2021 AIME I Problems/Problem 15

Problem

Let $S$ be the set of positive integers $k$ such that the two parabolas $y=x^2-k~~\text{and}~~x=2(y-20)^2-k$ intersect in four distinct points, and these four points lie on a circle with radius at most $21$ . Find the sum of the least element of $S$ and the greatest element of $S$ .

Solution

Solution 1

With binary search you can narrow down the k value. Newton raphson method let you narrow down the x and y solution for that specific k value. With 3 (x,y) pairs you can find radius of the circle.

You end up finding the bounds of 5 and 280. The sum is 285.

~Lopkiloinm

Solution 2

Make the translation $x \rightarrow x+20$ to obtain $20+x=y^2-k , y=2x^2-k$ . Multiply the second equation by 2 and sum, we see that $2(x^2+y^2)=3k+40+2x+y$ . Completing the square gives us $(x- \frac{1}{2})^2+(y - \frac{1}{4})^2 = \frac{325+24k}{16}$ ; this explains why the two parabolas intersect at four points that lie on a circle*. For the upper bound, observe that $LHS \leq 21^2=441 \rightarrow 24k \leq 6731$ , so $k \leq 280$ .

For the lower bound, we need to ensure there are 4 intersections to begin with. A quick check shows k=5 works while k=4 does not. Therefore, the answer is 5+280=285.

In general, this problem tells us that the intersection points of two conics without xy terms lie on a circle.

-Ross Gao

2021 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Last problem
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

Page

Toolbox

Search

2021 AIME I Problems/Problem 15

Contents

Problem

Solution

Solution 1

Solution 2

See also