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Difference between revisions of "2021 AMC 12A Problems/Problem 20"

(Created page with "==Problem== These problems will not be posted until the 2021 AMC12A is released on Thursday, February 4, 2021. ==Solution== The solutions will be posted once the problems are...")
 
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==Problem==
 
==Problem==
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These problems will not be posted until the 2021 AMC12A is released on Thursday, February 4, 2021.
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Suppose that on a parabola with vertex <math>V</math> and a focus <math>F</math> there exists a point <math>A</math> such that <math>AF=20</math> and <math>AV=21</math>. What is the sum of all possible values of the length <math>FV?</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==
−
The solutions will be posted once the problems are posted.
+
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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==Note==
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−
See [[2021 AMC 12A Problems/Problem 1|problem 1]].
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<asy>
 +
import olympiad;
 +
size(230);
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defaultpen(linewidth(0.8)+fontsize(11pt));
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real d = 1.1, edge = 2.5, Ax = 1.6;
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real f(real x)
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{
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return 1/(4*d) * x * x;
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}
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pair V = origin, F = (0,d), la = (-edge,-d), lb = (edge,-d), A = (Ax, f(Ax));
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pair P = foot(F,la,lb), Q = foot(A,la,lb), X = foot(F,A,Q), Y = foot(V,A,Q);
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draw(P--F--A--V--Y^^F--X--Q^^rightanglemark(F,P,la,4)^^rightanglemark(A,Q,lb,4)^^rightanglemark(A,X,F,4)^^rightanglemark(A,Y,V,4),red);
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draw(graph(f,-2.5,2.5));
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draw(la -- lb);
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dot(F^^A^^V);
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label("$F$",F,NW);
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label("$V$",V,SW);
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label("$A$",A,dir(F--A));
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label("$P$",P,S,red);
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label("$Q$",Q,S,red);
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label("$X$",X,E,red);
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label("$Y$",Y,E,red);
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</asy>
 +
<asy>
 +
import olympiad;
 +
size(200);
 +
defaultpen(linewidth(0.8)+fontsize(11pt));
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real d = 0.7, edge = 2.5, Ax = 1.9;
 +
real f(real x)
 +
{
 +
return 1/(4*d) * x * x;
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}
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pair V = origin, F = (0,d), la = (-edge,-d), lb = (edge,-d), A = (Ax, f(Ax));
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pair P = foot(F,la,lb), Q = foot(A,la,lb), X = foot(F,A,Q), Y = foot(V,A,Q);
 +
draw(Q--A--F--P^^F--X^^A--V--Y^^rightanglemark(F,P,la,4)^^rightanglemark(A,Q,lb,4)^^rightanglemark(A,X,F,4)^^rightanglemark(A,Y,V,4),red);
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draw(la -- lb);
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draw(graph(f,-2.5,2.5));
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dot(F^^A^^V);
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label("$F$",F,NW);
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label("$V$",V,SW);
 +
label("$A$",A,dir(F--A));
 +
label("$P$",P,S,red);
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label("$Q$",Q,S,red);
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label("$X$",X,E,red);
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label("$Y$",Y,E,red);
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</asy>
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Set <math>d = FV</math>, which by the definition of a parabola also equals <math>VP</math>. Then as <math>AQ = AF = 20</math>, we have <math>AY = 20 - d</math> and <math>AX = |20 - 2d|</math>. Since <math>FXYV</math> is a rectangle, <math>FX = VY</math>, so by Pythagorean Theorem on triangles <math>AFX</math> and <math>AVY</math>,
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\begin{align*}
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21^2 - (20 - d)^2 &= AV^2 - AY^2 = VY^2\\
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&= FX^2 = AF^2 - AX^2 = 20^2 - (20 - 2d)^2.
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\end{align*}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 right configuration and the larger solution with the left configuration - and so the requested answer is <math>\boxed{\tfrac{40}3}</math>.
 +
 
 
==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}}

Revision as of 15:09, 11 February 2021

Problem

Suppose that on a parabola with vertex $V$ and a focus $F$ there exists a point $A$ such that $AF=20$ and $AV=21$. What is the sum of all possible values of the length $FV?$

$\textbf{(A) }13 \qquad \textbf{(B) }\frac{40}3 \qquad \textbf{(C) }\frac{41}3 \qquad \textbf{(D) }14\qquad \textbf{(E) }\frac{43}3$

Solution

Let $\ell$ be the directrix of $\mathcal P$; recall that $\mathcal P$ is the set of points $T$ such that the distance from $T$ to $\ell$ is equal to $TF$. Let $P$ and $Q$ be the orthogonal projections of $F$ and $A$ onto $\ell$, and further let $X$ and $Y$ be the orthogonal projections of $F$ and $V$ onto $AQ$. Because $AF < AV$, there are two possible configurations which may arise, and they are shown below.

[asy] import olympiad; size(230); defaultpen(linewidth(0.8)+fontsize(11pt)); real d = 1.1, edge = 2.5, Ax = 1.6; real f(real x) { return 1/(4*d) * x * x; } pair V = origin, F = (0,d), la = (-edge,-d), lb = (edge,-d), A = (Ax, f(Ax)); pair P = foot(F,la,lb), Q = foot(A,la,lb), X = foot(F,A,Q), Y = foot(V,A,Q); draw(P--F--A--V--Y^^F--X--Q^^rightanglemark(F,P,la,4)^^rightanglemark(A,Q,lb,4)^^rightanglemark(A,X,F,4)^^rightanglemark(A,Y,V,4),red); draw(graph(f,-2.5,2.5)); draw(la -- lb); dot(F^^A^^V); label("$F$",F,NW); label("$V$",V,SW); label("$A$",A,dir(F--A)); label("$P$",P,S,red); label("$Q$",Q,S,red); label("$X$",X,E,red); label("$Y$",Y,E,red); [/asy] [asy] import olympiad; size(200); defaultpen(linewidth(0.8)+fontsize(11pt)); real d = 0.7, edge = 2.5, Ax = 1.9; real f(real x) { return 1/(4*d) * x * x; } pair V = origin, F = (0,d), la = (-edge,-d), lb = (edge,-d), A = (Ax, f(Ax)); pair P = foot(F,la,lb), Q = foot(A,la,lb), X = foot(F,A,Q), Y = foot(V,A,Q); draw(Q--A--F--P^^F--X^^A--V--Y^^rightanglemark(F,P,la,4)^^rightanglemark(A,Q,lb,4)^^rightanglemark(A,X,F,4)^^rightanglemark(A,Y,V,4),red); draw(la -- lb); draw(graph(f,-2.5,2.5)); dot(F^^A^^V); label("$F$",F,NW); label("$V$",V,SW); label("$A$",A,dir(F--A)); label("$P$",P,S,red); label("$Q$",Q,S,red); label("$X$",X,E,red); label("$Y$",Y,E,red); [/asy] Set $d = FV$, which by the definition of a parabola also equals $VP$. Then as $AQ = AF = 20$, we have $AY = 20 - d$ and $AX = |20 - 2d|$. Since $FXYV$ is a rectangle, $FX = VY$, so by Pythagorean Theorem on triangles $AFX$ and $AVY$, \begin{align*} 21^2 - (20 - d)^2 &= AV^2 - AY^2 = VY^2\\ &= FX^2 = AF^2 - AX^2 = 20^2 - (20 - 2d)^2. \end{align*}This equation simplifies to $3d^2 - 40d + 41 = 0$, which has solutions $d = \tfrac{20\pm\sqrt{277}}3$. Both values of $d$ work - the smaller solution with the right configuration and the larger solution with the left configuration - and so the requested answer is $\boxed{\tfrac{40}3}$.

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

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