Difference between revisions of "2014 AMC 10A Problems/Problem 22"
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+ | ==Solution 8 (Trigonometry)== | ||
+ | All trigonometric functions in this solution are in degrees. We know <cmath>\sin\left(a+b\right)=\sin\left(a\right)\cos\left(b\right)+\sin\left(b\right)\cos\left(a\right)</cmath> so <cmath>\sin\left(15\right)=\sin\left(45-30\right)=\sin\left(45\right)\cos\left(-30\right)+\sin\left(-30\right)\cos\left(45\right)</cmath> | ||
+ | <cmath>=\frac{\sqrt{2}}{2}\cdot\left(-\frac{\sqrt{3}}{2}\right)+\frac{1}{2}\cdot\frac{\sqrt{2}}{2}=\frac{-\sqrt{6}}{4}+\frac{\sqrt{2}}{4}=\frac{\sqrt{2}-\sqrt{6}}{4}</cmath> | ||
+ | <cmath>=\frac{\sqrt{2}-\sqrt{6}}{4}</cmath> | ||
+ | Let <math>EC=x</math>, then <math>BE=\sqrt{x^{2}+100}</math>. By the definition of sine, | ||
+ | <cmath>\frac{x}{\sqrt{x^{2}+100}}=\frac{\sqrt{2}-\sqrt{6}}{4}</cmath> | ||
+ | Squaring both sides, | ||
+ | <cmath>\frac{x^{2}}{x^{2}+100}=\frac{\left(\sqrt{2}-\sqrt{6}\right)^{2}}{16}=\frac{2-2\sqrt{12}+6}{16}=\frac{8-4\sqrt{3}}{16}=\frac{2-\sqrt{3}}{4}</cmath> | ||
+ | Cross-multiplying, | ||
+ | <cmath>4x^{2}=\left(2-\sqrt{3}\right)\left(x^{2}+100\right)=2x^{2}+200-\sqrt{3}x^{2}-100\sqrt{3}</cmath> | ||
+ | Simplifying, | ||
+ | <cmath>\left(2+\sqrt{3}\right)x^{2}=200-100\sqrt{3}</cmath> | ||
+ | <cmath>x^{2}=\frac{200-100\sqrt{3}}{2+\sqrt{3}}=\frac{100\left(2-\sqrt{3}\right)}{2+\sqrt{3}}=100\cdot\frac{2-\sqrt{3}}{2+\sqrt{3}}</cmath> | ||
+ | Let <math>\frac{2-\sqrt{3}}{2+\sqrt{3}}=x</math>. Notice that <math>\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)=2^{2}-\sqrt{3}^{2}=1</math> so <math>2-\sqrt{3}=\frac{1}{2+\sqrt{3}}</math>. <math>x</math> is then <cmath>\frac{2-\sqrt{3}}{2+\sqrt{3}}=\frac{\frac{1}{2+\sqrt{3}}}{2+\sqrt{3}}=\frac{1}{\left(2+\sqrt{3}\right)^{2}}</cmath> | ||
+ | Recall that | ||
+ | <cmath>x^{2}=100\cdot\frac{2-\sqrt{3}}{2+\sqrt{3}}</cmath> which we now know is <cmath>100\cdot\frac{1}{\left(2+\sqrt{3}\right)^{2}}=\frac{100}{\left(2+\sqrt{3}\right)^{2}}=\left(\frac{10}{2+\sqrt{3}}\right)^{2} | ||
+ | Therefore </cmath>x=\frac{10}{2+\sqrt{3}}<cmath> | ||
+ | Rationalizing the denominator, | ||
+ | </cmath>\frac{10}{2+\sqrt{3}}\cdot\frac{2-\sqrt{3}}{2-\sqrt{3}}=\frac{20-10\sqrt{3}}{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}<cmath> | ||
+ | Which by difference of squares reduces to | ||
+ | </cmath>20-10\sqrt{3}<math></math> | ||
+ | so <math>EC=20-10\sqrt{3}</math>. <math>ED</math> is then <math>20-\left(20-10\sqrt{3}\right)=10\sqrt{3}</math> and since we know <math>AD=10</math>, by the Pythagorean theorem, <math>AE = 20</math>. The answer is <math>\boxed{\textbf{(E)}~20}</math> | ||
+ | |||
+ | An alternate way to finish: since we know the lengths of <math>AD</math> and <math>DE</math>, we can figure out that <math>m\angle AED=30^{\circ}</math> and therefore <math>m\angle BEA=75^{\circ}</math>. Hence <math>\triangle ABE</math> is isosceles and <math>AE=AB=\boxed{\textbf{(E)}~20}</math>. | ||
+ | |||
+ | ~JH. L | ||
== Video Solution by Richard Rusczyk == | == Video Solution by Richard Rusczyk == |
Revision as of 02:00, 20 June 2022
Contents
- 1 Problem
- 2 Solution 1 (Trigonometry)
- 3 Solution 2 (No Trigonometry)
- 4 Solution 3 Quick Construction (No Trigonometry)
- 5 Solution 4 (No Trigonometry)
- 6 Solution 5
- 7 Solution 6 (Pure Euclidian Geometry)
- 8 Solution 7 (Pure Euclidian Geometry)
- 9 Solution 8 (Trigonometry)
- 10 Video Solution by Richard Rusczyk
- 11 See Also
Problem
In rectangle , and . Let be a point on such that . What is ?
Solution 1 (Trigonometry)
Note that . (It is important to memorize the sin, cos, and tan values of and .) 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 ripkobe_745
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
First, divide all side lengths by to make things easier. We’ll multiply our answer by at the end. Call side length . Using the Pythagorean Theorem, we can get side is .
The double angle identity for sine states that: So, We know . In triangle , and . Substituting these in, we get our equation: which simplifies to
Now, using the quadratic formula to solve for . Because the length must be close to one, the value of will be . We can now find = and use it to find . . To find , we can use the Pythagorean Theorem with sides and , OR we can notice that, based on the two side lengths we know, is a triangle. So .
Finally, we must multiply our answer by , . .
~AWCHEN01
Solution 6 (Pure Euclidian Geometry)
We are going to use pure Euclidian geometry to prove .
Reflect rectangle along line . Let the square be as shown. Construct equilateral triangle .
Because , , and , by .
So, , .
Because , , , .
by .
So, . By the reflection, .
This solution is inspired by AoPS "Introduction to Geometry" page 226 problem 8.22, and page 433 problem 16.42.
Solution 7 (Pure Euclidian Geometry)
We are going to use pure Euclidian geometry to prove .
Construct equilateral triangle , and let be the height of .
, , , .
by .
, , , by .
So, .
, , , , .
by .
So,
Solution 8 (Trigonometry)
All trigonometric functions in this solution are in degrees. We know so Let , then . By the definition of sine, Squaring both sides, Cross-multiplying, Simplifying, Let . Notice that so . is then Recall that which we now know is x=\frac{10}{2+\sqrt{3}}\frac{10}{2+\sqrt{3}}\cdot\frac{2-\sqrt{3}}{2-\sqrt{3}}=\frac{20-10\sqrt{3}}{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}20-10\sqrt{3}$$ (Error compiling LaTeX. Unknown error_msg) so . is then and since we know , by the Pythagorean theorem, . The answer is
An alternate way to finish: since we know the lengths of and , we can figure out that and therefore . Hence is isosceles and .
~JH. L
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.