2021 AMC 12A Problems/Problem 20
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
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 .
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.
Video Solution by OmegaLearn (Using parabola properties and system of equations)
Video Solution by TheBeautyofMath
|2021 AMC 12A (Problems • Answer Key • Resources)|
|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|