The Mango Gates – Puzzler

From Car Talk:

RAY: This Puzzler came from Kevin Tobin. A maharajah owned a mango tree that had very sweet, abundant fruit. People would come from miles around to pick the fruit. So the maharajah decided to protect the tree by erecting a series of seven concentric fences around it. Each fence had a gate with a guard. To get to the tree a person would have had to pass through seven gates and pass seven guards. One day, a man approached the guard at the first gate and said, “If you let me pass, when I come back I’ll bring you some mangoes. I will give you half of the mangoes I have, but you must give me one back.” For example, if he brings back 10 mangoes, he’ll give five to the guard and the guard will give one back. The guard let the man in through the gate, and the man proceeded to make the same deal with the other six guards. The question is: How many mangoes did the man have to get to pass back through the seven gates, giving half to each guard and the guard giving one back?

Similar to Gone Fishin’ puzzler.

3d Representation of Dog Chase

Not what it sounds like.
The four dog problem from Steve Strogatz:


Here is how it was used in my class.

Halfway Dogs

If the dogs blindly travel half way towards the next dog before re-aiming, then you get this picture, where each vertex represents the dog’s position:
2015-11-13_14h30_36
If you raise each square up in the z-dimension and print the resulting 3d shape then you can get this:
FullSizeRender

12% Dogs

If the dogs travel 12% of the way before re-aiming then you get this picture, and this 3d shape:

2015-11-13_14h30_20

IMG_2116

Interactive 2d Version

The generalized interactive 2d version of this concept is a lot of fun to play with. In this demo, your mouse horizontal controls how much to spin each square, and your mouse vertical controls how much to scale each square. Matt Enlow was the driving force behind this super fun demo. Some samples:

2015-11-13_14h49_43 2015-11-13_14h50_01 2015-11-13_14h52_05

Code

Here’s the code for both of these shapes in madeup:

Halfway dogs:

thickness = 3
size = 10.0
theta = 0
z = 0
repeat 15
  t = 0
  while t <= 360
    x = size * cos (t + theta)
    y = size * sin (t + theta)
    t = t + 90
    moveto x,y,z
  end
  extrude 0,0,1,thickness
  theta = theta + 45
  thickness = thickness * (0.5)^0.5
  z = z + thickness
  size = size * (0.5)^0.5
end

12% dogs:

thickness = 1
size = 10.0
theta = 0
z = 0
repeat 35
  t = 0
  while t <= 360
    x = size * cos (t + theta)
    y = size * sin (t + theta)
    t = t + 90
    moveto x,y,z
  end
  extrude 0,0,1,thickness
  theta = theta + 8.775
  thickness = thickness * 0.84
  z = z + thickness
  size = size * 0.8775
end

Linear Functions, Cardiods, and the Mandelbrot Set

This tweet was retweeted by Steve Strogatz:


What a great video, linking linear functions, cardiods, and the family of Mandelbrot sets. Rang my bell.

Here’s some of my previous lessons and experiments that involved these things:

  • The Mandelbrot Set viewed through Pre-Calculus. Presentation, Lesson Plans, and Interactives. Links to the various blog posts are at the bottom of the page.
  • Processing Experiments:
    • Waning Moon – Steps through linear functions with slope 1 to 4.
      2015-11-09_08h31_23
    • Interactive Waning Moon – Uses the mouse to control the slope and the y-intercept of the linear function.
      2015-11-09_08h29_26
    • Linear Waning Moon – Interactive. Same process, but instead of a circle, inputs are in a line at the top and outputs are in a line at the bottom.
      2015-11-09_08h29_54
    • Linear Mod Art – Interactive, similar setup as above, but useful for teaching how it works.
      2015-11-09_08h30_52
    • Sine Mod Art – Same as above, but using a sine function instead of a linear function.
      2015-11-09_08h32_37
    • Waning Sine Moon – Interactive circular version using sine.
      2015-11-09_08h33_58
    • Waning Exponential Moon – Interactive circular version using an exponential function.
      2015-11-09_08h35_05

The nice thing about the processing versions is you can pretty easily create large image files that you can print:
IMG_2083
IMG_2084

 

The Last Seat : Car Talk Puzzler

Yet another great car talk puzzler, The Last Seat:

RAY: You’re one of a hundred people standing in line to get onto an airplane that has 100 seats. There’s a seat for every person who’s in line, and each of you has a boarding pass for your assigned a seat. The first person to walk onto the plane drops his boarding pass and, instead of picking it up, decides, “I’m just going to sit anyplace.” He takes a seat at random.

Now, every other passenger will take either his assigned seat or, if that seat is taken, that passenger will take any seat at random.

TOM: I’ve been on that flight!

RAY: You are the last passenger to walk onto the plane. Obviously, there’s going to be one seat left, because everyone else is sitting in his correct seat, or not.

The question is: What are the chances that you get to sit in your assigned seat? I’m going make this multiple choice.

A: 1 out of 2.
B: 1 out of 10.
C: 2 out of 50.
D: 1 out of 100
E: Zero.

What strategies could you give students to help them reason their way through this puzzle?

Visualization of a Function Through Time

(This post demonstrates the math behind the creation of some unusual math art. I got the idea from this tumblr post which I don’t even think works like this. But it was the inspiration behind the pictures.)

Take this function: f(x) = x^2 + a*cos(ax). When you take values of a between 0 and 8 and back to 0, you get the following:

gifsmos

I’d like to make an image that represents the entire family of this function. Problem one: the function values get pretty large in this window, so find the remainder of the function output after dividing by 5. Here’s the gif of that output:

gifsmos (1)

Problem two: How do we represent this in one static picture? If you take the first image (where a = 0)

2015-10-15_10h49_09

and take the function values from 0 to 5 and make the pixels for row 1 colored based on the function value (0 being black, and 5 being white), then you get this row of pixels (expanded to 20 pixels high for ease of viewing):

blahfirstline

So this row of pixels represents a = 0. If you set a to be 0 at the top of the picture, a to be 8 at the middle of the picture, and a to be back to 0 at the bottom of the picture, stitch all these lines together you get:

smaller
Click on image for full 4k by 4k resolution.

Here’s another with the function y=(10a)/(1+x^2) with a going from 8 to 0 to 8.

type2s
Click on image for full 4k by 4k resolution.

Here’s another with the function y=ceil(ax) – ax + floor(ax) with a going from 0 to 5 to 0.

type3s
Click on image for full 4k by 4k resolution.

And lastly, here’s the last one, but colorized. The 0 to 255 now controls the hue of the color (with full saturation and balance, using HSB color).
type3cs
Click on image for full 4k by 4k resolution.

Here’s the code for those who are interested (or view it live on openprocessing)!

int max_a;

float mod(float a, float b)
{
  //desmos mod and processing mod act differently.
  //desmos mod always returns a positive value
  return (abs(a % b));
}

float func(float x, float a)
{
  //type 1
  max_a = 8;
  return (mod(x*x - a * cos(a*x),5));
  
  //type 2 this counts from a= 8 to 0 to 8
  //max_a = 8;
  //a = max_a - a;
  //return (mod((10*a)/(1+x*x) ,5));
  
  //type 3
  //max_a = 5;
  //return(mod(ceil(a*x)-a*x+floor(a*x) ,5));
}


void setup()
{
  size(1000,1000);
  colorMode(HSB,255);
  int w,h;
  float x,a,f,hue;
  max_a = 5;
  loadPixels();
  for (int i = 0; i < pixels.length; i++)
  {
    w = i % width;
    h = int(i / width);
    
    x = map(w,0,width,-6,6);
    
    if (h < height/2)
    {
      a = map(h,0,height/2,0,max_a);
    }
    else
    {
      a = map(h,height/2,height,max_a,0);
    }

    f = func(x,a);
    hue = map(f,0,5,0,255);
    
    pixels[i] = color(hue,255,255);
  }
  updatePixels();
  //save("type3c.png");
}

edit 10/16/15:
Here are a couple more pictures for you (the code on openprocessing is up to date with these new types):
type5s

type4s

Madeup and Calculus

The madeup language for creating 3d models came out a week ago for the kickstarter backers. There is more information about the computer programming language found at the kickstarter page. It’s been a lot of fun to play with but looking at it through the lens of being able to create calculus 3d solids has been really exciting.

Rotational Volumes

Here is the code to sketch out this 2d shape on the x-y plane and rotate it around the x-axis:
Screenshot 2015-10-09 at 2.10.41 PM

to func x
    out = -1 * (x - 1)^2 + 5
    out
end

moveto 0,0,0
x = 0
xmax = 3.0
numPoints = 100

while (x < xmax)
    out = func(x)
    moveto x,out,0
    x = x + xmax//numPoints
end

moveto xmax,0,0
moveto 0,0,0

nsides = 100
revolve 1,0,0,270

When you click the solidify you get this 3d shape (only rotated 3/4 of the way around for sake of visulization):
Screenshot 2015-10-09 at 2.04.11 PM

Here’s the magic step. Click Download and open the .obj file with your 3d printer software and hit print:

IMG_1866 (1)

Want to rotate around the y-axis instead and practice shells? No problem.

2015-10-15_08h47_13

And print!

IMG_1906

 

 

Cross-sectional Volumes

The AP (but not the IB) curriculum has students find the volume of a solid created by extruding out a known cross-section from a given area. For instance, on the 2010 AP Calculus AB exam, they asked the following question, zone in on part (c):
Screenshot 2015-10-09 at 1.33.40 PM
This type of question is difficult for students to visualize. Madeup can make some great models (that can be printed) for the students. While I think the programming is tricky enough that I wouldn’t encourage you to bring the code directly to novices; it is pretty clean and easy to modify. For example: take the same area as above and find the volume by taking cross sectional squares with one side on the xy plane and perpendicular to the x-axis.

In the madeup world, the code looks like this:

to func x
    out = x ^ 0.5 + 1
    out
end

moveto 0,0,0
x = 0
xmax = 3
numRect = 14

while (x < xmax)
    out = func(x)
    moveto x,out,0
    moveto x + xmax//numRect,out,0
    moveto x + xmax//numRect,0,0
    moveto x,0,0
    x = x + xmax//numRect
    extrude 0,0,1,out
end

moveto xmax,0,0
moveto 0,0,0


When you hit the extrude button, it takes the 14 rectangle slices and brings it up the z-axis to make a square. The result is the 3d shape:

2015-10-13_09h02_28

Here’s the shape (approximated) with 4 rectangles:

2015-10-13_09h02_18

And here’s the shape with 100 rectangles:

2015-10-13_09h02_41

Print!

FullSizeRender

 

Enjoy! As always, please fire away with questions/comments/etc.

Programming and Anna Weltman’s This Is Not A Math Book and Dan Meyer’s Contest

PSA: check out Anna Weltman’s new book, This Is Not A Math Book. Looks great.
Also check out Dan Meyer’s fall contest (ends 10/6/15) that asks the students to be creative in making their own Loop-de-loop mathematical art.
Programming connection: have the students make their loop-de-loop with python and turtles, here is a 3-5-2 loop-de-loop:
2015-09-29_07h50_35
or with Scratch, here is a 2-4-5: 2015-09-29_07h56_53
Enjoy!