf{(C)}</math>. | ||
+ | |||
+ | |||
+ | ==Solution2== | ||
Let <math>BE = x</math>. Let <math>JA = y</math>. Because <math>\angle FKH = \angle EJK = \angle AGJ = \angle DHG</math> and <math>\angle FHK = \angle EKJ = \angle AJG = \angle DGH</math>, <math>\triangle KEJ, \triangle JAG, \triangle GDH, \triangle HFK</math> are all similar. Using proportions and the pythagorean theorem, we find | Let <math>BE = x</math>. Let <math>JA = y</math>. Because <math>\angle FKH = \angle EJK = \angle AGJ = \angle DHG</math> and <math>\angle FHK = \angle EKJ = \angle AJG = \angle DGH</math>, <math>\triangle KEJ, \triangle JAG, \triangle GDH, \triangle HFK</math> are all similar. Using proportions and the pythagorean theorem, we find | ||
<cmath>EK = xy</cmath> | <cmath>EK = xy</cmath> |
Revision as of 18:15, 21 February 2014
Problem 21
In the figure, is a square of side length . The rectangles and are congruent. What is ?
Solution1
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 $BE=\frac{1}{2}-EJ=\frac{4-2\sqrt{3}{2}=2-\sqrt{3}$ (Error compiling LaTeX. Unknown error_msg), so the answer is .
Solution2
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
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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