Difference between revisions of "2014 AMC 12B Problems/Problem 21"
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Plugging into the previous equation with <math>x</math>, we get | Plugging into the previous equation with <math>x</math>, we get | ||
<cmath>x= 2\left(1-\sqrt{1-\frac{1}{4}}\right) = 2\left(\frac{2-\sqrt{3}}{2} \right) = \boxed{\textbf{(C)}\ 2-\sqrt{3}}</cmath> | <cmath>x= 2\left(1-\sqrt{1-\frac{1}{4}}\right) = 2\left(\frac{2-\sqrt{3}}{2} \right) = \boxed{\textbf{(C)}\ 2-\sqrt{3}}</cmath> | ||
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+ | ==Solution 3== | ||
+ | Let <math>BE = x</math>, <math>EK = a</math>, and <math>EJ = b</math>. Then <math>x^2 = a^2 + b^2</math> and because <math>\triangle KEJ \cong GDH</math> and <math>\triangle KEJ \sim \triangle JAG</math>, <math>\frac{GA}{1} = 1 - a = \frac{b}{x}</math>. Furthermore, the area of the four triangles and the two rectangles sums to 1: | ||
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+ | <cmath>1 = 2x + GA\cdot JA + ab</cmath> | ||
+ | |||
+ | <cmath>1 = 2x + (1 - a)(1 - (x + b)) + ab</cmath> | ||
+ | |||
+ | <cmath>1 = 2x + \frac{b}{x}(1 - x - b) + \left(1 - \frac{b}{x}\right)b</cmath> | ||
+ | |||
+ | <cmath>1 = 2x + \frac{b}{x} - b - \frac{b^2}{x} + b - \frac{b^2}{x}</cmath> | ||
+ | |||
+ | <cmath>x = 2x^2 + b - 2b^2</cmath> | ||
+ | |||
+ | <cmath>x - b = 2(x - b)(x + b)</cmath> | ||
+ | |||
+ | <cmath>x + b = \frac{1}{2}</cmath> | ||
+ | |||
+ | <cmath>b = \frac{1}{2} - x</cmath> | ||
+ | |||
+ | <cmath>a = 1 - \frac{b}{x} = 2 - \frac{1}{2x}</cmath> | ||
+ | |||
+ | By the Pythagorean theorem: <math>x^2 = a^2 + b^2</math> | ||
+ | |||
+ | <cmath>x^2 = \left(2 - \frac{1}{2x}\right)^2 + \left(\frac{1}{2} - x\right)^2</cmath> | ||
+ | |||
+ | <cmath>x^2 = 4 - \frac{2}{x} + \frac{1}{4x^2} + \frac{1}{4} - x + x^2</cmath> | ||
+ | |||
+ | <cmath>0 = \frac{1}{4x^2} - \frac{2}{x} + \frac{17}{4} - x</cmath> | ||
+ | |||
+ | <cmath>0 = 1 - 8x + 17x^2 - 4x^3</cmath>. | ||
+ | |||
+ | Then by the rational root theorem, this has roots <math>\frac{1}{4}</math>, <math>2 - \sqrt{3}</math>, and <math>2 + \sqrt{3}</math>. The first and last roots are extraneous because they imply <math>a = 0</math> and <math>x > 1</math>, respectively, thus <math>x = \boxed{\textbf{(C)}\ 2-\sqrt{3}}</math>. | ||
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== See also == | == See also == | ||
{{AMC12 box|year=2014|ab=B|num-b=20|num-a=22}} | {{AMC12 box|year=2014|ab=B|num-b=20|num-a=22}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 17:43, 9 March 2014
Contents
[hide]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 | |
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