Difference between revisions of "2021 AIME I Problems/Problem 15"

Revision as of 11:45, 11 August 2021

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 1 (Inequalities and Circles)

Note that $y=x^2-k$ is an upward-opening parabola with the vertex at $(0,-k),$ and $x=2(y-20)^2-k$ is a rightward-opening parabola with the vertex at $(-k,20).$ We consider each condition separately:

The two parabolas intersect at four distinct points.

By a quick sketch, we have two subconditions:

The point $(-k,20)$ is on or below the parabola $y=x^2-k.$
We need $20\leq(-k)^2-k,$ from which $k\geq5.$
Moreover, the point $(-k,20)$ is on the parabola $y=x^2-k$ when $k=5.$ We will prove that the two parabolas intersect at four distinct points at this value of $k:$
Substituting $y=x^2-5$ into $x=2(y-20)^2-5,$ we get $x=2\left(\left(x^2-5\right)-20\right)^2-5.$ Expanding and rearranging give $2x^4-100x^2-x+1245=0. \hspace{20mm}(\bigstar)$ By either the graphs of the parabolas or the Rational Root Theorem, we conclude that $x=-5$ is a root of $(\bigstar).$ So, we factor its left side: $(x+5)\left(2x^3-10x^2-50x+249\right)=0.$ By either the graphs of the parabolas or Descartes' Rule of Signs, we conclude that $2x^3-10x^2-50x+249=0$ has two positive roots and one negative root other than $x=-5.$ So, $(\bigstar)$ has four distinct real roots, or the two parabolas intersect at four distinct points.
The point $(0,-k)$ is on or below the parabola $x=2(y-20)^2-k.$

The four points of intersection lie on a circle with radius at most $21.$

SOLUTION IN PROGRESS. NO EDIT PLEASE, THANKS.

~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

@@ Line 16: / Line 16: @@
 We need <math>20\leq(-k)^2-k,</math> from which <math>k\geq5.</math> <p>
 Moreover, the point <math>(-k,20)</math> is on the parabola <math>y=x^2-k</math> when <math>k=5.</math> We will prove that the two parabolas intersect at four distinct points at this value of <math>k:</math><p>
-Substituting <math>y=x^2-5</math> into <math>x=2(y-20)^2-5,</math> we get <math>x=2\left(\left(x^2-5\right)-20\right)^2-5.</math> Expanding and rearranging give <cmath>2x^4-100x^2-x+1245=0.</cmath>
+Substituting <math>y=x^2-5</math> into <math>x=2(y-20)^2-5,</math> we get <math>x=2\left(\left(x^2-5\right)-20\right)^2-5.</math> Expanding and rearranging give <cmath>2x^4-100x^2-x+1245=0. \hspace{20mm}(\bigstar)</cmath>
+By either the graphs of the parabolas or the Rational Root Theorem, we conclude that <math>x=-5</math> is a root of <math>(\bigstar).</math> So, we factor its left side: <cmath>(x+5)\left(2x^3-10x^2-50x+249\right)=0.</cmath>
+By either the graphs of the parabolas or Descartes' Rule of Signs, we conclude that <math>2x^3-10x^2-50x+249=0</math> has two positive roots and one negative root other than <math>x=-5.</math> So, <math>(\bigstar)</math> has four distinct real roots, or the two parabolas intersect at four distinct points.
 </li>
    <li>The point <math>(0,-k)</math> is on or below the parabola <math>x=2(y-20)^2-k.</math><p>

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