Difference between revisions of "User:Afly"
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MathLeague International * 1 | MathLeague International * 1 | ||
+ | ==Visual proofs== | ||
+ | (Demonstrates HI IQ) | ||
+ | ===FAQ: What is a triangular number?=== | ||
+ | n=3 | ||
+ | ┌─┐ 1 | ||
+ | ├─┴─┐ 2 | ||
+ | ├───┴─┐ 3 | ||
+ | └─────┘ | ||
+ | 2 of them are: | ||
+ | ┌─┐ ┌─┐ | ||
+ | ├─┴─┐ ├─┴─┐ | ||
+ | ├───┴─┐ ├───┴─┐ | ||
+ | └─────┘ └─────┘ | ||
+ | Which are: | ||
+ | ┌─┐ ┌─────┐ | ||
+ | ├─┴─┐ └─┬───┤ | ||
+ | ├───┴─┐ └─┬─┤ | ||
+ | └─────┘ └─┘ | ||
+ | |||
+ | And... | ||
+ | |||
+ | ┌─┬─────┐ | ||
+ | ├─┴─┬───┤ n | ||
+ | ├───┴─┬─┤ | ||
+ | └─────┴─┘ | ||
+ | n+1 | ||
+ | so 2(1+2+3...+n)=n(n+1) | ||
+ | |||
+ | They're special characters in unicode. deal with it. | ||
==Some of my asymptote creations== | ==Some of my asymptote creations== | ||
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draw(shift(dir(270)/4)*E--shift(dir(270)/24)*E); | draw(shift(dir(270)/4)*E--shift(dir(270)/24)*E); | ||
label("1",E--F,SW); | label("1",E--F,SW); | ||
− | label("$\tan \theta$", | + | label("$\tan \theta$",D--E); |
draw(shift(dir(degrees(d)+90)/4)*brace(O,F)); | draw(shift(dir(degrees(d)+90)/4)*brace(O,F)); | ||
− | label("$\sec \theta$",shift(dir(d+90)/4)*brace(O,F),dir(d+90)); | + | label("$\sec \theta$",shift(dir(degrees(d)+90)/4)*brace(O,F),dir(d+90)); |
draw(shift(dir(degrees(d)+90)/4)*O--shift(dir(degrees(d)+90)/24)*O); | draw(shift(dir(degrees(d)+90)/4)*O--shift(dir(degrees(d)+90)/24)*O); | ||
draw(shift(dir(degrees(d)+90)/4)*F--shift(dir(degrees(d)+90)/24)*F); | draw(shift(dir(degrees(d)+90)/4)*F--shift(dir(degrees(d)+90)/24)*F); | ||
Line 1,196: | Line 1,225: | ||
==Solutions created== | ==Solutions created== | ||
− | ===2019 AIME II # | + | ===2019 AIME II #4 Solution 7 (Generating functions)=== |
+ | [[https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_4#Solution_7.28Generating_functions.29 Actual solution]] | ||
Let's look at the prime factorization of some of these rolls: | Let's look at the prime factorization of some of these rolls: |
Latest revision as of 16:21, 29 July 2024
Contents
- 1 Afly's notes
- 1.1 About Afly
- 1.2 Visual proofs
- 1.3 Some of my asymptote creations
- 1.4 Created files
- 1.5 LaTeX Creations (coming soon!)
- 1.6 Created pages
- 1.7 Solutions created
- 2 Yrock's Notes
- 3 ConfidentFlamingo's notes
- 4 Other Users' notes
Afly's notes
About Afly
Friends with Yrock and ConfidentFlamingo
User # 1013218 Profile
Competitions I have participated in:
AMC8 * 2
AMC10A * 1
AMC10B * 1
MathLeague Qualifications * 7
MathLeague District * 1
MathLeague District Countdown * 1
MathLeague State * 1
MathLeague State Countdown * 1
MathLeague National * 1
MathLeague National Countdown * 1
MathLeague International * 1
Visual proofs
(Demonstrates HI IQ)
FAQ: What is a triangular number?
n=3
┌─┐ 1 ├─┴─┐ 2 ├───┴─┐ 3 └─────┘
2 of them are:
┌─┐ ┌─┐ ├─┴─┐ ├─┴─┐ ├───┴─┐ ├───┴─┐ └─────┘ └─────┘
Which are:
┌─┐ ┌─────┐ ├─┴─┐ └─┬───┤ ├───┴─┐ └─┬─┤ └─────┘ └─┘
And...
┌─┬─────┐ ├─┴─┬───┤ n ├───┴─┬─┤ └─────┴─┘ n+1
so 2(1+2+3...+n)=n(n+1)
They're special characters in unicode. deal with it.
Some of my asymptote creations
These are a random collection of diagrams and text that I have made. YOU ARE ALLOWED TO USE THESE IMAGES AND/OR TEXT IN YOUR PAGES OR USE THEM FOR REFERENCE.
I try to put them in the order of how much technique & technical points there are in them
Enough talking. Let's get into the pictures.
Triangle
Starring: Basic path drawing and LaTeX
ANSWER: LOOK ONLY WHEN THE PROBLEM IS COMPLETED >>> THIS IS ONLY TO CHECK YOUR ANSWER
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Pythagorean
Starring: Colored paths and line width
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Trapezoid
Starring: Filling regions
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Hexagon (In AMC 2023 10A #24)
Starring: polygon(), shift(), dir(), unitsize, labels
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Tiny 2D experiments
Grid with numbers
Starring: Custom colors
Tiny 3D experiments
Starring: Using cleverly placed colors and lines to make a illusion (just pure skill, not really Asymptote)
Cube
Smaller cube
Cube with a cube hole in it
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Pascal's triangle mod
Starring: Loops and arrays
mod 2
mod 2 big
int[][] pascal1={{},{},{},{},{},{},{},{},{},{},{},{},{},{},{},{},{},{},{},{},{},{},{},{},{},{},{},{},{},{},{},{},{},{},{},{},{},{},{},{},{},{},{},{},{},{},{},{},{},{},{},{},{},{},{},{},{},{},{},{},{},{},{},{}}; pen p1=rgb(1/8,1/8,7/8); pen p2=rgb(1/8,7/8,1/8); for (int i=0; i<64; ++i) { for (int j=0; j<i+1; ++j) { if (j==0 || i==j) { pascal1[i].push(1); } else { pascal1[i].push(pascal1[i-1][j-1]+pascal1[i-1][j]); } if (pascal1[i][j]%2==0){ fill(shift(dir(60)*7)*shift(dir(30))*shift(dir(240)*i*sqrt(3))*shift(dir(0)*j*sqrt(3))*rotate(30)*polygon(6),p1); } else { fill(shift(dir(60)*7)*shift(dir(30))*shift(dir(240)*i*sqrt(3))*shift(dir(0)*j*sqrt(3))*rotate(30)*polygon(6),p2); } } } int[] ref1=pascal1[63]; int[][] pascal2={{},{},{},{},{},{},{},{},{},{},{},{},{},{},{},{},{},{},{},{},{},{},{},{},{},{},{},{},{},{},{},{},{},{},{},{},{},{},{},{},{},{},{},{},{},{},{},{},{},{},{},{},{},{},{},{},{},{},{},{},{},{},{},{}}; pen p1=rgb(1/8,1/8,7/8); pen p2=rgb(1/8,7/8,1/8); for (int i=0; i<64; ++i) { for (int j=0; j<i+65; ++j) { if (j==0 || i==j) { pascal2[i].push(1); } else { if (i==0){ pascal2[i].push(ref[j-1]+ref[j]); } else { pascal2[i].push(pascal2[i-1][j-1]+pascal2[i-1][j]); } } if (pascal1[i][j]%2==0){ fill(shift(dir(60)*7)*shift(dir(30))*shift(dir(240)*(i+64)*sqrt(3))*shift(dir(0)*j*sqrt(3))*rotate(30)*polygon(6),p1); } else { fill(shift(dir(60)*7)*shift(dir(30))*shift(dir(240)*(i+64)*sqrt(3))*shift(dir(0)*j*sqrt(3))*rotate(30)*polygon(6),p2); } } } (Error making remote request. Unknown error_msg)
mod 3
mod 4
mod 5
mod 6
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How trig works
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Rotating cube
Starring: Asymptote three
Perspective
Orthographic
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Circle diagrams
A collection of stuff having to do with circles
Intersection points
Starring: Fancy labels, Troubleshooting
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Function ![$f(x) = 3\tanh (x)+\sin (x)$](//latex.artofproblemsolving.com/c/c/4/cc40869cc372d4f77637dced2f9eb91297b105e3.png)
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Sheet
Starring: Bezier surfaces
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Another cube
Starring: Knowledge*
* How to put everything together in a nice example, why the only good thing about asymptote three is bezier surfaces. If you're bored, look at the code and try to figure out how it all works.
Created files
(Using desmos)
Colored dodecagon
Original file:
Pentagon 2023 12B Q25 dissmo
Original size:
LaTeX Creations (coming soon!)
Created pages
Solutions created
2019 AIME II #4 Solution 7 (Generating functions)
Let's look at the prime factorization of some of these rolls:
01 => 2^0*3^0*5^0 02 => 2^1*3^0*5^0 03 => 2^0*3^1*5^0 04 => 2^2*3^0*5^0 05 => 2^0*3^0*5^1 06 => 2^1*3^1*5^0
Now, using multi-variable generating functions, we get:
f(x,y,z)=1+x+y+x^2+z+xy | | \| |/ 'v' f(x,y,z)=1+x+y+z+x^2+xy =(for our purposes) =2+x+y+z+xy
Let's square that!
4+2x+2y+2z+2xy+2x+x^2+xy+xz+x^2y+2y+xy+y^2+yz+xy^2+2z+xz+yz+z^2+xyz+2xy+x^2y+xy^2+xyz+x^2y^2 Combining like terms . . . 4+4x+4y+4z+x^2+y^2+z^2+6xy+2xz+2yz+2xyz+2x^2y+2xy^2+x^2y^2
Since we only want the parity of each of the exponents, we can collapse it again.
8+6x+6y+4z+6xy+2xz+2yz+2xyz
Last simplification: factor out a factor of two and save it for later.
4+3x+3y+2z+3xy+xz+yz+xyz
Let's take a better approach this time.
___ term :4*4+3*3+3*3+2*2+3*3+1*1+1*1+1*1=16+09+09+04+09+01+01+01=50 __x term :4*3+3*4+3*3+2*1+3*3+1*2+1*1+1*1=12+12+09+02+09+02+01+01=48 __y term :4*3+3*3+3*4+2*1+3*3+1*1+1*2+1*1=12+09+12+02+09+01+02+01=48 __z term :4*2+3*1+3*1+2*4+3*1+1*3+1*3+1*3=08+03+03+08+03+03+03+03=34 _xy term :4*3+3*3+3*3+2*1+3*4+1*1+1*1+1*2=12+09+09+02+12+01+01+02=48 _xz term :4*1+3*2+3*1+2*3+3*1+1*4+1*3+1*3=04+06+03+06+03+04+03+03=32 _yz term :4*1+3*1+3*2+2*3+3*1+1*3+1*4+1*3=04+03+06+06+03+03+04+03=32 xyz term :4*1+3*1+3*1+2*3+3*2+1*3+1*3+1+4=04+03+03+06+06+03+03+04=32 Result: 50+48x+48y+34z+48xy+32xz+32yz+32xyz
I know we could use vectors and dot products to make it look neater but come on. It already looks neat enough. Also, we didn't need the other parts, but it just looks nicer. Now let's stick back the two that turned into a four.
200+192x+192y+136z+192xy+128xz+128yz+128xyz
We seek the constant term which is 200. 200/1296=100/648=50/324=25/162, 25+162=187
Yrock's Notes
hello
ConfidentFlamingo's notes
Other Users' notes
Note from afly: If you leave a note, write who it is from