Difference between revisions of "2021 AMC 12B Problems/Problem 11"
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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> | 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)} | + | 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\sqrt2}</math>. |
==Solution 2== | ==Solution 2== | ||
− | Using Stewart's Theorem we find <math>BP = 8\sqrt{2}</math>. From the similar triangles <math>BPA\sim DPC</math> and <math>BPC\sim EPA</math> we have | + | Using Stewart's Theorem we find <math>BP = 8\sqrt{2}</math>. From the similar triangles <math>BPA\sim DPC</math> and <math>BPC\sim EPA</math>, we have |
<cmath>DP = BP\cdot\frac{PC}{PA} = 2BP</cmath> | <cmath>DP = BP\cdot\frac{PC}{PA} = 2BP</cmath> | ||
<cmath>EP = BP\cdot\frac{PA}{PC} = \frac12 BP</cmath> | <cmath>EP = BP\cdot\frac{PA}{PC} = \frac12 BP</cmath> | ||
− | So | + | So, <cmath>DE = \frac{3}{2}BP = \boxed{\textbf{(D) }12\sqrt2}.</cmath> |
− | <cmath>DE = \frac{3}{2}BP = \boxed{\textbf{(D) }12\sqrt2}</cmath> | ||
==Solution 3== | ==Solution 3== | ||
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&= 512 | &= 512 | ||
\end{align*}</cmath> | \end{align*}</cmath> | ||
− | And <math>x = 16\sqrt2</math>. The answer is then <math>\frac34x = \boxed{\textbf{(D) }12\sqrt2}</math> | + | And <math>x = 16\sqrt2</math>. The answer is then <math>\frac34x = \boxed{\textbf{(D) }12\sqrt2}</math>. |
==Solution 4 (Heron's Formula, Pythagorean Theorem, Similar Triangles)== | ==Solution 4 (Heron's Formula, Pythagorean Theorem, Similar Triangles)== | ||
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We can find <math>P</math> in barycentric coordinates as <math>\Bigr(\frac{2}{3},0,\frac{1}{3}\Bigr)</math>. We can then write <math>\overline{BP}</math> as <math>x-2z=0</math>, where <math>(x,y,z)</math> defines a point in barycentric coordinates. We have <math>\overline{AD}\parallel \overline{BC}</math> as <math>y+z=0</math> and <math>\overline{CE}\parallel \overline{AB}</math> as <math>x+y=0</math>. We can then compute <math>D</math> and <math>E</math> by intersecting lines: | We can find <math>P</math> in barycentric coordinates as <math>\Bigr(\frac{2}{3},0,\frac{1}{3}\Bigr)</math>. We can then write <math>\overline{BP}</math> as <math>x-2z=0</math>, where <math>(x,y,z)</math> defines a point in barycentric coordinates. We have <math>\overline{AD}\parallel \overline{BC}</math> as <math>y+z=0</math> and <math>\overline{CE}\parallel \overline{AB}</math> as <math>x+y=0</math>. We can then compute <math>D</math> and <math>E</math> by intersecting lines: | ||
− | |||
<cmath>\begin{cases} | <cmath>\begin{cases} | ||
x-2z=0\\ | x-2z=0\\ | ||
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x+y+z=1 | x+y+z=1 | ||
\end{cases}</cmath> | \end{cases}</cmath> | ||
− | + | Which gives us <math>D=\left(1, -\frac{1}{2}, \frac{1}{2}\right)</math>. We can get <math>E</math> with: | |
− | Which gives us <math>D=(1, -\frac{1}{2}, \frac{1}{2})</math>. We can get <math>E</math> with: | ||
− | |||
<cmath> | <cmath> | ||
\begin{cases} | \begin{cases} | ||
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\end{cases} | \end{cases} | ||
</cmath> | </cmath> | ||
− | + | Which gives us <math>E=(2, -2, 1)</math>. Then, finding the displacement vector, we have <math>\overrightarrow{ED}=\left(1,-\frac{3}{2},\frac{1}{2}\right)</math>. Using the barycentric distance formula: | |
− | Which gives us <math>E=(2, -2, 1)</math>. Then, finding the displacement vector, we have <math>\overrightarrow{ED}=(1,-\frac{3}{2},\frac{1}{2})</math>. Using the barycentric distance formula: | + | <cmath>\begin{align*} |
− | + | \text{dist}(D,E)&=\sqrt{-a^2yx-b^2zx-c^2xy} \\ | |
− | <cmath>\text{dist}(D,E)=\sqrt{-a^2yx-b^2zx-c^2xy} | + | &=\sqrt{-14^2\Bigr(-\frac{3}{2}\Bigr)\Bigr(\frac{1}{2}\Bigr)-15^2\Bigr(1\Bigr)\Bigr(\frac{1}{2}\Bigr)-13^2\Bigr(-\frac{3}{2}\Bigr)\Bigr(\frac{1}{2}\Bigr)} \\ |
− | + | &=\boxed{\textbf{(D) }12\sqrt2}. | |
− | + | \end{align*}</cmath> | |
==Video Solution by Punxsutawney Phil== | ==Video Solution by Punxsutawney Phil== |
Latest revision as of 19:31, 17 June 2021
Contents
- 1 Problem
- 2 Diagram
- 3 Solution 1 (Analytic Geometry)
- 4 Solution 2
- 5 Solution 3
- 6 Solution 4 (Heron's Formula, Pythagorean Theorem, Similar Triangles)
- 7 Solution 5 (Barycentric Coordinates)
- 8 Video Solution by Punxsutawney Phil
- 9 Video Solution by OmegaLearn (Properties of 13-14-15 Triangle)
- 10 Video Solution by Hawk Math
- 11 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, Pythagorean Theorem, Similar Triangles)
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 right we get By the AA Similarity, we have with the ratio of similitude It follows that Since is a rectangle, we get By the Pythagorean Theorem on right we get
By again, we get and Also, by the AA Similarity, we have with the ratio of similitude It follows that
Finally, we obtain
Remark
If you memorize that the area of a triangle is then the Heron's Formula part will be done instantly.
~MRENTHUSIASM
Solution 5 (Barycentric Coordinates)
(For those unfamiliar with barycentric coordinates, consider reading the barycentric coordinates article written by Evan Chen and Max Schindler here: https://artofproblemsolving.com/wiki/index.php/Resources_for_mathematics_competitions#Articles)
We can find in barycentric coordinates as . We can then write as , where defines a point in barycentric coordinates. We have as and as . We can then compute and by intersecting lines: Which gives us . We can get with: Which gives us . Then, finding the displacement vector, we have . Using the barycentric distance formula:
Video Solution by Punxsutawney Phil
https://YouTube.com/watch?v=yxt8-rUUosI&t=450s
Video Solution by OmegaLearn (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.