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We are attempting to measure roots of quintic equations in Powerpoint. This is a mathematical curiosity inspired by Dr. Zye's recent video on making flags in Powerpoint. If you are interested in the rules of construction in Powerpoint (recommended), see below. If not, skip to the next section.

Dr. Zye's video on making flags in Powerpoint

Rules of Powerpoint construction

The rules of construction are:

  1. No inputting numbers allowed. For example, to make a 2 by 1 rectangle, you are not allowed to input the dimensions as 2 inches by 1 inch.
  2. Assume a perfect, axiomatic Powerpoint. This means that circles are assumed to be perfect circles, even though internally they are constructed from Béziers.
  3. No eyeballing.
  4. A number $x$ is considered constructible if you can make a square with side length $x$, relative to another unit square of side length 1.

An example result from this is that rational numbers are constructible. Any integer can be constructed by snapping unit squares next to each other in a line. To construct the inverses of those numbers, group those squares together and resize them such that their total width is the size of a unit square. For more results, see Dr. Zye's video, as well as Arglin's video on constructing square roots.

Solving quintics has now become useful for one of the flags: Croatia. It requires 5-secting an angle, and the value $\tan \frac{\arctan\frac12}5$, which is a root of the quintic $2x^5-5x^4-20x^3+10x^2+10x-1$, will be useful. We will talk about intersections in the next section, which are useful because we can take advantage of Powerpoint's Merge Shapes option to obtain a snappable vertex where the intersection is.

Solving Quintics

Our current approach is as follows.

Let our desired quintic be $Q(x)$ and our desired root be $n$.

Method

  1. Find the formulaic intersection of two builtin shapes. I chose to use waves (cubic, $y = 2x^3 - 3x^2 + x$) and shifted/scaled ellipses (quadratic, $\frac{(x - a)^2}b + \frac{(y - c)^2}d = 1$, where $b,d > 0$).
  2. The intersection's x-coordinates are the roots of a sextic polynomial $f(x)$.
  3. Construct a target sextic $g(x) = (x - p)Q(x - u)$. This has a root of $n + u$.
  4. Match the coefficients of $vf(x)$ and $g(x)$ and solve for $a,b,c,d,p,u,v$ (7 unknowns, 7 equations). We need $vf(x)$ instead of $f(x)$ because the coefficients of the sextics only have to differ by a constant.

I wrote some Python code in SymPy to solve this system (for the Croatia quintic), but unfortunately, all of the solutions did not satisfy $b, d > 0$. If I need to link the code, please drop a comment.

Questions

  • Is there a better choice for $g(x)$?
  • Can we use different figures for intersection? I tried rotating a wave by 90 degrees and intersecting it with another wave, but that resulted in a nonic polynomial that was too scary for SymPy to handle.
  • What other approaches could we use? Could we add an additional unknown somewhere so that the system is more flexible?
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  • $\begingroup$ Not an answer but this question was made obsolete due to mistakes made on my part. The actual construction is a lot harder, and if you want to see the progress you should go over to Arglin's Discord Server discord.gg/uNdpFzeNeY . $\endgroup$ Commented Nov 23 at 10:23
  • $\begingroup$ @Sunny well, even if it doesn't have an application, it would still be nice if we could get roots of arbitrary quintics (and hence get quintic roots) $\endgroup$ Commented Nov 23 at 13:38
  • $\begingroup$ Ahh, you're right, my bad. But we no longer care about the specific quintic... I think. $\endgroup$ Commented Nov 23 at 13:49
  • $\begingroup$ I'm not familiar with the Powerpoint constructions, but be aware that general quintics cannot be solved by compositions of addition, subtraction, multiplication, division or n-th roots. $\endgroup$ Commented Nov 24 at 3:26
  • $\begingroup$ @TobErnack that's something I do know about, but I don't have to express the actual value of the quintic, just need to measure it exactly. If you're not familiar with Powerpoint construction, that's alright. My question is essentially asking how I can take a specific cubic ($y=2x^3-3x^2+x$) and intersect it with an ellipse that can be stretched or shifted such that the x/y coordinate of the intersection is the root of a quintic. Just Cartesian math. Otherwise, we've actually found out how to get arbitrary quintic roots with some new Powerpoint tech, but this is still an interesting problem. $\endgroup$ Commented Nov 26 at 14:37

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