MHB Solving Cubic Equations using Origami

  • Thread starter Thread starter greenglasses
  • Start date Start date
  • Tags Tags
    Cubic
AI Thread Summary
The discussion revolves around using origami to solve cubic equations, specifically the equation x^3 + ax^2 + bx + c = 0. It explains how to create a parabola using origami folds, with points P1 and P2 representing specific coordinates and lines L1 and L2 defining the framework for the folds. The confusion arises from the derivation of the equations (x1-a)^2 = 4y1 and (x1-a)(x-x1) = 2(y-y1), which relate to the properties of the parabola and the crease formed during the folding process. The thread also references resources on origami's applications in computational problems and its evolution as both an art and engineering tool. Understanding these concepts can clarify the mathematical principles behind solving cubic equations with origami.
greenglasses
Messages
23
Reaction score
0
I have to write a research paper on a mathematical topic for my class; I chose the above topic.

I understand that a parabola can be formed using a focus and directrix, both created by origami folds, and that Axiom 6 of Origami-Folding (Given two points p1 and p2 and two lines l1 and l2, there is a fold that places p1 onto l1 and p2 onto l2) can be used to solve a cubic equation. But some of this explanation of why confuses me:

"Now, let's solve the cubic equation x^3+ax^2+bx+c=0 with origami. Let two points P1 and P2 have the coordinates (a,1) and (c,b), respectively. Also let two lines L1 and L2 have the equations y+1=0 and x+c=0, respectively. Fold a line placing P1 onto L1 and placing P2 onto L2, and the slope of the crease is the solution of x^3+ax^2+bx+c=0.

I will explain why. Let p1 be a parabola having the focus P1 and the directrix L1. Since the crease is not parallel to the y-axis, we can let the crease have the equation y=tx+u. Let the crease be tangent to p1 at (x1,y1), and (x1-a)^2=4y1. Because the crease has the equation (x1-a)(x-x1)=2(y-y1), we get t=(x1-a)/2 and u=y1-x1(x1-a)/2. From these equations, we get u=-t2-at."

Specifically, I do not understand where the equations (x1-a)^2=4y1 and (x1-a)(x-x1)=2(y-y1) are coming from and w hat they mean.

I would greatly appreciate someone helping to explain.

[this explanation comes from K's Origami : Origami Construction if you want a look at the entire thing]
 
Mathematics news on Phys.org
I found a reference that describes how origami is used in a variety of computational problems. In particular, the classic trisecting an angle and doubling the cube from ancient Greece which are based on cubic equations.

https://plus.maths.org/content/power-origami

and to pique your origami interest further there is the PBS documentary Between the Folds that shows how far Origami has come as an art and as an engineering tool:



There are many other resources on Origami as books and videos from Robert Lang and Eric DeMaine



 
  • Like
Likes Wrichik Basu
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Is it possible to arrange six pencils such that each one touches the other five? If so, how? This is an adaption of a Martin Gardner puzzle only I changed it from cigarettes to pencils and left out the clues because PF folks don’t need clues. From the book “My Best Mathematical and Logic Puzzles”. Dover, 1994.
Back
Top