2021 AIME I Problems/Problem 15

Revision as of 08:24, 11 August 2021 by MRENTHUSIASM (talk | contribs) (Solution 1: Minor fixes in the coding; also added the title.)

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 $S$.

Diagram

Graph in Desmos: https://www.desmos.com/calculator/gz8igmkykn

~MRENTHUSIASM

Solution 2 (Translations, Inequalities, Circles)

Make the translation $y \rightarrow y+20$ to obtain $20+y=x^2-k$ and $x=2y^2-k$. Multiply the first equation by $2$ and sum, we see that $2(x^2+y^2)=3k+40+2y+x$. Completing the square gives us $\left(y- \frac{1}{2}\right)^2+\left(x - \frac{1}{4}\right)^2 = \frac{325+24k}{16}$; 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. (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 $y=x^2-k$. As we increase the value of $k$, two more intersections appear on the "left branch":

$k=4$ does not work because the "leftmost" point of $x=2(y-20)^2-4$ is $(-4,20)$ which lies to the right of $\left(-\sqrt{24}, 20\right)$, which is on the graph $y=x^2-4$. 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, $k<4$ does not work.

$k=5$ does work because the two graphs intersect at $(-5,20)$, 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 $5+280=\boxed{285}$.

  • In general (assuming four intersections exist), when two conics intersect, if one conic can be written as $ax^2+by^2=f(x,y)$ and the other as $cx^2+dy^2=g(x,y)$ for polynomials $f$ and $g$ of degree at most $1$, whenever $(a,b),(c,d)$ are linearly independent (L.I.), we can combine the two equations and then complete the square to achieve $(x-p)^2+(y-q)^2=r^2$. We can also combine these two equations to form a parabola, or a hyperbola, or an ellipse. When $(a,b),(c,d)$ 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 point(s), the statement that all these points lie on a circle is trivially true.

-Ross Gao

See Also

2021 AIME I (ProblemsAnswer KeyResources)
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Problem 14
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