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Difference between revisions of "User:Afly"

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Friends with [https://artofproblemsolving.com/wiki/index.php/User:Yrock Yrock] and [https://artofproblemsolving.com/wiki/index.php/User:ConfidentFlamingo ConfidentFlamingo]
 
Friends with [https://artofproblemsolving.com/wiki/index.php/User:Yrock Yrock] and [https://artofproblemsolving.com/wiki/index.php/User:ConfidentFlamingo ConfidentFlamingo]
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User # 1013218 [https://artofproblemsolving.com/community/user/1013218 Profile]
  
 
==Some of my asymptote creations==
 
==Some of my asymptote creations==

Revision as of 13:08, 9 December 2023

About Afly

Friends with Yrock and ConfidentFlamingo

User # 1013218 Profile

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.

Triangle

[asy] pair A=(0,0); pair B=(2,0); pair C=(1,sqrt(3)); draw(A--B--C); draw(A--C--B); [/asy]

$\text{That is a equilateral triangle with side length 2!}$

$\text{The height is } \sqrt{2^2-1^2} = \sqrt{3}$

$\text{The area is } 2 \cdot \sqrt{3} \cdot \frac{1}{2} = \boxed{\sqrt{3}}$

$\dots$

[asy] pair A=(0,0); pair B=(0,2); pair C=(1,1); draw(A--B--C); draw(A--C--B); [/asy]

$\text{Homework: calculate the area}$

ANSWER: LOOK ONLY WHEN THE PROBLEM IS COMPLETED >>> THIS IS ONLY TO CHECK YOUR ANSWER

$\text{This is an isosceles right triangle. It has side length }\sqrt{2}$

$\text{It has an height of } \frac{\sqrt{2}}{\sqrt{2}} = 1 \text{ and a base of } \sqrt{2} \cdot \sqrt{2} = 2$

$\text{The area is } 1 \cdot 2 \cdot \frac{1}{2} = \boxed{1}$ ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

Pythagorean

[asy] pair A=(0,0); pair B=(0,3); pair C=(4,0); draw(A--B--C--A,black+linewidth(1)); pair D=(-3,0); pair E=(-3,3); draw(A--B--E--D--A,red+linewidth(1)); pair F=(0,-4); pair G=(4,-4); draw(A--C--G--F--A,orange+linewidth(1)); pair H=(7,4); pair I=(3,7); draw(B--C--H--I--B,yellow+linewidth(1)); draw(A--B--C--A,black+linewidth(1)); [/asy] $\text{The pythagorean theorem: red plus orange equals yellow}$

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Isosceles Trapezoid

[asy] pair A=(0,0); pair B=(10,0); pair C=(6,8); pair D=(2,8); pair E=(30/7,40/7); draw(A--B--C--D--A,black+linewidth(5)); fill(A--E--D--cycle,red); fill(B--E--C--cycle,yellow); dot(A); dot(B); dot(C); dot(D); dot(E); label(A,'A',SW); label(B,'B',SE); label(C,'C',NE); label(D,'D',NW); label(E,'E',S); draw(A--E--C,blue+linewidth(2)); draw(B--E--D,blue+linewidth(2)); [/asy] $\text{Given that } \overline{AB} \parallel \overline{DC} \text{ , prove that } A_\triangle AED = A_\triangle BEC .$

++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

Hexagon (In AMC 2023 10A #24)

[asy] unitsize(1cm); unitsize(5cm); draw(scale(3)*polygon(6)); filldraw(shift(dir(0)*2+dir(120)*3/7)*polygon(6), lightgray); filldraw(shift(dir(60)*2+dir(180)*3/7)*polygon(6), lightgray); filldraw(shift(dir(120)*2+dir(240)*3/7)*polygon(6), lightgray); filldraw(shift(dir(180)*2+dir(300)*3/7)*polygon(6), lightgray); filldraw(shift(dir(240)*2+dir(0)*3/7)*polygon(6), lightgray); filldraw(shift(dir(300)*2+dir(60)*3/7)*polygon(6), yellow); pair A = (0,0) + 3 * dir(300); pair B = A + 3/7 *dir(60); pair C = B + 1 * dir(180); pair D = C + 3/7 * dir(240); pair E = C + 4/7 * dir(120); pair F = E + 3/7 * dir(240); pair G = F + 4/7 * dir(240); pen p = red+linewidth(6); draw(A--B--C--D--cycle,p); draw(C--E--F--D--cycle,p); draw(F--G--D--cycle,p); label("1",(B + C)/2,dir(90)); label("1",(A + D)/2,dir(270)); label("3/7",(A + B)/2,dir(330)); label("3/7",(C + D)/2,dir(330)); label("4/7",(C+E)/2,dir(30)); label("3/7",(E+F)/2,dir(150)); label("4/7",(F+D)/2,dir(30)); label("4/7",(F+G)/2,dir(150)); label("4/7",(G+D)/2,dir(270)); [/asy]

++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

Tiny 3D experiment

[asy] unitsize(60); pair A = (0,0); pair B = A + dir(0); pair C = A + dir(60); pair D = A + dir(120); pair E = A + dir(180); pair F = A + dir(240); pair G = A + dir(300); filldraw(A--B--C--D--cycle,orange); filldraw(A--D--E--F--cycle,blue); filldraw(A--F--G--B--cycle,gray); [/asy]

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