2021 AMC 12A Problems/Problem 20
Contents
[hide]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.