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

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~MRENTHUSIASM
 
~MRENTHUSIASM
  
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==Solution 2 (Translations, Inequalities, Circles) ==
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== Solution 1 (Inequalities and Circles) ==
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Note that <math>y=x^2-k</math> is an upward-opening parabola with the vertex at <math>(0,-k),</math> and <math>x=2(y-20)^2-k</math> is a rightward-opening parabola with the vertex at <math>(-k,20).</math> We consider each condition separately:
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<ol style="margin-left: 1.5em;">
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  <li>The two parabolas intersect at four distinct points.<p></li>
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  <li>The four points of intersection lie on a circle with radius at most <math>21.</math><p></li>
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</ol>
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<b>SOLUTION IN PROGRESS. NO EDIT PLEASE, THANKS.
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~MRENTHUSIASM
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== Solution 2 (Translations, Inequalities, Circles) ==
  
 
Make the translation <math>y \rightarrow y+20</math> to obtain <math>20+y=x^2-k</math> and <math>x=2y^2-k</math>. Multiply the first equation by <math>2</math> and sum, we see that <math>2(x^2+y^2)=3k+40+2y+x</math>. Completing the square gives us <math>\left(y- \frac{1}{2}\right)^2+\left(x - \frac{1}{4}\right)^2 = \frac{325+24k}{16}</math>; this explains why the two parabolas intersect at four points that lie on a circle*. For the upper bound, observe that <math>LHS \leq 21^2=441 \rightarrow 24k \leq 6731</math>, so <math>k \leq 280</math>.  
 
Make the translation <math>y \rightarrow y+20</math> to obtain <math>20+y=x^2-k</math> and <math>x=2y^2-k</math>. Multiply the first equation by <math>2</math> and sum, we see that <math>2(x^2+y^2)=3k+40+2y+x</math>. Completing the square gives us <math>\left(y- \frac{1}{2}\right)^2+\left(x - \frac{1}{4}\right)^2 = \frac{325+24k}{16}</math>; this explains why the two parabolas intersect at four points that lie on a circle*. For the upper bound, observe that <math>LHS \leq 21^2=441 \rightarrow 24k \leq 6731</math>, so <math>k \leq 280</math>.  

Revision as of 09:14, 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:

  1. The two parabolas intersect at four distinct points.

  2. 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

See Also

2021 AIME I (Problems • Answer Key • Resources)
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