2021 AMC 12A Problems/Problem 20
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
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
,
\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
, which has solutions
. Both values of
work - the smaller solution with the right configuration and the larger solution with the left configuration - and so the requested answer is
.
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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