Difference between revisions of "2014 AMC 12B Problems/Problem 21"
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<cmath>a = \frac{1 - \frac{\sqrt 3}{2}}{\frac{1}{2}} = 2 - \sqrt 3</cmath> | <cmath>a = \frac{1 - \frac{\sqrt 3}{2}}{\frac{1}{2}} = 2 - \sqrt 3</cmath> | ||
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+ | ==Solution 5== | ||
+ | Note that <math>HJ</math> is a diagonal of <math>JKHG</math>, so it must be equal in length to <math>FB</math>. Therefore, quadrilateral <math>FHJH</math> has <math>FH\parallel BJ</math>, and <math>FB=HJ</math>, so it must be either an isosceles trapezoid or a parallelogram. But due to the slope of <math>FB</math> and <math>HJ</math>, we see that it must be a parallelogram. Therefore, <math>FH=BJ</math>. But by the symmetry in rectangle <math>FEAD</math>, we see that <math>FH=JA</math>. Therefore, <math>BJ=FH=JA</math>. We also know that <math>BJ+JA=1</math>, hence <math>BJ=JA=\frac12</math>. | ||
+ | |||
+ | As <math>JG=1</math> and <math>JA=\frac12</math>, and as <math>\triangle GJA</math> is right, we know that <math>\triangle GJA</math> must be a 30-60-90 triangle. Therefore, <math>GA=\sqrt{3}/2</math> and <math>DG=1-\sqrt{3}/2</math>. But by similarity, <math>\triangle DHG</math> is also a 30-60-90 triangle, hence <math>DH=\sqrt{3}-3/2</math>. But <math>\triangle DHG\cong\triangle EJK</math>, hence <math>EJ=\sqrt{3}-3/2</math>. As <math>BJ=1/2</math>, this implies that <math>BE=BJ-EJ=1/2-\sqrt{3}+3/2=2-\sqrt{3}</math>. Thus the answer is <math>\boxed{\textbf{(C)}}</math>. | ||
== 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 14:51, 7 December 2016
Problem 21
In the figure, is a square of side length . The rectangles and are congruent. What is ?
Solution 1
Draw the altitude from to and call the foot . Then . Consider . It is the hypotenuse of both right triangles and , and we know , so we must have by Hypotenuse-Leg congruence. From this congruence we have .
Notice that all four triangles in this picture are similar. Also, we have . So set and . Now . This means , so 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 .
Solution 4
Let = and = . It is shown that all four triangles in the picture are similar. From the square side lengths:
Solving for we get:
Solution 5
Note that is a diagonal of , so it must be equal in length to . Therefore, quadrilateral has , and , so it must be either an isosceles trapezoid or a parallelogram. But due to the slope of and , we see that it must be a parallelogram. Therefore, . But by the symmetry in rectangle , we see that . Therefore, . We also know that , hence .
As and , and as is right, we know that must be a 30-60-90 triangle. Therefore, and . But by similarity, is also a 30-60-90 triangle, hence . But , hence . As , this implies that . Thus the answer is .
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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