Difference between revisions of "2014 AMC 10A Problems/Problem 22"
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[[Category: Introductory Geometry Problems]] | [[Category: Introductory Geometry Problems]] | ||
− | ==Solution (Trigonometry)== | + | ==Solution 1 (Trigonometry)== |
Note that <math>\tan 15^\circ=\frac{EC}{10} \Rightarrow EC=20-10 \sqrt 3</math>. (If you do not know the tangent half-angle formula, it is <math>\tan \frac{\theta}2= \frac{1-\cos \theta}{\sin \theta}</math>). Therefore, we have <math>DE=10\sqrt 3</math>. Since <math>ADE</math> is a <math>30-60-90</math> triangle, <math>AE=2 \cdot AD=2 \cdot 10=\boxed{\textbf{(E)} \: 20}</math> | Note that <math>\tan 15^\circ=\frac{EC}{10} \Rightarrow EC=20-10 \sqrt 3</math>. (If you do not know the tangent half-angle formula, it is <math>\tan \frac{\theta}2= \frac{1-\cos \theta}{\sin \theta}</math>). Therefore, we have <math>DE=10\sqrt 3</math>. Since <math>ADE</math> is a <math>30-60-90</math> triangle, <math>AE=2 \cdot AD=2 \cdot 10=\boxed{\textbf{(E)} \: 20}</math> | ||
− | ==Solution 2 ( | + | ==Solution 2 (No Trigonometry)== |
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>. | ||
− | + | ~edited by dolphin7 | |
− | |||
− | + | ==Solution 3 Quick Construction (No Trigonometry)== | |
− | + | Reflect <math>\triangle{ECB}</math> over line segment <math>\overline{CD}</math>. Let the point <math>F</math> be the point where the right angle is of our newly reflected triangle. By subtracting <math>90 - (15+15) = 60</math> to find <math>\angle ABF</math>, we see that <math>\triangle{ABC}</math> is a <math>30-60-90</math> right triangle. By using complementary angles once more, we can see that <math>\angle{EAD}</math> is a <math>60^\circ</math> angle, and we've found that <math>\triangle{EAD}</math> is a <math>30-60-90</math> right triangle. From here, we can use the <math>1-2-\sqrt{3}</math> properties of a <math>30-60-90</math> right triangle to see that <math>\overline{AE}=\boxed{\textbf{(E)}~20}.</math> | |
− | |||
− | |||
− | + | == Solution 4 (No Trigonometry) == | |
+ | |||
+ | Let <math>F</math> be a point on <math>BC</math> such that <math>\angle{FEC}=60^{\circ}</math>. Then <cmath>\angle{BEF}=\angle{BEC}-\angle{FEC}=15^{\circ}</cmath> Since <math>\angle{BEF}=\angle{EBF}</math>, <math>\bigtriangleup{BFE}</math> is isosceles. | ||
+ | |||
+ | Let <math>CF=x</math>. Since <math>\bigtriangleup{FEC}</math> is <math>60^{\circ}-90^{\circ}-30^{\circ}</math>, we have <math>EF=\frac{2}{\sqrt{3}}x</math> | ||
+ | |||
+ | Since <math>\bigtriangleup{BFE}</math> is isosceles, we have <math>BF=EF=\frac{2}{\sqrt{3}}x</math>. Since <math>BF+FC=BF</math>, we have <cmath>\frac{2}{\sqrt{3}}x+x=10 \Longrightarrow x=20\sqrt{3}-30</cmath> Thus <math>EC=\frac{1}{\sqrt{3}}BC=20-10\sqrt{3}</math> and <math>DE=DC-EC=20-EC=10\sqrt{3}</math>. | ||
+ | |||
+ | Finally, by the Pythagorean Theorem, we have <cmath>AE=\sqrt{AD^2+DE^2}=\sqrt{10^2+(10\sqrt{3})^2}=20 \boxed{\mathrm{(E)}}</cmath> | ||
+ | |||
+ | ~ 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 | ||
+ | |||
+ | ==Solution 5(15-75-90 Triangle)== | ||
+ | We notice a 15-75-90 triangle on <math>\triangle{ADE}</math>. | ||
+ | Implying the formula, we will continue as follows to get <math>\boxed{\mathrm{(E)}}</math>. ~mathboy282 | ||
+ | ===Note=== | ||
+ | Some people may not know this relationship; and this method is only for those who know it. If you do not know this relationship I highly suggest NOT using this solution and/or searching it up. | ||
+ | Here it one https://www.quora.com/What-are-the-side-relationships-of-a-15-75-90-triangle | ||
+ | |||
+ | == Video Solution by Richard Rusczyk == | ||
+ | |||
+ | https://www.youtube.com/watch?v=-GBvCLSfTuo | ||
==See Also== | ==See Also== |
Revision as of 01:05, 7 December 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
Solution 5(15-75-90 Triangle)
We notice a 15-75-90 triangle on . Implying the formula, we will continue as follows to get . ~mathboy282
Note
Some people may not know this relationship; and this method is only for those who know it. If you do not know this relationship I highly suggest NOT using this solution and/or searching it up. Here it one https://www.quora.com/What-are-the-side-relationships-of-a-15-75-90-triangle
Video Solution by Richard Rusczyk
https://www.youtube.com/watch?v=-GBvCLSfTuo
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.