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

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-Ross Gao
 
-Ross Gao
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==Solution 3 (Parabola propertyes)==
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<pre style="color: blue">
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Lemma
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</pre>
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Let the axes of two parabolas be perpendicular, their focal parameters be  <math>p_1</math> and <math>p_2</math> and the distances from the foci to the point of intersection of the axes be  <math>x_2</math> and <math>y_1</math>. Suppose that these parabolas intersect at four points. Prove that these points lie on the circle and find its center and radius.
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<pre style="color: blue">
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Proof:
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</pre>
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Let's introduce a coordinate system with the center at the point of intersection of the axes. Let the first parabola have axis <math>x = 0,</math> focal parameter  <math>p_1</math> and focus at point <math>(0, –y_1), y_1 > 0.</math> Let second parabola have axis <math>y = 0,</math> focal parameter <math>p_2</math> and focus at point <math>(–x_2,0), x_2  > 0.</math>
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Let us denote the angle between the vector connecting the focus of the first parabola and its point and the positive direction of the ordinate axis <math>2\theta,</math> its length <math>\rho_1(\theta),</math> the angle between the vector connecting the focus of the second parabola and its point and the positive direction of the abscissa axis <math>2\phi,</math> its length <math>\rho(\phi).</math> Then
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<cmath>\begin{align*} \rho_1(\theta) = \frac{p_1}{1 - \cos(2\theta)},\end{align*}</cmath>
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<cmath>\begin{align*} \rho_2(\phi) = \frac{p_2}{1 - \cos(2\phi)}.\end{align*}</cmath>
  
 
==See Also==
 
==See Also==

Revision as of 11:57, 29 May 2022

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 $k\geq5.$

      Remark for Subcondition A

      Recall that if $1\leq k\leq 4,$ then the point $(-k,20)$ is above the parabola $y=x^2-k.$ It follows that for $-k\leq x\leq0:$

      • The maximum value of $y$ for the parabola $y=x^2-k$ occurs at $x=-k,$ from which $y=k^2-k\leq12.$
      • The minimum value of $y$ for the parabola $x=2(y-20)^2-k$ occurs at $x=0,$ from which $y=20-\sqrt{\frac k2}>18.$

      Clearly, the parabola $x=2(y-20)^2-k$ and the left half of the parabola $y=x^2-k$ do not intersect. Therefore, the two parabolas do not intersect at four distinct points.

    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

Solution 3 (Parabola propertyes)

Lemma

Let the axes of two parabolas be perpendicular, their focal parameters be $p_1$ and $p_2$ and the distances from the foci to the point of intersection of the axes be $x_2$ and $y_1$. Suppose that these parabolas intersect at four points. Prove that these points lie on the circle and find its center and radius.

Proof:

Let's introduce a coordinate system with the center at the point of intersection of the axes. Let the first parabola have axis $x = 0,$ focal parameter $p_1$ and focus at point $(0, –y_1), y_1 > 0.$ Let second parabola have axis $y = 0,$ focal parameter $p_2$ and focus at point $(–x_2,0), x_2  > 0.$ Let us denote the angle between the vector connecting the focus of the first parabola and its point and the positive direction of the ordinate axis $2\theta,$ its length $\rho_1(\theta),$ the angle between the vector connecting the focus of the second parabola and its point and the positive direction of the abscissa axis $2\phi,$ its length $\rho(\phi).$ Then \begin{align*} \rho_1(\theta) = \frac{p_1}{1 - \cos(2\theta)},\end{align*} \begin{align*} \rho_2(\phi) = \frac{p_2}{1 - \cos(2\phi)}.\end{align*}

See Also

2021 AIME I (ProblemsAnswer KeyResources)
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Problem 14
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