2021 AIME I Problems/Problem 15
Problem
Let be the set of positive integers
such that the two parabolas
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. (Here I'm using the un-translated coordinates.)
does not work because the "leftmost" point of
is
which lies to the right of
, which is on the graph
. Clearly, no k less than 4 works either.
does work because the two graphs intersect at
, and by drawing the graph, you realize this is not a tangent point and there is in fact another intersection, due to slope. 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.
-Ross Gao
See also
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