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>HG=BE</math> and <math>JG=1</math>, so we must 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>HG=BE</math> and <math>JG=1</math>, so we must have <math>LJ=HG=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>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= | + | Notice that all four triangles in this picture are similar and thus we have <math>LA=HD=EJ=1-BE</math>. This means <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>. |
==Solution2== | ==Solution2== |
Revision as of 18:18, 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 , 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 |
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