Difference between revisions of "2021 AMC 12B Problems/Problem 11"
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− | ==Solution 1 ( | + | ==Solution 1 (analytic geometry)== |
+ | Toss on the Cartesian plane with <math>A=(5, 12), B=(0, 0),</math> and <math>C=(14, 0)</math>. Then <math>\overline{AD}\parallel\overline{BC}, \overline{AB}\parallel\overline{CE}</math> by the trapezoid condition, where <math>D, E\in\overline{BP}</math>. Since <math>PC=10</math>, point <math>P</math> is <math>\tfrac{10}{15}=\tfrac{2}{3}</math> of the way from <math>C</math> to <math>A</math> and is located at <math>(8, 8)</math>. Thus line <math>BP</math> has equation <math>y=x</math>. Since <math>\overline{AD}\parallel\overline{BC}</math> and <math>\overline{BC}</math> is parallel to the ground, we know <math>D</math> has the same <math>y</math>-coordinate as <math>A</math>, except it'll also lie on the line <math>y=x</math>. Therefore, <math>D=(12, 12). \, \blacksquare</math> | ||
− | + | To find the location of point <math>E</math>, we need to find the intersection of <math>y=x</math> with a line parallel to <math>\overline{AB}</math> passing through <math>C</math>. The slope of this line is the same as the slope of <math>\overline{AB}</math>, or <math>\tfrac{12}{5}</math>, and has equation <math>y=\tfrac{12}{5}x-\tfrac{168}{5}</math>. The intersection of this line with <math>y=x</math> is <math>(24, 24)</math>. Therefore point <math>E</math> is located at <math>(24, 24). \, \blacksquare</math> | |
+ | |||
+ | The distance <math>DE</math> is equal to the distance between <math>(12, 12)</math> and <math>(24, 24)</math>, which is <math>\boxed{\textbf{(D)} ~12\sqrt{2}}</math> | ||
==Solution 2== | ==Solution 2== |
Revision as of 16:09, 16 February 2021
Contents
[hide]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
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.