Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(This can only be done with a computer. This is not feasible to do by hand.)
(→‎Solution 1)
Line 11: Line 11:
  
 
-Ross Gao
 
-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 <math>P1: y=x^2-k</math> is first parabola with axis <math>x=0</math> and vertex at <math>(0,k)</math> and <math>P2: x=2(y-20)^2-k</math> be the second parabola with axis at <math>y=20</math> with vertex at <math>(-k,20)</math>.
 +
 +
Vertex for <math>k=0</math> the <math>P2</math> vertex is at <math>(0,20)</math>, so <math>P2</math> is intersecting <math>P1</math> only at two points. As we increase <math>k</math>, <math>P2</math>'s vertex gets closer to <math>P1</math>. It intersects <math>P1</math> at <math>(-k,20)</math>.
 +
 +
Plugging in <math>(-k,20)</math> in <math>P1</math>.<cmath>20=(-k)^2 -k \implies (k-5)(k+4)=0 \implies k=5 \textrm{ given } k \in \mathbb{Z}^{+}</cmath>
 +
 +
Note <math>k=5</math> gives exactly <math>3</math> intersections between <math>P1</math> and <math>P2</math>. For <math>P1</math> and <math>P2</math> to have 4 intersections, the smallest <math>k_{\textrm{min}}</math> needs to be <math>6</math> 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 <math>k</math> increases beyond <math>6</math> the number of intersections remain <math>4</math> but the radius of the common intersecting circle will increase. Consider the largest circle of radius <math>21</math> and test if an integer <math>k</math> that satisfies the common intersection between <math>P1, P2</math>.
 +
 +
The circle will be symmetric about y-axis and line <math>y=20</math> with center at <math>(0,20)</math>. So the general equation of circle <cmath>C1(r): x^2+(y-20)^2=r^2</cmath> Using <math>P1+\frac{1}{2}P2</math> and <math>C1(r)</math> we get a line equation <cmath>L1: y+\frac{1}{2}x = r^2-\frac{3k}{2}</cmath>
 +
 +
Solving for <math>k</math> using largest circle <math>C1(21)</math> and <math>P1,P2</math>:
 +
 +
<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>
 +
 +
 +
<cmath>C1(21): x^2+(y-20)^2=441 -(1)</cmath>
 +
<cmath>P1: x=2(y-20)^2-k -(2)</cmath>
 +
<cmath>P2: y=x^2-k -(3)</cmath>
 +
<cmath>(3)+\frac{1}{2}*(2)+(1): y+\frac{1}{2}x = 441-\frac{3}{2}k -(4)</cmath>
 +
<cmath>(1)+\frac{1}{2}*(2): x^2+\frac{x}{2}-(441-\frac{k}{2})=0 -(5)</cmath>
 +
<cmath>(1)+(3): (y-20)^2+y-(441-k)=0 -(6)</cmath>
 +
Solving <math>(5)</math> we get:
 +
<cmath>x = -\frac{1}{4}\pm \frac{\sqrt{7057 - 8k}}{4}</cmath>
 +
Solving <math>(6)</math> we get:
 +
<cmath>y = \frac{39}{2}\pm \frac{\sqrt{1685 - 4k}}{2}</cmath>
 +
 +
Plugging for <math>(x,y)</math> pairs in <math>(4)</math> we get <math>k</math> = <math>279, 283</math>; the value of <math>k</math> satisfies (1) is <math>279</math> meaning <math>k_{\textrm{max}}=279</math>
 +
 +
Hence <math>k_{\textrm{min}}+k_{\textrm{max}} \forall k \in S \implies 6+279= \boxed{285}</math>
 +
 +
~Math_Genius_164
  
 
==See also==
 
==See also==

Revision as of 16:47, 13 March 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 $x \rightarrow x+20$ 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

See also

2021 AIME I (Problems • Answer Key • Resources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2021_AIME_I_Problems/Problem_15&oldid=149424"