User:Afly

Afly's notes

About Afly

Friends with Yrock and ConfidentFlamingo

User # 1013218 Profile

Competitions I have participated in:

AMC8 * 1

AMC10A * 1

AMC10B * 1

MathLeague Qualifications * 7

MathLeague District * 2

MathLeague District Countdown * 1

MathLeague State * 1

MathLeague State Countdown * 1

MathLeague National * 1

MathLeague National Countdown * 1

MathLeague International * 1

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

How trig works

draw(unitcirle)
pair O = origin
dot(O)
label("O",O,dir(197.5))
label("$\theta$",O,dir(17.5))
pair G = (0,1)
label("G",G)
pair A = (Cos(35),0)
label("A",A)
 (Error making remote request. Unknown error_msg)

Rotating cube

Starring: Asymptote three

Perspective

Orthographic

Circle diagrams

A collection of stuff having to do with circles

Intersection points

Starring: Fancy labels, Troubleshooting

Created files

(Using desmos)

Colored dodecagon

Original file:

File

Pentagon 2023 12B Q25 dissmo

Original size:

File

Solutions created

2019 AIME II #5 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

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