2021 AIME I Problems/Problem 15

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 $x \rightarrow x+20$ to obtain $20+x=y^2-k , y=2x^2-k$ . Multiply the first equation by 2 and sum, we see that $2(x^2+y^2)=3k+40+2x+y$ . Completing the square gives us $(x- \frac{1}{2})^2+(y - \frac{1}{4})^2 = \frac{325+24k}{16}$ ; 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. A quick check shows k=5 works while k=4 does not. 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

Solution 2

$[asy] import graph; size(300); Label f; f.p=fontsize(6); xaxis(-20,20,Ticks(f, 5.0)); yaxis(-10,30,Ticks(f, 5.0)); real p1(real x) { return x^2-5; } pair P2(real t) { return (2*(t-20)^2-5,t); } real p2_axis(real x) { return 20; } path p2 = graph(P2, 16,24); draw(graph(p1,-6,6,n=400),linewidth(1)); draw(p2,linewidth(1)); //draw(p2_axis,linewidth(1)+dashed); [/asy]$

Let $P1: y=x^2-k$ is first parabola with axis $x=0$ and vertex at $(0,k)$ and $P2: x=2(y-20)^2-k$ be the second parabola with axis at $y=20$ with vertex at $(-k,20)$ .

Vertex for $k=0$ the $P2$ vertex is at $(0,20)$ , so $P2$ is intersecting $P1$ only at two points. As we increase $k$ , $P2$ 's vertex gets closer to $P1$ . It intersects $P1$ at $(-k,20)$ .

Plugging in $(-k,20)$ in $P1$ . $20=(-k)^2 -k \implies (k-5)(k+4)=0 \implies k=5 \textrm{ given } k \in \mathbb{Z}^{+}$

Note $k=5$ gives exactly $3$ intersections between $P1$ and $P2$ . For $P1$ and $P2$ to have 4 intersections, the smallest $k_{\textrm{min}}$ needs to be $6$ and corresponding circle will be the smallest circle.

$[asy] import graph; size(300); Label f; f.p=fontsize(6); xaxis(-20,20,Ticks(f, 5.0)); yaxis(-10,30,Ticks(f, 5.0)); real k = 6; real p1(real x) { return x^2-k; } pair P2(real t) { return (2*(t-20)^2-k,t); } pair c1(real t) { real r = 5.35; return (r*cos(t),20+r*sin(t)); } real p2_axis(real x) { return 20; } path p2 = graph(P2, 16,24); draw(graph(p1,-6,6,n=400),linewidth(1)); draw(p2,linewidth(1)); draw(graph(c1,0,6.28,n=200),linewidth(1)); [/asy]$

We do realize that as $k$ increases beyond $6$ the number of intersections remain $4$ but the radius of the common intersecting circle will increase. Consider the largest circle of radius $21$ and test if an integer $k$ that satisfies the common intersection between $P1, P2$ .

The circle will be symmetric about y-axis and line $y=20$ with center at $(0,20)$ . So the general equation of circle $C1(r): x^2+(y-20)^2=r^2$ Using $P1+\frac{1}{2}P2$ and $C1(r)$ we get a line equation $L1: y+\frac{1}{2}x = r^2-\frac{3k}{2}$

Solving for $k$ using largest circle $C1(21)$ and $P1,P2$ :

$[asy] import graph; size(300); Label f; f.p=fontsize(6); xaxis(-300,100,Ticks(f, 50.0)); yaxis(-300,100,Ticks(f, 50.0)); real k = 279; real p1(real x) { return x^2-k; } pair P2(real t) { return (2*(t-20)^2-k,t); } pair c1(real t) { real r = 21; return (r*cos(t),20+r*sin(t)); } real p2_axis(real x) { return 20; } path p2 = graph(P2, 20-15,20+15); draw(graph(p1,-20,20,n=400),linewidth(1)); draw(p2,linewidth(1)); draw(graph(c1,0,6.28,n=400),linewidth(1)); [/asy]$

$C1(21): x^2+(y-20)^2=441 -(1)$ $P1: x=2(y-20)^2-k -(2)$ $P2: y=x^2-k -(3)$ $(3)+\frac{1}{2}*(2)+(1): y+\frac{1}{2}x = 441-\frac{3}{2}k -(4)$ $(1)+\frac{1}{2}*(2): x^2+\frac{x}{2}-(441-\frac{k}{2})=0 -(5)$ $(1)+(3): (y-20)^2+y-(441-k)=0 -(6)$ Solving $(5)$ we get: $x = -\frac{1}{4}\pm \frac{\sqrt{7057 - 8k}}{4}$ Solving $(6)$ we get: $y = \frac{39}{2}\pm \frac{\sqrt{1685 - 4k}}{2}$

Plugging for $(x,y)$ pairs in $(4)$ we get $k$ = $279, 283$ ; the value of $k$ satisfies (1) is $279$ meaning $k_{\textrm{max}}=279$

Hence $k_{\textrm{min}}+k_{\textrm{max}} \forall k \in S \implies 6+279= \boxed{285}$

~Math_Genius_164

2021 AIME I (Problems • Answer Key • Resources)
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2021 AIME I Problems/Problem 15

Contents

Problem

Solution 1

Solution 2

See also