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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==Diagram== | ==Diagram== | ||
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\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, Similar Triangles, Pythagorean Theorem)== | ||
+ | Let the brackets denote areas. By Heron's Formula, we have | ||
+ | <cmath>\begin{align*} | ||
+ | [ABC]&=\sqrt{\frac{13+14+15}{2}\left(\frac{13+14+15}{2}-13\right)\left(\frac{13+14+15}{2}-14\right)\left(\frac{13+14+15}{2}-15\right)} \\ | ||
+ | &=\sqrt{21\left(21-13\right)\left(21-14\right)\left(21-15\right)} \\ | ||
+ | &=\sqrt{21\left(8\right)\left(7\right)\left(6\right)} \\ | ||
+ | &=\sqrt{\left(3\cdot7\right)\left(2^3\right)\left(7\right)\left(2\cdot3\right)} \\ | ||
+ | &=2^2\cdot3\cdot7 \\ | ||
+ | &=84. | ||
+ | \end{align*}</cmath> | ||
+ | It follows that the height of <math>ABCD</math> is <math>\frac{2[ABC]}{14}=12.</math> | ||
+ | |||
+ | Next, we drop the altitudes <math>\overline{AF}</math> and <math>\overline{DG}</math> of <math>ABCD.</math> By the Pythagorean Theorem on <math>\triangle AFB,</math> we get <math>BF=5.</math> By the Pythagorean Theorem on <math>\triangle DGB,</math> we have <math>BD=12\sqrt2.</math> | ||
+ | |||
+ | By the AA Similarity, <math>\triangle ADP\sim\triangle CBP,</math> with the ratio of similitude <math>1:2.</math> It follows that <math>BP=8\sqrt2</math> and <math>DP=4\sqrt2.</math> Also, <math>\triangle ABP\sim\triangle CEP,</math> with the ratio of similitude <math>1:2.</math> It follows that <math>EP=16\sqrt2.</math> | ||
+ | |||
+ | Finally, <math>DE=EP-DP=\boxed{\textbf{(D) }12\sqrt2}.</math> | ||
+ | |||
+ | ~MRENTHUSIASM | ||
==Video Solution by Punxsutawney Phil== | ==Video Solution by Punxsutawney Phil== |
Revision as of 19:49, 18 February 2021
Contents
- 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,
~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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.