2021 AMC 12B Problems/Problem 11
Contents
[hide]- 1 Problem
- 2 Diagram
- 3 Solution 1 (analytic geometry)
- 4 Solution 2
- 5 Solution 3
- 6 Solution 4 (Heron's Formula, Similar Triangles, Pythagorean Theorem)
- 7 Video Solution by Punxsutawney Phil
- 8 Video Solution by OmegaLearn (Using properties of 13-14-15 triangle)
- 9 Video Solution by Hawk Math
- 10 See Also
Problem
Triangle has and . Let be the point on such that . There are exactly two points and on line such that quadrilaterals and are trapezoids. What is the distance
Diagram
Solution 1 (analytic geometry)
Toss on the Cartesian plane with and . Then by the trapezoid condition, where . Since , point is of the way from to and is located at . Thus line has equation . Since and is parallel to the ground, we know has the same -coordinate as , except it'll also lie on the line . Therefore,
To find the location of point , we need to find the intersection of with a line parallel to passing through . The slope of this line is the same as the slope of , or , and has equation . The intersection of this line with is . Therefore point is located at
The distance is equal to the distance between and , which is
Solution 2
Using Stewart's Theorem we find . From the similar triangles and we have So
Solution 3
Let be the length . From the similar triangles and we have Therefore . Now extend line to the point on , forming parallelogram . As we also have so .
We now use the Law of Cosines to find (the length of ): As , we have (by Law of Cosines on triangle ) Therefore And . The answer is then
Solution 4 (Heron's Formula, Similar Triangles, Pythagorean Theorem)
Let the brackets denote areas. By Heron's Formula, we have It follows that the height of is
Next, we drop the altitudes and of By the Pythagorean Theorem on we get By the Pythagorean Theorem on we have
By the AA Similarity, with the ratio of similitude It follows that and Also, with the ratio of similitude It follows that
Finally,
PS: If you memorize that the area of a -- triangle is the Heron's Formula part will be done instantly.
~MRENTHUSIASM
Video Solution by Punxsutawney Phil
https://YouTube.com/watch?v=yxt8-rUUosI&t=450s
Video Solution by OmegaLearn (Using properties of 13-14-15 triangle)
~ pi_is_3.14
Video Solution by Hawk Math
https://www.youtube.com/watch?v=p4iCAZRUESs
See Also
2021 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 10 |
Followed by Problem 12 |
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 12 Problems and Solutions |
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