Difference between revisions of "2014 AMC 12B Problems/Problem 21"
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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>JG=HL=1</math>, so we must have <math>\triangle HGJ\cong\triangle JLH</math> by Hypotenuse-Leg congruence. From this congruence we have <math>LJ=HG=BE</math>. | 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>JG=HL=1</math>, so we must have <math>\triangle HGJ\cong\triangle JLH</math> by Hypotenuse-Leg congruence. From this congruence we have <math>LJ=HG=BE</math>. | ||
− | Notice that all four triangles in this picture are similar. Also, we have <math>LA=HD=EJ</math>. So set <math>x=LJ=HG=BE</math> and <math>y=LA=HD=EJ</math>. Now <math>BE+EJ+JL+LA | + | Notice that all four triangles in this picture are similar. Also, we have <math>LA=HD=EJ</math>. So set <math>x=LJ=HG=BE</math> and <math>y=LA=HD=EJ</math>. Now <math>BE+EJ+JL+LA=2(x+y)=1</math>. This means <math>x+y=\frac{1}{2}=BE+EJ=BJ</math>, so <math>J</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=\sqrt{3} \cdot AJ=\sqrt{3}/2</math> and subsequently <math>GD=\frac{2-\sqrt{3}}{2}=KE</math>. This means <math>EJ=\sqrt{3} \cdot KE=\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>. |
==Solution 2== | ==Solution 2== |
Revision as of 22:47, 6 December 2015
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 , 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 .
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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