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| | -Ross Gao | | -Ross Gao |
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| − | ===Solution 2===
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| − | <asy>
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| − | import graph; size(300); Label f; f.p=fontsize(6);
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| − | xaxis(-20,20,Ticks(f, 5.0));
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| − | yaxis(-10,30,Ticks(f, 5.0));
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| − | real p1(real x) { return x^2-5; }
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| − | pair P2(real t) {
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| − | return (2*(t-20)^2-5,t);
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| − | }
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| − | real p2_axis(real x) { return 20; }
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| − |
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| − | path p2 = graph(P2, 16,24);
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| − | draw(graph(p1,-6,6,n=400),linewidth(1));
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| − | draw(p2,linewidth(1));
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| − | //draw(p2_axis,linewidth(1)+dashed);
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| − | </asy>
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| − | 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>.
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| − | 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>.
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| − | 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>
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| − | 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.
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| − | <asy>
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| − | import graph; size(300); Label f; f.p=fontsize(6);
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| − | xaxis(-20,20,Ticks(f, 5.0));
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| − | yaxis(-10,30,Ticks(f, 5.0));
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| − | real k = 6;
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| − | real p1(real x) { return x^2-k; }
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| − | pair P2(real t) {
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| − | return (2*(t-20)^2-k,t);
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| − | }
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| − | pair c1(real t) {
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| − | real r = 5.35;
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| − | return (r*cos(t),20+r*sin(t));
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| − | }
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| − | real p2_axis(real x) { return 20; }
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| − | path p2 = graph(P2, 16,24);
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| − |
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| − | draw(graph(p1,-6,6,n=400),linewidth(1));
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| − | draw(p2,linewidth(1));
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| − | draw(graph(c1,0,6.28,n=200),linewidth(1));
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| − | </asy>
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| − | 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>.
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| − | 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>
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| − | Solving for <math>k</math> using largest circle <math>C1(21)</math> and <math>P1,P2</math>:
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| − | <asy>
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| − | import graph; size(300); Label f; f.p=fontsize(6);
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| − | xaxis(-300,100,Ticks(f, 50.0));
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| − | yaxis(-300,100,Ticks(f, 50.0));
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| − | real k = 279;
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| − | real p1(real x) { return x^2-k; }
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| − | pair P2(real t) {
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| − | return (2*(t-20)^2-k,t);
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| − | }
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| − | pair c1(real t) {
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| − | real r = 21;
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| − | return (r*cos(t),20+r*sin(t));
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| − | }
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| − | real p2_axis(real x) { return 20; }
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| − | path p2 = graph(P2, 20-15,20+15);
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| − | draw(graph(p1,-20,20,n=400),linewidth(1));
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| − | draw(p2,linewidth(1));
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| − | draw(graph(c1,0,6.28,n=400),linewidth(1));
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| − | </asy>
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| − | <cmath>C1(21): x^2+(y-20)^2=441 -(1)</cmath>
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| − | <cmath>P1: x=2(y-20)^2-k -(2)</cmath>
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| − | <cmath>P2: y=x^2-k -(3)</cmath>
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| − | <cmath>(3)+\frac{1}{2}*(2)+(1): y+\frac{1}{2}x = 441-\frac{3}{2}k -(4)</cmath>
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| − | <cmath>(1)+\frac{1}{2}*(2): x^2+\frac{x}{2}-(441-\frac{k}{2})=0 -(5)</cmath>
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| − | <cmath>(1)+(3): (y-20)^2+y-(441-k)=0 -(6)</cmath>
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| − | Solving <math>(5)</math> we get:
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| − | <cmath>x = -\frac{1}{4}\pm \frac{\sqrt{7057 - 8k}}{4}</cmath>
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| − | Solving <math>(6)</math> we get:
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| − | <cmath>y = \frac{39}{2}\pm \frac{\sqrt{1685 - 4k}}{2}</cmath>
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| − | 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>
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| − | Hence <math>k_{\textrm{min}}+k_{\textrm{max}} \forall k \in S \implies 6+279= \boxed{285}</math>
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| − |
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| − | ~Math_Genius_164
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| | ==See also== | | ==See also== |
Problem
Let 
be the set of positive integers 
such that the two parabolas![\[y=x^2-k~~\text{and}~~x=2(y-20)^2-k\]](//latex.artofproblemsolving.com/6/f/9/6f9e25390a9657644dc69c3cd2ca6d154cd6e9be.png)
intersect in four distinct points, and these four points lie on a circle with radius at most 
. Find the sum of the least element of and the greatest element of .
Solution 1
Make the translation 
to obtain 
. Multiply the first equation by 2 and sum, we see that 
. Completing the square gives us 
; this explains why the two parabolas intersect at four points that lie on a circle*. For the upper bound, observe that 
, so 
.
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
and the other as 
for f,g polynomials of degree at most 1, whenever 
are linearly independent, we can combine the two equations and then complete the square to achieve 
. We can also combine these two equations to form a parabola, or a hyperbola, or an ellipse. When 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
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
be the set of positive integers 
such that the two parabolas![\[y=x^2-k~~\text{and}~~x=2(y-20)^2-k\]](http://latex.artofproblemsolving.com/6/f/9/6f9e25390a9657644dc69c3cd2ca6d154cd6e9be.png)
intersect in four distinct points, and these four points lie on a circle with radius at most 
. Find the sum of the least element of 
to obtain 
. Multiply the first equation by 2 and sum, we see that 
. Completing the square gives us 
; this explains why the two parabolas intersect at four points that lie on a circle*. For the upper bound, observe that 
, so 
.
and the other as 
for f,g polynomials of degree at most 1, whenever 
are linearly independent, we can combine the two equations and then complete the square to achieve 
. We can also combine these two equations to form a parabola, or a hyperbola, or an ellipse. When 
