Refer to the attached diagram sketch to help visualize the equation.

I am requesting help with an interesting math problem. Basically, I am diagraming infinity using three planes. These planes connect with each other at either side at 120 degree angles. These connection lines never really connect at the center, as they go on into infinity, but appear to (think repeating 9). Three perfect circles are placed on each of the planes. When viewed 2 dimensionally, the circles appear to be ellipses.

The Problem: What is the width/height ratio of these illusional ellipses? For computational purposes, put the three perfect circles on each plane touching each other. The actual width of the minor axis for the smallest illusional ellipse can be set to 1 for computational purposes (or really any number you fee like).

The Diagram: I have used the 1:3 ratio to make the circles in this diagram (120 by 360, 360 by 1080, and 1080 by 3240). I suspect that this ratio is not correct; however, the answer should be fairly close to this ratio. I am not entirely sure if the ratio will be different for each of the circles, but it could be.

Thank you in advance for your time working on this problem. If you have seen it before, great! If not, hopefully you will enjoy the challenge. Solving this problem in needed to complete a logo for a non-profit organization. Credit will be given to whoever solves this.

enter image description here

  • $\begingroup$ Ok, so maybe I'm off base here, but this is what comes to mind. First off, I see you're splitting things up as if they were on a plane. The way my mind conceptualized the problem was that the three planes are actually X,Y,Z in 3D space. So if that's the case, we can treat each "side" as it's own plane and do the calculations before the projection from 3D to 2D. That projection is going to depend on how much foreshortening you want. For that, I recommend looking at Graphical Projections. Does this jive with what you're asking? $\endgroup$ – J Trana Dec 24 '13 at 7:45

As @JTrana suggests in a comment, the answer depends on how much foreshortening you want to do. If you look down on a tripod, you'll see the legs make $120^\circ$ angles, no matter how "open" the tripod is, so that part is unchanging; circles fitted between the legs of the tripod, however, will appear as ellipses of varying eccentricities for different degrees of tripod open-ness.

Since there's no inherent foreshortening to go along with the apparent $120^\circ$ separation of the legs, then, you may use any ellipse you like in your logo, secure in the knowledge that it corresponds to some amount of foreshortening. This gives you the artistic freedom to use ellipses you find aesthetically pleasing (aspect ratio = golden ratio, perhaps?).

If, however, you really-really want the image to represent the shadow of a particular 3D configuration, then we can do that, too. I'll walk you through it.

Let $OABC$ be a translucent triangular pyramid (ie, a tetrahedron) with equilateral base $\triangle ABC$ and identical isosceles lateral faces such as $\triangle OBC$. (The lateral edges of the pyramid are like the legs of the tripod.) A circle painted on $\triangle OBC$ will cast an elliptical shadow onto the base; one dimension of this ellipse (the one parallel to $\overline{BC}$ will match the diameter of the circle; the perpendicular dimension will undergo the same deformation as the shadow cast by face $\triangle OBC$ itself.

Now, the shadow of $\triangle OBC$ is $\triangle PBC$, where $P$, the shadow of $O$, is the center of $\triangle ABC$. Observe that here, too, that one dimension of the shadow (namely $\overline{BC}$) matches that of the face, while the perpendicular dimension has been deformed. Writing $M$ for the midpoint of $\overline{BC}$, we can express the deformation factor as $$k = \frac{|\overline{PM}|}{|\overline{OM}|}$$ so that the shadow ellipses have aspect ratios $1:k$.

Since $\triangle ABC$ is equilateral, we have that $|\overline{PM}| = \frac{1}{3}|\overline{AM}|$; moreover, $|\overline{AM}| = \frac{\sqrt{3}}{2}|\overline{BC}|$. Thus, we can write $$k = \frac{1}{2\sqrt{3}}\;\;\frac{|\overline{BC}|}{|\overline{OM}|} = \frac{\sqrt{3}}{6}\;\frac{\text{base of } \triangle OBC}{\text{altitude of } \triangle OBC}$$

If you like, we can express $k$ in terms of other aspects of the tetrahedron.

  • If the edges of base $\triangle ABC$ have length $b$, while the lateral edges have length $\ell$, then applying the Pythagorean Theorem to $\triangle OMB$ yields $$|\overline{OM}|^2 + \left(\frac{1}{2} b\right)^2 = \ell^2 \qquad \text{so that} \qquad k = \frac{b\;\sqrt{3}}{3\sqrt{4\ell^2 - b^2}}$$

  • If the lateral edges make an angle of $\theta$ with each other, then $|\overline{OM}| = \ell \cos(\theta/2)$ and $|\overline{BC}| = 2 \ell \sin(\theta/2)$, so that $$k = \frac{\sqrt{3}}{3}\;\frac{\sin(\theta/2)}{\cos(\theta/2)} = \frac{\sqrt{3}}{3}\;\tan\frac{\theta}{2}$$

  • If the pyramid is in fact a regular tetrahedron ---with all edges equal and making $60^\circ$ angles with each other--- then either of the above reduces to $$k = \frac{1}{3}$$

  • If the pyramid is a right-corner tetrahedron ---with lateral edges at right angles to each other--- then we have $$k = \frac{\sqrt{3}}{3}$$

... and so forth.

  • $\begingroup$ THANK YOU!! I was so happy when I saw this reply, your time is much appreciated!!!!!! $\endgroup$ – Douglas Jan 13 '14 at 3:23

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