2021 AMC 12A Problems/Problem 20
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 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 right configuration and the larger solution with the left 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 |
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All AMC 12 Problems and Solutions |
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