2021 AIME I Problems/Problem 15
Contents
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
.
Diagram
Graph in Desmos: https://www.desmos.com/calculator/gz8igmkykn
~MRENTHUSIASM
Solution 1 (Inequalities and Circles)
Note that is an upward-opening parabola with the vertex at
and
is a rightward-opening parabola with the vertex at
We consider each condition separately:
- The two parabolas intersect at four distinct points.
- The point
is on or below the parabola
We need
from which
Moreover, the point
is on the parabola
when
We will prove that the two parabolas intersect at four distinct points at this value of
Substituting
into
we get
Expanding and rearranging give
By either the graphs of the parabolas or the Rational Root Theorem, we conclude that
is a root of
So, we factor its left side:
By either the graphs of the parabolas or Descartes' Rule of Signs, we conclude that
has two positive roots and one negative root such that
So,
has four distinct real roots, or the two parabolas intersect at four distinct points.
For Subcondition A, we deduce that
Remark for Subcondition A
Recall that if
then the point
is above the parabola
It follows that for
- The maximum value of
for the parabola
occurs at
from which
- The minimum value of
for the parabola
occurs at
from which
Clearly, the parabola
and the left half of the parabola
do not intersect. Therefore, the two parabolas do not intersect at four distinct points.
- The maximum value of
- The point
is on or below the parabola
The lower half of the parabola
is
We need
which holds for all values of
For Subcondition B, we deduce that
can be any positive integer.
- The four points of intersection lie on a circle with radius at most
For equations of circles, the coefficients of
and
must be the same. So, we add the equation
to half the equation
We expand, rearrange, and complete the squares:
We need
from which
For Condition 2, we obtain
By a quick sketch, we have two subconditions:
For Condition 1, we obtain by taking the intersection of Subconditions A and B.
Taking the intersection of Conditions 1 and 2 produces Therefore, the answer is
~MRENTHUSIASM
Solution 2 (Translations, Inequalities, Circles)
Make the translation to obtain
and
. Multiply the first equation by
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 intersections to begin with. (Here I'm using the un-translated coordinates.) Draw up a graph, and realize that two intersections are guaranteed, on the so called "right branch" of
. As we increase the value of
, two more intersections appear on the "left branch":
does not work because the "leftmost" point of
is
which lies to the right of
, which is on the graph
. While technically speaking this doesn't prove that there are no intersections (why?), drawing the graph should convince you that this is the case. Clearly,
does not work.
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 nearby, 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 polynomials
and
of degree at most
, whenever
are linearly independent (L.I.), 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
or
intersection point(s), the statement that all these points lie on a circle is trivially true.
-Ross Gao
Solution 3 (Parabola propertyes)
Lemma
Let the axes of two parabolas be perpendicular, their focal parameters be and
and the distances from the foci to the point of intersection of the axes be
and
. Suppose that these parabolas intersect at four points. Prove that these points lie on the circle and find its center and radius.
Proof:
Let's introduce a coordinate system with the center at the point of intersection of the axes. Let the first parabola have axis focal parameter
and focus at point
Let second parabola have axis
focal parameter
and focus at point
Let us denote the angle between the vector connecting the focus of the first parabola and its point and the positive direction of the ordinate axis
its length
the angle between the vector connecting the focus of the second parabola and its point and the positive direction of the abscissa axis
its length
Then
See Also
2021 AIME I (Problems • Answer Key • Resources) | ||
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