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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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− | 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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− | 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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− | ~Math_Genius_164
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| ==See also== | | ==See also== |