Difference between revisions of "2014 AMC 10A Problems/Problem 22"
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Let <math>F</math> be a point on line <math>\overline{CD}</math> such that points <math>C</math> and <math>F</math> are distinct and that <math>\angle EBF = 15^\circ</math>. By the angle bisector theorem, <math>\frac{\overline{BC}}{\overline{BF}} = \frac{\overline{CE}}{\overline{EF}}</math>. Since <math>\triangle BFC</math> is a <math>30-60-90</math> right triangle, <math>\overline{CF} = \frac{10\sqrt{3}}{3}</math> and <math>\overline{BF} = \frac{20\sqrt{3}}{3}</math>. Additionally, <cmath>\overline{CE} + \overline{EF} = \overline{CF} = \frac{10\sqrt{3}}{3}</cmath>Now, substituting in the obtained values, we get <math>\frac{10}{\frac{20\sqrt{3}}{3}} = \frac{\overline{CE}}{\overline{EF}} \Rightarrow \frac{2\sqrt{3}}{3}\overline{CE} = \overline{EF}</math> and <math>\overline{CE} + \overline{EF} = \frac{10\sqrt{3}}{3}</math>. Substituting the first equation into the second yields <math>\frac{2\sqrt{3}}{3}\overline{CE} + \overline{CE} = \frac{10\sqrt{3}}{3} \Rightarrow \overline{CE} = 20 - 10\sqrt{3}</math>, so <math>\overline{DE} = 10\sqrt{3}</math>. Because <math>\triangle ADE</math> is a <math>30-60-90</math> triangle, <math>\overline{AE} = \boxed{\textbf{(E)}~20}</math>. | Let <math>F</math> be a point on line <math>\overline{CD}</math> such that points <math>C</math> and <math>F</math> are distinct and that <math>\angle EBF = 15^\circ</math>. By the angle bisector theorem, <math>\frac{\overline{BC}}{\overline{BF}} = \frac{\overline{CE}}{\overline{EF}}</math>. Since <math>\triangle BFC</math> is a <math>30-60-90</math> right triangle, <math>\overline{CF} = \frac{10\sqrt{3}}{3}</math> and <math>\overline{BF} = \frac{20\sqrt{3}}{3}</math>. Additionally, <cmath>\overline{CE} + \overline{EF} = \overline{CF} = \frac{10\sqrt{3}}{3}</cmath>Now, substituting in the obtained values, we get <math>\frac{10}{\frac{20\sqrt{3}}{3}} = \frac{\overline{CE}}{\overline{EF}} \Rightarrow \frac{2\sqrt{3}}{3}\overline{CE} = \overline{EF}</math> and <math>\overline{CE} + \overline{EF} = \frac{10\sqrt{3}}{3}</math>. Substituting the first equation into the second yields <math>\frac{2\sqrt{3}}{3}\overline{CE} + \overline{CE} = \frac{10\sqrt{3}}{3} \Rightarrow \overline{CE} = 20 - 10\sqrt{3}</math>, so <math>\overline{DE} = 10\sqrt{3}</math>. Because <math>\triangle ADE</math> is a <math>30-60-90</math> triangle, <math>\overline{AE} = \boxed{\textbf{(E)}~20}</math>. | ||
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+ | ~edited by dolphin7 | ||
==Solution 3 Quick Construction (No Trigonometry)== | ==Solution 3 Quick Construction (No Trigonometry)== |
Revision as of 14:29, 3 April 2020
Contents
Problem
In rectangle , and . Let be a point on such that . What is ?
Solution 1 (Trigonometry)
Note that . (If you do not know the tangent half-angle formula, it is ). Therefore, we have . Since is a triangle,
Solution 2 (No Trigonometry)
Let be a point on line such that points and are distinct and that . By the angle bisector theorem, . Since is a right triangle, and . Additionally, Now, substituting in the obtained values, we get and . Substituting the first equation into the second yields , so . Because is a triangle, .
~edited by dolphin7
Solution 3 Quick Construction (No Trigonometry)
Reflect over line segment . Let the point be the point where the right angle is of our newly reflected triangle. By subtracting to find , we see that is a right triangle. By using complementary angles once more, we can see that is a angle, and we've found that is a right triangle. From here, we can use the properties of a right triangle to see that
Solution 4 (No Trigonometry)
Let be a point on such that . Then Since , is isosceles.
Let . Since is , we have
Since is isosceles, we have . Since , we have Thus and .
Finally, by the Pythagorean Theorem, we have
~ Solution by Nafer
~ Edited by TheBeast5520
Note from williamgolly: When you find DE, note how ADE is congruent to a 30-60-90 triangle and you can easily find AE from there
Video Solution
https://www.youtube.com/watch?v=-GBvCLSfTuo
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See Also
2014 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 21 |
Followed by Problem 23 | |
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 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.