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

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==Problem==
 
==Problem==
These problems will not be available until the 2021 AIME I is released on Wednesday, March 10, 2021.
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Let <math>S</math> be the set of positive integers <math>k</math> such that the two parabolas<cmath>y=x^2-k~~\text{and}~~x=2(y-20)^2-k</cmath>intersect in four distinct points, and these four points lie on a circle with radius at most <math>21</math>. Find the sum of the least element of <math>S</math> and the greatest element of <math>S</math>.
  
==Solution==  
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==Solution 1==
Uh, don't cheat! AOPS isn't made for people to cheat. Please come back after you've officially taken AIME on March 10th
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And stop catching us up [[User:Zixuan12|Zixuan12]]<sup>[[User talk:Zixuan12|你好]]</sup> 13:49, 14 February 2021 (EST)
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Make the translation <math>y \rightarrow y+20</math> to obtain <math>20+y=x^2-k , x=2y^2-k</math>. Multiply the first equation by 2 and sum, we see that <math>2(x^2+y^2)=3k+40+2y+x</math>. Completing the square gives us <math>(y- \frac{1}{2})^2+(x - \frac{1}{4})^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>.
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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 <math>y=x^2-k</math>. As we increase the value of k, two more intersections appear on the "left branch."
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<math>k=4</math> does not work because the "leftmost" point of <math>x=2(y-20)^2-4</math> is <math>(-4,20)</math> which lies to the right of <math>(-\sqrt{24}, 20)</math>, which is on the graph <math>y=x^2-4</math>. 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, no k less than 4 works either.
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<math>k=5</math> does work because the two graphs intersect at <math>(-5,20)</math>, 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 <math>5+280=285</math>.
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*In general, (Assuming four intersections exist) when two conics intersect, if one conic can be written as <math>ax^2+by^2=f(x,y)</math> and the other as <math>cx^2+dy^2=g(x,y)</math> for f,g polynomials of degree at most 1, whenever <math>(a,b),(c,d)</math> are linearly independent, we can combine the two equations and then complete the square to achieve <math>(x-p)^2+(y-q)^2=r^2</math>. We can also combine these two equations to form a parabola, or a hyperbola, or an ellipse. When <math>(a,b),(c,d)</math> 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.
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-Ross Gao
  
 
==See also==
 
==See also==
 
{{AIME box|year=2021|n=I|num-b=14|after=Last problem}}
 
{{AIME box|year=2021|n=I|num-b=14|after=Last problem}}
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[[Category:Intermediate Algebra Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 20:20, 17 April 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 $S$.

Solution 1

Make the translation $y \rightarrow y+20$ to obtain $20+y=x^2-k , x=2y^2-k$. 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 $(y- \frac{1}{2})^2+(x - \frac{1}{4})^2 = \frac{325+24k}{16}$; 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 $(-\sqrt{24}, 20)$, 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, no k less than 4 works either.

$k=5$ 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 $5+280=285$.


  • 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 f,g polynomials of degree at most 1, whenever $(a,b),(c,d)$ are linearly independent, 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 points, the statement that all these points lie on a circle is trivially true.

-Ross Gao

See also

2021 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
Last problem
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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