Difference between revisions of "2021 AMC 12A Problems/Problem 20"
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+ | ==Video Solution by TheBeautyofMath== | ||
+ | https://youtu.be/WidS_IOjkQo | ||
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+ | ~IceMatrix | ||
==See also== | ==See also== | ||
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Revision as of 06:27, 19 February 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
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 right configuration and the larger solution with the left configuration - and so the requested answer is
.
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.