2014 AMC 10A Problems/Problem 22
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 Solution 9
- 11 Video Solution by Richard Rusczyk
- 12 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,
Note: Similar to previous Solution
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 Therefore Rationalizing the denominator, Which by difference of squares reduces to 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
Solution 9
Similar to the others except that you find the base of the 15-75-90 triangle knowing that the side opposite side to 75 is 10, finding that you can subtract to find the base of the rectangle of the diagonal we are trying to find.
~YBSuburbanTea
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 | |
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