In response to: http://mathhombre.tumblr.com/post/138365834664/half-of-a-page-simon-gregg-was-planning-a-lesson
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In response to: http://mathhombre.tumblr.com/post/138365834664/half-of-a-page-simon-gregg-was-planning-a-lesson
@mathhombre
Here’s a picture of the boyfriend figuring out a new proof to show that every other Fibonacci number is even, and my illustrated story about how I kept pronouncing (ra+sb) as ‘ratsby’ in every GCD-related proof. Yes, this is how we hang out.
But… The first part is from 3 to 5.., that’s probably the exception to the rule right?
1-1-2-3-5-8-13-21… Nope. Maybe every 3rd number!!
21-34-55-89…. Yeah, looks like every 3rd number… Idk
Exactly right, pardon the errata. Speaking of errata, we got new books today and the following conversation just took place: Him: “Which of your new books do you want me to read to you?” Me: “Aw, it would be adorable if you read me a math book! I’m kind of feeling fiction right now.” Him: “Well, I could read you Lang…” I was commensurately offended.
The matrices haven't been multiplied correctly. The bottom right entry in the answer should be a 1, not 0.
Do you want to build a pentagonal dodecahedron?
It doesn’t have to be a pentagonal dodecahedron.
Just for Fun !
Some fun #wtfmath.
Fractal
Hexbox Math!
I was chatting with foldawayorigami about the hex box base, and he brought up some interesting ideas as far as using this as a base for different height/radius ratios for the same size of paper, using different shapes of paper to begin with (such as hexagon), and using this as a base for other shapes (such as a dodecagon). So, I’ll work through the math as I’m trying these out and take you on my journey! Now, mind you, I’m no mathologist, so brilliantandblonde and railehatesfun please check me (and, really anyone who’s better at math, feel free to tell me I’m wrong and how to do it right). Some of the math I’ll have to research or remember from what I was taught oh lord I’m old a decade ago, but some of it is going to be the makeshift intuitive kind. Hey, I’m the kind of guy who refuses to measure out divisions of paper with a ruler. If I can’t figure out the divisions by folding them, they’re obviously unimportant.
Before I begin with the math, although normally you would need to fold a completed model to see the emergent properties, we’ve all folded enough boxes that I should be safe with only folding two or three units. Buuuut we both know that I’ll probably end up completing the model.
So, the first thing I’m going talk about is the use of pleats to create interior angles. I don’t know the actual formula to figure out the interior angle of regular polygons, so I use the formula I came up with drunk: 180 - 360/n, where n is the number of sides, of course. The reason I like this is that, essentially, you’re attempting to take flat paper, which would be 180 degrees (how do I make the little circle thingy?) and subtracting the correct angle from half of it, which makes the paper stand up and have the correct interior angle. So for a hexagon, the interior angle is 120 and we’re subtracting 60, for an octagon, 135 is the interior and we’re subtracting 45. For a dodecagon, 150 and 30. If you’ve made Tomoko Fuse’s octagon box, I’m sure you’ve noticed the pleats. (note: I forgot that there was a specific word for the angle we’re subtracting - supplement! But, I told you it was gonna be rough, I’m just gonna leave it with the fumbling terminology.)
Now, of course, since we’re making pleats, we need to halve the angle we’re subtracting because the pleat takes 2x whatever the angle that is taken out. For Fuse, we see 22.5 degree angles and pleats, for my box, we have 30 degrees. I’m gonna try to use the same general format to make a dodecagon box, so I’ll need to make 15 degree pleats.
Now! Let’s talk radius. I was VERY LUCKY that I chose a hexagon while drunk. Hexagons are composed of equilateral triangles, which means that the radius is LITERALLY the side length. No special math needed. With the dodecahedron, using the sine rule, I got that the side:radius ratio is 1:~1.93 which means that I need to have the paper divided into fourths, use the top two horizontal sections as the rim and use the horizontal fourths each as a section (making 1:2 rectangles (not exact, but lets try it)). I should only need four pieces of paper. Hold for a moment while I fold it, please.
And it worked perfectly. So now, I’m going to try to make a hexbox with a radius:height ratio of 1:1.5. For this, I’m going to to divide the paper in vertical fourths, and fold the horizontal ½ line. I’ll take the bottom half of the paper and fold it to the middle (that’ll create the four squares which I’ll use to make the bottom of the box) and then fold the top edge to the newly created ¼ mark. That’ll make the edge. I should only need two pieces of paper.
And that worked perfectly as well. Now the only thing left is to try it with hexagonal paper - which I don’t have right now so I’ll have to fold and cut. And it’s almost time for work - I’ll do that tonight.
So far, though, what we’ve discovered is that the box will work for hypothetically any polygon as long as: 1.) the rectangles that you’re going to use for the base have one side that is equal to the edge of the polygon and one side that is equal to the radius of the polygon and 2.) you can fold a pleat for the supplement of the inside angle of the polygon.
I’ll work on the hex paper tonight.
Larry Phillips originally shared:
The Brachistochrone This animation is about one of the most significant problems in the history of mathematics: The Brachistochrone Challenge: If a ball is to roll down a ramp which connects two points, what must be the shape of the ramp’s curve be, such that the descent time is a minimum? Intuition says that it should be a straight line. That would minimize the distance, but the minimum time happens when the ramp curve is the one shown: a cycloid. Johann Bernoulli posed the problem to the mathematicians of Europe in 1696, and ultimately, several found the solution. However, a new branch of mathematics,Calculus of Variations, had to be invented to deal with such problems. Today, calculus of variations is vital in Quantum Mechanics and other fields.
Straight Lines…
It’s funny how our sense of sight can so easily be fooled
Mathematically, all you need to do is subtract or add +1 to the position of each ball and you get a perfect circle roll. Cool
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Repeated barycentric subdivision results in a gorgeous fractal-ish pattern.
Would like to see this as an animation.
Repeated barycentric subdivision results in a gorgeous fractal-ish pattern.
Golden Ratio φ = (1+sqrt(5))/2 = 1.6180339887498948482…
In mathematics, two quantities are in the golden ratio if their ratio is the same as the ratio of their sum to the larger of the two quantities. Expressed algebraically, for quantities a and b with a > b > 0. Two quantities a and b are said to be in the golden ratio φ if
(a+b)/a = a/b = φ
One method for finding the value of φ is to start with the left fraction. Through simplifying the fraction and substituting in b/a = 1/φ:
(a+b)/a = 1+ b/a = 1+1/φ
Therefore: 1+1/φ = φ Multiplying by φ gives: φ^2 - φ - 1 = 0
Using the quadratic formula, two solutions are obtained::
φ = (1- sqrt(5))/2 or φ = (1+sqrt(5))/2
Because φ is the ratio between positive quantities φ is necessarily positive:
φ = (1+sqrt(5))/2 = 1.6180339887498948482…
See more at Golden Ratio.
Image: Phi (golden number) by Steve Lewis.
I feel like this is a construction I haven’t seen before. And yet… it looks so familiar…
Golden Ratio φ = (1+sqrt(5))/2 = 1.6180339887498948482…
In mathematics, two quantities are in the golden ratio if their ratio is the same as the ratio of their sum to the larger of the two quantities. Expressed algebraically, for quantities a and b with a > b > 0. Two quantities a and b are said to be in the golden ratio φ if
(a+b)/a = a/b = φ
One method for finding the value of φ is to start with the left fraction. Through simplifying the fraction and substituting in b/a = 1/φ:
(a+b)/a = 1+ b/a = 1+1/φ
Therefore: 1+1/φ = φ Multiplying by φ gives: φ^2 - φ - 1 = 0
Using the quadratic formula, two solutions are obtained::
φ = (1- sqrt(5))/2 or φ = (1+sqrt(5))/2
Because φ is the ratio between positive quantities φ is necessarily positive:
φ = (1+sqrt(5))/2 = 1.6180339887498948482…
See more at Golden Ratio.
Image: Phi (golden number) by Steve Lewis.
I feel like this is a construction I haven’t seen before. And yet… it looks so familiar…