2021 AIME I Problems/Problem 15
Problem
Let 
be the set of positive integers 
such that the two parabolas![\[y=x^2-k~~\text{and}~~x=2(y-20)^2-k\]](http://latex.artofproblemsolving.com/6/f/9/6f9e25390a9657644dc69c3cd2ca6d154cd6e9be.png)
intersect in four distinct points, and these four points lie on a circle with radius at most 
. Find the sum of the least element of and the greatest element of .
Solution 1
Make the translation 
to obtain 
. Multiply the first equation by 2 and sum, we see that 
. Completing the square gives us 
; this explains why the two parabolas intersect at four points that lie on a circle*. For the upper bound, observe that 
, so 
.
For the lower bound, we need to ensure there are 4 intersections to begin with.
does not work because the "leftmost" point of 
is 
while lies to the right of 
, which on the graph 
.
does work because they intersect at 
and are NOT tangent at that point because the slopes are 
and undefined respectively. Therefore, the answer is 
.
- In general, (Assuming four intersections exist) when two conics intersect, if one conic can be written as
and the other as 
for f,g polynomials of degree at most 1, whenever 
are linearly independent, we can combine the two equations and then complete the square to achieve 
. We can also combine these two equations to form a parabola, or a hyperbola, or an ellipse. When are not L.I., the intersection points instead lie on a line, which is a circle of radius infinity. When the two conics only have 3,2 or 1 intersection points, the statement that all these points lie on a circle is trivially true.
and the other as 
for f,g polynomials of degree at most 1, whenever 
are linearly independent, we can combine the two equations and then complete the square to achieve 
. We can also combine these two equations to form a parabola, or a hyperbola, or an ellipse. When -Ross Gao
See also
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