What is the slope of the third type of slice? (Try 2 -- wrong value for minor radius)

Algebra Level 4

This problem’s question: {\color{#D61F06}\text{This problem's question:}} What is the slope of the third type of slice?

The tori are centered on the origin with the circular path enclosing the z-axis and in the x-y plane.

The slope is the tangent of the acute angle between the cut plan and the x-y plane.

The radius of the circular path is 29. The radius of the circular cross-section is 21.

The wrong value for the minor radius is the reason for try 2.

Thank you, Jon Haussmann for catching this..

There are three ways of slicing a torus with a circular path and a circular cross-section to get pairs of circles on the cut plane which also passes through the origin:

  1. by a plane which passes through the z-axis, that is it has an infinite slope with respect to the x-y plane

  1. by the x-y plane, that is, it has a zero slope respect to itself.

  1. by some other plane that is not parallel to any of the planes above

No illustration here. That will be in the solution! \text{No illustration here. That will be in the solution!}


The answer is 1.05.

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1 solution

The unsliced torus:

The torus sliced vertically with the cut plane passing through the z z -axis:

The torus sliced horizontally on the x-y plane:

As promised. two illustrations for the third cut plane case with the two cut circles in purple (one is solid and one is dashed):

The blue plane is at a fixed slope of and passes through the y y -axis. In the first of the two illustrations, the plane is edge on to the viewer and you can see a small amount of the torus in front of the plane at the top of the torus and a small amount of open space at the bottom of the torus; the slope is greater than 1.

In the second of the two illustration, the view is more into the openings of the cut torus and you can see a small amount of the torus in front of the plane at the top of the torus and none at the bottom of the torus

I have verified that the two cut edges are circles, that they are on the surface of torus by solution of the intersection parameters and their radius is the same as the larger torus radius.

The parameters: a = 21 a=21 , b = 20 b=20 and c = 29 c=29 . Yes, the parameters of a Pythagorean triangle.

Slope is defined as rise run \frac{\text{rise}}{\text{run}} . The slope is 21 20 = 1.05 \frac{21}{20}= 1.05 . Yes, I have my a a and b b reversed from the normal convention. I did not want to go back an swap the two variables in the program.

The Mathematica line that produced the third case illustration is at the end of this solution.

Brilliant requires that a code block be the last item in a solution. The back-slashes have been replaced by $ as the Brilliant code processor does not yet handle backslashes well.

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Show[
 {
  ParametricPlot3D[
   If[b (a Sin[v]) <= 
     a ((c + a Cos[v]) Cos[u]), {(c + a Cos[v]) Cos[
       u], (c + a Cos[v]) Sin[u], a Sin[v]}], {u, 0, 2 $[Pi]}, {v, 0, 
    2 $[Pi]}, AspectRatio -> Automatic, ImageSize -> Large, 
   MaxRecursion -> 3, PlotPoints -> 317, AxesLabel -> {"X", "Y", "Z"},
    AxesOrigin -> {0, 0, 0}, PlotStyle -> {{Orange, Opacity[.6]}},
   PlotRange -> Automatic, 
   PlotLabel -> {"The blue plane is at a slope of 1.$nThe cut plane is at a slope of " <> ToString[N[a/b, 4]] <> ".$n" <>
      "The radius of the cut circles is " <> ToString[c] <> 
      ".$nThe smaller torus radius is " <> ToString[a] <> ".$n" <>
      " and the larger torus circle radius is " <> ToString[c] <> 
      "."}, Boxed -> False],
  ParametricPlot3D[{{b Sin[w], c Cos[w] - a, a Sin[w]}, {b Sin[w], 
     c Cos[w] + a, a Sin[w]}}, {w, 0, 2 $[Pi]}, 
   PlotStyle -> {{Purple, Thick}, {Purple, Thick, Dashed}}, 
   AxesOrigin -> {0, 0, 0}, AxesLabel -> {"X", "Y", "Z"}],
  ParametricPlot3D[c {Cos[u], Sin[u], 0}, {u, 0, 2 $[Pi]}, 
   PlotStyle -> {{Red}}],
  ParametricPlot3D[{0, -c + a Cos[v], a Sin[v]}, {v, 0, 2 $[Pi]}, 
   PlotStyle -> {{Green, Dashed}}],
  Graphics3D[{Blue, Opacity[.4], 
    InfinitePlane[{0, 0, 0}, {{0, 1, 0}, {1, 0, 1}}]}, Axes -> True, 
   AxesOrigin -> {0, 0, 0}]
  }
 ]

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