Difference between revisions of "2021 AMC 12A Problems/Problem 20"
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<math>\textbf{(A) }13 \qquad \textbf{(B) }\frac{40}3 \qquad \textbf{(C) }\frac{41}3 \qquad \textbf{(D) }14\qquad \textbf{(E) }\frac{43}3</math> | <math>\textbf{(A) }13 \qquad \textbf{(B) }\frac{40}3 \qquad \textbf{(C) }\frac{41}3 \qquad \textbf{(D) }14\qquad \textbf{(E) }\frac{43}3</math> | ||
− | ==Solution== | + | ==Solution 1== |
Let <math>\ell</math> be the directrix of <math>\mathcal P</math>; recall that <math>\mathcal P</math> is the set of points <math>T</math> such that the distance from <math>T</math> to <math>\ell</math> is equal to <math>TF</math>. Let <math>P</math> and <math>Q</math> be the orthogonal projections of <math>F</math> and <math>A</math> onto <math>\ell</math>, and further let <math>X</math> and <math>Y</math> be the orthogonal projections of <math>F</math> and <math>V</math> onto <math>AQ</math>. Because <math>AF < AV</math>, there are two possible configurations which may arise, and they are shown below. | Let <math>\ell</math> be the directrix of <math>\mathcal P</math>; recall that <math>\mathcal P</math> is the set of points <math>T</math> such that the distance from <math>T</math> to <math>\ell</math> is equal to <math>TF</math>. Let <math>P</math> and <math>Q</math> be the orthogonal projections of <math>F</math> and <math>A</math> onto <math>\ell</math>, and further let <math>X</math> and <math>Y</math> be the orthogonal projections of <math>F</math> and <math>V</math> onto <math>AQ</math>. Because <math>AF < AV</math>, there are two possible configurations which may arise, and they are shown below. | ||
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&= FX^2 = AF^2 - AX^2 = 20^2 - (20 - 2d)^2 | &= FX^2 = AF^2 - AX^2 = 20^2 - (20 - 2d)^2 | ||
\end{align*}</cmath> | \end{align*}</cmath> | ||
− | This equation simplifies to <math>3d^2 - 40d + 41 = 0</math>, which has solutions <math>d = \tfrac{20\pm\sqrt{277}}3</math>. Both values of <math>d</math> work - the smaller solution with the | + | This equation simplifies to <math>3d^2 - 40d + 41 = 0</math>, which has solutions <math>d = \tfrac{20\pm\sqrt{277}}3</math>. Both values of <math>d</math> work - the smaller solution with the bottom configuration and the larger solution with the top configuration - and so the requested answer is <math>\boxed{\textbf{(B)}\ \tfrac{40}{3}}</math>. |
+ | |||
+ | ==Solution 2== | ||
+ | Let <math>\mathcal{P}</math> be the parabola, let <math>V</math> be the origin, <math>F</math> lie on the positive <math>y</math> axis, and <math>d=FV</math>. The equation of the parabola is then <math>x^{2}=4dy</math>. If the coordinates of <math>A</math> are <math>(p, q),</math> then <math>p^{2}+q^{2}=441</math> since the distance from the origin to <math>A</math> is <math>21</math>. Note also that the parabola is the set of all points equidistant from <math>F</math> and a line known as its directrix, which in this case is a horizontal line <math>d</math> units below the origin. Since the distance from <math>A</math> to its directrix is equal to <math>AF,</math> then <math>A</math> is <math>20</math> units above this line and therefore <math>q=20-d</math>. Substituting for <math>p</math> and <math>q</math> yields <math>4d(20-d)+(20-d)^{2}=441</math> or <math>(20-d)(20+3d)=441,</math> which simplifies to <math>3d^{2}-40d+41=0</math>. Therefore the sum of all possible values of <math>d</math> is <math>\tfrac{-b}{a}=\boxed{\textbf{(B)} ~\tfrac{40}{3}}</math> by Viète's Formulas. | ||
+ | |||
+ | ~sugar_rush | ||
== Video Solution by OmegaLearn (Using parabola properties and system of equations) == | == Video Solution by OmegaLearn (Using parabola properties and system of equations) == | ||
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~ pi_is_3.14 | ~ pi_is_3.14 | ||
+ | |||
+ | ==Video Solution by TheBeautyofMath== | ||
+ | https://youtu.be/WidS_IOjkQo | ||
+ | |||
+ | ~IceMatrix | ||
==See also== | ==See also== | ||
{{AMC12 box|year=2021|ab=A|num-b=19|num-a=21}} | {{AMC12 box|year=2021|ab=A|num-b=19|num-a=21}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 17:33, 13 July 2021
Contents
Problem
Suppose that on a parabola with vertex and a focus
there exists a point
such that
and
. What is the sum of all possible values of the length
Solution 1
Let be the directrix of
; recall that
is the set of points
such that the distance from
to
is equal to
. Let
and
be the orthogonal projections of
and
onto
, and further let
and
be the orthogonal projections of
and
onto
. Because
, there are two possible configurations which may arise, and they are shown below.
Set
, which by the definition of a parabola also equals
. Then as
, we have
and
. Since
is a rectangle,
, so by Pythagorean Theorem on triangles
and
,
This equation simplifies to
, which has solutions
. Both values of
work - the smaller solution with the bottom configuration and the larger solution with the top configuration - and so the requested answer is
.
Solution 2
Let be the parabola, let
be the origin,
lie on the positive
axis, and
. The equation of the parabola is then
. If the coordinates of
are
then
since the distance from the origin to
is
. Note also that the parabola is the set of all points equidistant from
and a line known as its directrix, which in this case is a horizontal line
units below the origin. Since the distance from
to its directrix is equal to
then
is
units above this line and therefore
. Substituting for
and
yields
or
which simplifies to
. Therefore the sum of all possible values of
is
by Viète's Formulas.
~sugar_rush
Video Solution by OmegaLearn (Using parabola properties and system of equations)
~ pi_is_3.14
Video Solution by TheBeautyofMath
~IceMatrix
See also
2021 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 19 |
Followed by Problem 21 |
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.