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

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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>.
 
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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==Diagram==
===Solution 1===
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Graph in Desmos: https://www.desmos.com/calculator/gz8igmkykn
With binary search you can narrow down the k value. Newton raphson method let you narrow down the x and y solution for that specific k value. With 3 (x,y) pairs you can find radius of the circle.  
 
  
You end up finding the bounds of 5 and 280. The sum is 285.
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~MRENTHUSIASM
  
~Lopkiloinm
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== Solution 1 (Inequalities and Circles) ==
 +
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:
 +
<ol style="margin-left: 1.5em;">
 +
  <li>The two parabolas intersect at four distinct points.<p></li>
 +
By a quick sketch, we have two subconditions:
 +
<ol style="margin-left: 1.5em;" type="A">
 +
  <li>The point <math>(-k,20)</math> is on or below the parabola <math>y=x^2-k.</math><p>
 +
We need <math>20\leq(-k)^2-k,</math> from which <math>k\geq5.</math> <p>
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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. \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 such that <math>x\neq-5.</math> So, <math>(\bigstar)</math> has four distinct real roots, or the two parabolas intersect at four distinct points.<p>
 +
For Subcondition A, we deduce that <math>k\geq5.</math>
 +
</li>
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  <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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The lower half of the parabola <math>x=2(y-20)^2-k</math> is <math>y=20-\sqrt{\frac{x+k}{2}}.</math> We need <math>-k\leq20-\sqrt{\frac k2},</math> which holds for all values of <math>k.</math><p>
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For Subcondition B, we deduce that <math>k</math> can be any positive integer.
 +
</li>
 +
</ol>
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<b>For Condition 1, we obtain <math>\boldsymbol{k\geq5}</math> by taking the intersection of Subconditions A and B.</b>
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  <li>The four points of intersection lie on a circle with radius at most <math>21.</math><p>
 +
For equations of circles, the coefficients of <math>x^2</math> and <math>y^2</math> must be the same. So, we add the equation <math>y=x^2-k</math> to half the equation <math>x=2(y-20)^2-k:</math> <cmath>y+\frac12x=x^2+(y-20)^2-\frac32k.</cmath>
 +
We expand, rearrange, and complete the squares:
 +
<cmath>\begin{align*}
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y+\frac12x&=x^2+y^2-40y+400-\frac32k \\
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\frac32k-400&=\left(x^2-\frac12x\right)+\left(y^2-41y\right) \\
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\frac32k-400+\frac{1}{16}+\frac{1681}{4}&=\left(x-\frac14\right)^2+\left(y-\frac{41}{2}\right)^2.
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\end{align*}</cmath>
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We need <math>\frac32k-400+\frac{1}{16}+\frac{1681}{4}\leq21^2,</math> from which <math>k\leq\left\lfloor\frac{6731}{24}\right\rfloor=280.</math><p>
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<b>For Condition 2, we obtain <math>\boldsymbol{k\leq280.}</math></b>
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</li>
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</ol>
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Taking the intersection of Conditions 1 and 2 produces <math>5\leq k\leq280.</math> Therefore, the answer is <math>5+280=\boxed{285}.</math>
  
===Solution 2===
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~MRENTHUSIASM
  
Make the translation <math>x \rightarrow x+20</math> to obtain <math>20+x=y^2-k , y=2x^2-k</math>. Multiply the first equation by 2 and sum, we see that <math>2(x^2+y^2)=3k+40+2x+y</math>. Completing the square gives us <math>(x- \frac{1}{2})^2+(y - \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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== Solution 2 (Translations, Inequalities, Circles) ==
  
For the lower bound, we need to ensure there are 4 intersections to begin with. A quick check shows k=5 works while k=4 does not. Therefore, the answer is 5+280=285.
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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>.  
  
*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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For the lower bound, we need to ensure there are <math>4</math> 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 <math>k</math>, two more intersections appear on the "left branch":
 +
 
 +
<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>\left(-\sqrt{24}, 20\right)</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, <math>k<4</math> does not work.
 +
 
 +
<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=\boxed{285}</math>.
 +
 
 +
*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 polynomials <math>f</math> and <math>g</math> of degree at most <math>1</math>, whenever <math>(a,b),(c,d)</math> are linearly independent (L.I.), 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 <math>3,2</math> or <math>1</math> intersection point(s), the statement that all these points lie on a circle is trivially true.  
  
 
-Ross Gao
 
-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}}
  
 
[[Category:Intermediate Algebra Problems]]
 
[[Category:Intermediate Algebra Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 17:39, 16 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. By a quick sketch, we have two subconditions:

    1. 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 such that $x\neq-5.$ So, $(\bigstar)$ has four distinct real roots, or the two parabolas intersect at four distinct points.

      For Subcondition A, we deduce that $k\geq5.$

    2. The point $(0,-k)$ is on or below the parabola $x=2(y-20)^2-k.$

      The lower half of the parabola $x=2(y-20)^2-k$ is $y=20-\sqrt{\frac{x+k}{2}}.$ We need $-k\leq20-\sqrt{\frac k2},$ which holds for all values of $k.$

      For Subcondition B, we deduce that $k$ can be any positive integer.

    For Condition 1, we obtain $\boldsymbol{k\geq5}$ by taking the intersection of Subconditions A and B.

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

    For equations of circles, the coefficients of $x^2$ and $y^2$ must be the same. So, we add the equation $y=x^2-k$ to half the equation $x=2(y-20)^2-k:$ \[y+\frac12x=x^2+(y-20)^2-\frac32k.\] We expand, rearrange, and complete the squares: \begin{align*} y+\frac12x&=x^2+y^2-40y+400-\frac32k \\ \frac32k-400&=\left(x^2-\frac12x\right)+\left(y^2-41y\right) \\ \frac32k-400+\frac{1}{16}+\frac{1681}{4}&=\left(x-\frac14\right)^2+\left(y-\frac{41}{2}\right)^2. \end{align*} We need $\frac32k-400+\frac{1}{16}+\frac{1681}{4}\leq21^2,$ from which $k\leq\left\lfloor\frac{6731}{24}\right\rfloor=280.$

    For Condition 2, we obtain $\boldsymbol{k\leq280.}$

Taking the intersection of Conditions 1 and 2 produces $5\leq k\leq280.$ Therefore, the answer is $5+280=\boxed{285}.$

~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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