Use Random Cut Theorem and Simple Probability

In Euclid's "Elements" are several versions of the Random Cut Theorem; in this article, you will see one of the simpler ones (Book II Proposition 4) to make a point about simple probability you can use.


The Tutorial

  1. Learn or recall from "Elements", Book II, Proposition 4, that "If a straight line be cut at random, the square on the whole is equal to the squares on the segments and twice the rectangle contained by the segments."

    • Grab/copy the yellow and orange diagram from this page onto the Clipboard.
    • TYpe Command and c for copy to make a copy of the diagram.
    • At the Desktop, click on the XL icon on the Dock to start Excel.
    • Open a New Workbook in Excel.
    • Holding down the Shift Key, do Edit Paste Picture into a new worksheet.
    • Save the file under an appropriate filename into a logical folder, and make notes from the steps below, below the diagram.
  2. Learn or recall that, per the above diagram, if segment AC = x and segment CB = y, that (x+y)^2 also equals x^2 + y^2 + 2xy. Stated another way, if x = y+z, then x^2 = (y+z)^2 = y^2 + 2yz + z^2. In the diagram then, square HF = y^2, square CK = z^2, rectangle AG = yz and rectangle GE = yz as well.

  3. Call the Area of the Whole square AE equal to 1 or 100% Probability.
  4. Guesstimate from the random cut in the diagram that it occurred at a value of .70 of length AB.
  5. Calculate from your guesstimate what a dart's chance's of landing in each square are, or where the next robbery will be, i.e. where the next random event will be.
    • There is a .3 * .7 = 21% chance the next event will be in one of the two rectangles, and so a 42% chance it will be in either rectangle (or in 2yz).
    • There is a .3 *.3 = 9% chance the next event will be in square CK (or in z^2).
    • There is a .7 * .7 = 49% chance the next event will be in square HF (or in y^2).
    • (.42 + .09 + .49 = 1.00 = 100% chance there will be a next event (presumably). This is an important assumption and not always the case. Or perhaps it's a presumption. It's in the sump somewhere, speaking etymologically ...

Explanatory Charts, Diagrams, Photos

  1. It's important to remember that Euclid solved this about 2.300 years ago! It's still very usable information today. A random cut of a card deck, a random cut in an accident, a random cut in Nature, etc., etc. -- all subject to analysis because of Euclid.

Helpful Guidance

  1. Make use of helper articles when proceeding through this tutorial:
    • See the article How to Describe a Square on a Given Line AB for a list of articles related to Euclid, Geometric and/or Trigonometric Art, Charting/Diagramming and Algebraic Formulation.
    • For more art charts and graphs, you might also want to click on Microsoft Excel Imagery, Mathematics, Spreadsheets or Graphics to view many Excel worksheets and charts where Trigonometry, Geometry and Calculus have been turned into Art, or simply click on the category as appears in the upper right white portion of this page, or at the bottom left of the page.


  • Note that if there's an equation like (2.47 + 6.03x)^2, the numbers can be summed and pro-rated to determine the likelihood of a result being a cause of one of the quantities. Thus, 2.47 + 6.03 = 8.5 and 2.47/8.5 = .2906 and 6.03/8.5 must be .7094; these we can combine as above into our areas of y^2, 2yz and z^2. A result following from this formula will fall somewhere into one of those 3 classes, according to the probability of each, but even more so with the x attached as a factor to one member, as that will skew the results further from what they'd otherwise be (unless x=1, or both variables equate to 0).


  • Do not forget to be reasonable. If the event was a "cut", like an axe cut, another one may soon fall quite nearby. Look for a CAUSE and EFFECT relationship before going hogwild with statistics. If the situation was that y was bound to z, or made fixed in an AND in some way, then maybe an extension at either end or again a branching at the cut point is probable. Learn to think outside the box in other words. And if y+z is determined to be part of a series, say z = y/(y-1) so that y+z = y*z, then look for the series to continue.

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