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

(Added in Solution 2.)
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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 right configuration and the larger solution with the left configuration - and so the requested answer is <math>\boxed{\textbf{(B)}\ \tfrac{40}{3}}</math>.
+
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==
 
==Solution 2==

Latest revision as of 16:33, 13 July 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 1

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 bottom configuration and the larger solution with the top configuration - and so the requested answer is $\boxed{\textbf{(B)}\ \tfrac{40}{3}}$.

Solution 2

Let $\mathcal{P}$ be the parabola, let $V$ be the origin, $F$ lie on the positive $y$ axis, and $d=FV$. The equation of the parabola is then $x^{2}=4dy$. If the coordinates of $A$ are $(p, q),$ then $p^{2}+q^{2}=441$ since the distance from the origin to $A$ is $21$. Note also that the parabola is the set of all points equidistant from $F$ and a line known as its directrix, which in this case is a horizontal line $d$ units below the origin. Since the distance from $A$ to its directrix is equal to $AF,$ then $A$ is $20$ units above this line and therefore $q=20-d$. Substituting for $p$ and $q$ yields $4d(20-d)+(20-d)^{2}=441$ or $(20-d)(20+3d)=441,$ which simplifies to $3d^{2}-40d+41=0$. Therefore the sum of all possible values of $d$ is $\tfrac{-b}{a}=\boxed{\textbf{(B)} ~\tfrac{40}{3}}$ by Viète's Formulas.

~sugar_rush

Video Solution by OmegaLearn (Using parabola properties and system of equations)

https://youtu.be/DcaD9vvcKL0

~ pi_is_3.14

Video Solution by TheBeautyofMath

https://youtu.be/WidS_IOjkQo

~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

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