Difference between revisions of "2021 AMC 12B Problems/Problem 11"
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<math>\textbf{(A) }\frac{42}5 \qquad \textbf{(B) }6\sqrt2 \qquad \textbf{(C) }\frac{84}5\qquad \textbf{(D) }12\sqrt2 \qquad \textbf{(E) }18</math> | <math>\textbf{(A) }\frac{42}5 \qquad \textbf{(B) }6\sqrt2 \qquad \textbf{(C) }\frac{84}5\qquad \textbf{(D) }12\sqrt2 \qquad \textbf{(E) }18</math> | ||
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− | + | ==Solution 1 (fakesolve)== | |
Using Stewart's Theorem of <math>man+dad=bmb+cnc</math> calculate the cevian to be <math>8\sqrt{2}</math>. It then follows that the answer must also have a factor of the <math>\sqrt{2}</math>. Having eliminated 3 answer choices, we then proceed to draw a rudimentary semiaccurate diagram of this figure. Drawing that, we realize that <math>6\sqrt2</math> is too small making out answer <math>\boxed{\textbf{(D) }12\sqrt2}</math> ~Lopkiloinm | Using Stewart's Theorem of <math>man+dad=bmb+cnc</math> calculate the cevian to be <math>8\sqrt{2}</math>. It then follows that the answer must also have a factor of the <math>\sqrt{2}</math>. Having eliminated 3 answer choices, we then proceed to draw a rudimentary semiaccurate diagram of this figure. Drawing that, we realize that <math>6\sqrt2</math> is too small making out answer <math>\boxed{\textbf{(D) }12\sqrt2}</math> ~Lopkiloinm | ||
− | + | ==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> | ||
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<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== | |
Let <math>x</math> be the length <math>PE</math>. From the similar triangles <math>BPA\sim DPC</math> and <math>BPC\sim EPA</math> we have | Let <math>x</math> be the length <math>PE</math>. From the similar triangles <math>BPA\sim DPC</math> and <math>BPC\sim EPA</math> we have | ||
<cmath>BP = \frac{PA}{PC}x = \frac12 x</cmath> | <cmath>BP = \frac{PA}{PC}x = \frac12 x</cmath> |
Revision as of 13:09, 12 February 2021
Contents
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
Solution 1 (fakesolve)
Using Stewart's Theorem of calculate the cevian to be
. It then follows that the answer must also have a factor of the
. Having eliminated 3 answer choices, we then proceed to draw a rudimentary semiaccurate diagram of this figure. Drawing that, we realize that
is too small making out answer
~Lopkiloinm
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.