Difference between revisions of "User:Negativebplusorminus"
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| − | A AoPS member. | + | A AoPS member, National MathCounts qualifier, and USAJMO qualifier. |
| − | + | ==Contest Results== | |
| − | == | + | ===MathCounts=== |
| − | I | + | In 2011, as a 7th grader, I qualified for the State Countdown Round. In 2012, as an 8th grader, I qualified for National MathCounts. Hopefully, this page will be updated when I know the results of the National competition. |
| + | ===AMCs=== | ||
| + | 2012: 117 on AMC 10A, 127.5 on AMC 10B, 8 on AIME, 207.5 index for USAJMO. The cutoff was a 204.5, so I qualified for the USAJMO. Hopefully, this page will be updated when I know the results of the USAJMO. | ||
==negativebplusorminus== | ==negativebplusorminus== | ||
My username is from the [[Quadratic formula | quadratic formula]], which states that the roots of the equation <math>ax^2+bx+c=0</math> are | My username is from the [[Quadratic formula | quadratic formula]], which states that the roots of the equation <math>ax^2+bx+c=0</math> are | ||
<cmath>x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}</cmath> | <cmath>x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}</cmath> | ||
which, when read aloud, is "negativebplusorminus..." | which, when read aloud, is "negativebplusorminus..." | ||
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| − | <cmath>\sqrt{a+bi}=\sqrt{\frac{a+\sqrt{a^2+b^2}}{2}}+i\sqrt{\frac{-a+\sqrt{a^2+b^2}}{2}}</cmath> | + | ==Equations for the Roots of the Complex== |
| − | + | <cmath>\sqrt{a+bi}=\sqrt{\frac{a+\sqrt{a^2+b^2}}{2}}+i\sqrt{\frac{-a+\sqrt{a^2+b^2}}{2}}</cmath> | |
| − | + | I derived that equation myself, and I am quite proud of it. I have a similar one for the fourth roots of <math>a+bi</math> which can be derived from inputting that equation into itself. I have also found various roots of unity in their radical forms during my spare time. | |
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| − | I derived that equation myself, and I am quite proud of it. I have a similar one for the fourth roots of <math>a+bi</math> which can be derived from inputting that equation into itself. | ||
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==Spirographs== | ==Spirographs== | ||
| − | I have created a great number of spirographs, each interesting and unique. More can be found | + | I have created a great number of spirographs, each interesting and unique. More can be found on my [http://www.artofproblemsolving.com/Forum/blog.php?u=93546& AoPS blog] (but you might have to look through a few pages of other stuff, too). To view the entire collection, please visit [http://www.negativebplusorminus.blogspot.com negativebplusorminus.blogspot.com]. Here are some samples: |
<asy> | <asy> | ||
string s="Spirograph by user negativebplusorminus of the Art of Problem Solving forum. Please to not plagiarize; it is illegal and insulting."; | string s="Spirograph by user negativebplusorminus of the Art of Problem Solving forum. Please to not plagiarize; it is illegal and insulting."; | ||
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path g=polargraph(f,-200pi,200pi,10000, operator --); | path g=polargraph(f,-200pi,200pi,10000, operator --); | ||
draw(g, orange);</asy> | draw(g, orange);</asy> | ||
| − | + | <asy> | |
| − | + | import graph; | |
| − | + | size(300); | |
| − | + | string s="for(real t,real u){return t^2u^3, store as f};"; | |
| − | + | real f(real t) {return floor(t);} | |
| − | + | int p=25; | |
| − | { | + | int n=45+p; |
| − | + | path g=polargraph(f,-100pi,100pi,281, operator --); | |
| − | + | draw(g, blue);</asy> | |
| − | + | ==Inspirographs== | |
| − | + | Another amazing creation of mine. More can be found [http://www.artofproblemsolving.com/Forum/blog.php?u=93546& here] (but you might have to look through a few pages of other stuff, too). | |
| − | + | To view the entire collection, please visit [http://www.negativebplusorminus.blogspot.com negativebplusorminus.blogspot.com]. Below are a few samples. | |
| − | + | <asy2> | |
| − | + | import graph3; | |
| − | + | import grid3; | |
| − | + | import palette; | |
| − | + | size(400,300,IgnoreAspect); | |
| − | + | defaultrender.merge=true; | |
| − | ( | + | real f(pair z) {return sin(z.y)*(z.x^2+1)^(0.1*log(z.y^2+1));} |
| − | + | surface s=surface(f,(-30,-30),(30,30),70,Spline); | |
| − | + | s.colors(palette(s.map(zpart),Rainbow())); | |
| − | + | draw(s,render(compression=Low,merge=true)); | |
| − | + | grid3(XYZgrid);</asy2> | |
| − | + | <asy2> | |
| − | + | import graph3; | |
| − | + | import grid3; | |
| − | + | import palette;currentprojection=orthographic(1,5,0.2); | |
| − | = | + | size(400,300,IgnoreAspect); |
| − | + | defaultrender.merge=true; | |
| − | + | real f(pair z) {return sin(z.x^2+z.y^2);} | |
| − | + | surface s=surface(f,(-2.95,-2.95),(2.95,2.95),70,Spline); | |
| − | + | s.colors(palette(s.map(zpart),Rainbow())); | |
| − | + | draw(s,render(compression=Low,merge=true)); | |
| − | + | grid3(XYZgrid);</asy2> | |
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Revision as of 08:21, 12 April 2012
A AoPS member, National MathCounts qualifier, and USAJMO qualifier.
Contents
Contest Results
MathCounts
In 2011, as a 7th grader, I qualified for the State Countdown Round. In 2012, as an 8th grader, I qualified for National MathCounts. Hopefully, this page will be updated when I know the results of the National competition.
AMCs
2012: 117 on AMC 10A, 127.5 on AMC 10B, 8 on AIME, 207.5 index for USAJMO. The cutoff was a 204.5, so I qualified for the USAJMO. Hopefully, this page will be updated when I know the results of the USAJMO.
negativebplusorminus
My username is from the quadratic formula, which states that the roots of the equation 
are
![\[x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]](http://latex.artofproblemsolving.com/d/0/1/d01fd749d6550b117a645c357f01b3c9377e5756.png)
which, when read aloud, is "negativebplusorminus..."
Equations for the Roots of the Complex
![\[\sqrt{a+bi}=\sqrt{\frac{a+\sqrt{a^2+b^2}}{2}}+i\sqrt{\frac{-a+\sqrt{a^2+b^2}}{2}}\]](http://latex.artofproblemsolving.com/1/1/5/115b79b994447d87745338438f7a8dc8bf2c25ad.png)
I derived that equation myself, and I am quite proud of it. I have a similar one for the fourth roots of 
which can be derived from inputting that equation into itself. I have also found various roots of unity in their radical forms during my spare time.
Spirographs
I have created a great number of spirographs, each interesting and unique. More can be found on my AoPS blog (but you might have to look through a few pages of other stuff, too). To view the entire collection, please visit negativebplusorminus.blogspot.com. Here are some samples:
![[asy] string s="Spirograph by user negativebplusorminus of the Art of Problem Solving forum. Please to not plagiarize; it is illegal and insulting."; import graph; size(300); string s="for(real t,real u){return t^2u^3, store as f};"; real f(real t) {return t+log(t^2+t^4+1);} int p=15; int n=45+p; path g=polargraph(f,-200pi,200pi,10000, operator --); draw(g, orange);[/asy]](http://latex.artofproblemsolving.com/a/5/3/a53f363cc7a774840a58ada7d8754d4f9c51fb23.png)
![[asy] import graph; size(300); string s="for(real t,real u){return t^2u^3, store as f};"; real f(real t) {return floor(t);} int p=25; int n=45+p; path g=polargraph(f,-100pi,100pi,281, operator --); draw(g, blue);[/asy]](http://latex.artofproblemsolving.com/9/9/1/991f1fbb461dc3aa2a412ddd6842ec73d7f9ecda.png)
Inspirographs
Another amazing creation of mine. More can be found here (but you might have to look through a few pages of other stuff, too). To view the entire collection, please visit negativebplusorminus.blogspot.com. Below are a few samples. <asy2> import graph3; import grid3; import palette; size(400,300,IgnoreAspect); defaultrender.merge=true; real f(pair z) {return sin(z.y)*(z.x^2+1)^(0.1*log(z.y^2+1));} surface s=surface(f,(-30,-30),(30,30),70,Spline); s.colors(palette(s.map(zpart),Rainbow())); draw(s,render(compression=Low,merge=true)); grid3(XYZgrid);</asy2> <asy2> import graph3; import grid3; import palette;currentprojection=orthographic(1,5,0.2); size(400,300,IgnoreAspect); defaultrender.merge=true; real f(pair z) {return sin(z.x^2+z.y^2);} surface s=surface(f,(-2.95,-2.95),(2.95,2.95),70,Spline); s.colors(palette(s.map(zpart),Rainbow())); draw(s,render(compression=Low,merge=true)); grid3(XYZgrid);</asy2>
