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> | ||
− | |||
==Diagram== | ==Diagram== | ||
<asy> | <asy> | ||
− | size( | + | /* Made by Brendanb4321; edited by MRENTHUSIASM */ |
+ | size(250); | ||
pair A = (5,12); | pair A = (5,12); | ||
pair B = (0,0); | pair B = (0,0); | ||
Line 14: | Line 14: | ||
pair D = 3/2*P; | pair D = 3/2*P; | ||
pair E = 3*P; | pair E = 3*P; | ||
− | draw(A--B--C-- | + | draw(A--B--C--cycle^^A--D^^C--E--B); |
− | |||
− | |||
− | dot("$A$",A,N); | + | dot("$A$",A,1.5*N); |
− | dot("$B$",B, | + | dot("$B$",B,1.5*SW); |
− | dot("$C$",C, | + | dot("$C$",C,1.5*SE); |
− | dot("$D$",D,N); | + | dot("$D$",D,1.5*N); |
− | dot("$P$",P,W); | + | dot("$P$",P,1.5*W); |
− | dot("$E$",E,N); | + | dot("$E$",E,1.5*N); |
− | + | label("$13$", (A+B)/2, 1.5*NW); | |
− | label("$13$", (A+B)/2, NW); | + | label("$14$", (B+C)/2, 1.5*S); |
− | label("$14$", (B+C)/2, S); | + | label("$5$", (A+P)/2, NE); |
− | label("$5$",(A+P)/2, NE); | ||
label("$10$", (C+P)/2, NE); | label("$10$", (C+P)/2, NE); | ||
</asy> | </asy> | ||
+ | ~Brendanb4321 | ||
− | ==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> | 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> | 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 (Stewart's Theorem)== |
− | 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} = | + | <cmath>DP = BP\cdot\frac{PC}{PA} = \frac{1}{2}BP</cmath> |
− | <cmath>EP = BP\cdot\frac{PA}{PC} = | + | <cmath>EP = BP\cdot\frac{PA}{PC} = 2BP</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== | ||
Line 62: | Line 59: | ||
&= 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)== | ||
+ | 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 right <math>\triangle AFB,</math> we get <math>BF=5.</math> By the AA Similarity, we have <math>\triangle ADP\sim\triangle CBP,</math> with the ratio of similitude <math>1:2.</math> It follows that <math>AD=7.</math> Since <math>ADGF</math> is a rectangle, we get <math>FG=AD=7.</math> By the Pythagorean Theorem on right <math>\triangle DGB,</math> we get <math>BD=12\sqrt2.</math> | ||
+ | |||
+ | By <math>\triangle ADP\sim\triangle CBP</math> again, we get <math>BP=8\sqrt2</math> and <math>DP=4\sqrt2.</math> Also, by the AA Similarity, we have <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, we obtain <math>DE=EP-DP=\boxed{\textbf{(D) }12\sqrt2}.</math> | ||
+ | |||
+ | <u><b>Remark</b></u> | ||
+ | |||
+ | It is well known that the area of a <math>13\text{-}14\text{-}15</math> triangle is <math>84.</math> | ||
+ | |||
+ | ~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 <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} | ||
+ | x-2z=0\\ | ||
+ | y+z=0\\ | ||
+ | x+y+z=1 | ||
+ | \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: | ||
+ | <cmath> | ||
+ | \begin{cases} | ||
+ | x-2z=0\\ | ||
+ | x+y=0\\ | ||
+ | x+y+z=1 | ||
+ | \end{cases} | ||
+ | </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: | ||
+ | <cmath>\begin{align*} | ||
+ | \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== | ||
https://YouTube.com/watch?v=yxt8-rUUosI&t=450s | https://YouTube.com/watch?v=yxt8-rUUosI&t=450s | ||
− | == Video Solution by OmegaLearn ( | + | == Video Solution by OmegaLearn (Properties of 13-14-15 Triangle) == |
https://youtu.be/mTcdKf5-FWg | https://youtu.be/mTcdKf5-FWg | ||
Latest revision as of 12:59, 24 July 2024
Contents
- 1 Problem
- 2 Diagram
- 3 Solution 1 (Analytic Geometry)
- 4 Solution 2 (Stewart's Theorem)
- 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
~Brendanb4321
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 (Stewart's Theorem)
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
It is well known that the area of a triangle is
~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.