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I made a cardboard triangular tile out of 6 other triangles. Here are the triplets shown in order of CAB, BAC, CAB, BAC, CAB, BAC. The slashes show sides-C joined and the vertical lines show sides-B joined. I found the arcsines of the sine values and my best calculations show that the exterior acute angles add up to $\pi$ radians. The problem is, unless this is proven, I cannot ’know’ that the peaks all meet at a central point and that the figure is 2D and not a pyramid. What approach can I take to prove my conjecture?

/ 565 403 396 | 396 1053 1125 / 1125 675 900 | 900 1925 2125 / 2125 2107 276 | 276 493 565 \

enter image description here

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    $\begingroup$ That the sum of angles in a triangle is equal to $\pi$ is equivalent to the parallel postulate, one of the axioms of Euclidean Geometry and which need not necessarily be true for example in Hyperbolic or Spherical Geometries. See https://en.wikipedia.org/wiki/Sum_of_angles_of_a_triangle. Showing that a particular sum of numbers is approximately equal to $\pi$ can be accomplished using basic arithmetic. Showing that those numbers adds up to exactly $\pi$ is impossible in your case since the numbers are inexact approximations. $\endgroup$ – JMoravitz Jun 8 '18 at 20:33
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Like @dxiv, I considered the central angles. Unlike @dxiv, I reached a positive verdict:

Your pieces should fit.

We know your pieces are perfect right triangles that fit together in pairs as follows:

enter image description here

To determine whether they fit together in one whole, we need only determine whether $\alpha+\beta+\gamma = 360^\circ$. Well,

$$\begin{align} \sin(\alpha+\beta+\gamma) &= \sin\alpha\cos\beta\cos\gamma + \cos\alpha\sin\beta\cos\gamma +\cos\alpha\cos\beta\sin\gamma - \sin\alpha\sin\beta\sin\gamma \\ \cos(\alpha+\beta+\gamma) &= \cos\alpha\cos\beta\cos\gamma - \cos\alpha\sin\beta\sin\gamma - \sin\alpha\cos\beta\sin\gamma - \sin\alpha\sin\beta\cos\gamma \end{align}$$

where

$$\cos\alpha = \frac{-a^2+q^2+r^2}{2qr} \qquad \cos\beta = \frac{-b^2+r^2+p^2}{2rp} \qquad \cos\gamma = \frac{-c^2+p^2+q^2}{2pq}$$ $$\sin\alpha = \frac{2|\triangle OBC|}{qr} = \frac{ad}{qr} \qquad \sin\beta = \frac{be}{rp} \qquad \sin\gamma = \frac{cf}{pq}$$

Substituting ... $$a \to 493 + 2107 \qquad b \to 675 + 1925 \qquad c \to 1053 + 403$$ $$d \to 276 \qquad e \to 900 \qquad f \to 396$$ $$p \to 1125 \qquad q \to 565 \qquad r \to 2125$$ ... Mathematica calculates (via precise integer arithmetic) ... $$\sin(\alpha+\beta+\gamma) = 0 \;\;\text{(exactly)} \qquad\qquad \cos(\alpha+\beta+\gamma) = 1\;\;\text{(exactly)}$$ ... so that (subject to the accuracy of your measurements) $\alpha+\beta+\gamma$ is definitely an integer multiple of $360^\circ$. By inspection, the integer multiplier is $1$. $\square$

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  • $\begingroup$ What drawing software do you use? I could use it for a paper I'm writing on this subject. $\endgroup$ – poetasis Jun 9 '18 at 18:00
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    $\begingroup$ @poetasis: I use GeoGebra for my drawings. $\endgroup$ – Blue Jun 9 '18 at 21:44
  • $\begingroup$ @Blue Thanks for the correction. Could you please enter the expression linked in my now deleted answer in the standalone Mathematica and see what it returns. I am puzzled that WA would screw up on such a straightforward calculation. $\endgroup$ – dxiv Jun 10 '18 at 5:27
  • $\begingroup$ @dxiv: Hmmm ... As it turns out, you seem to have made a simple typo: $2125/276$ should be $2107/276$. $\endgroup$ – Blue Jun 10 '18 at 5:37
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    $\begingroup$ @Blue Silly typo, indeed. Fixed now, thanks for catching that. $\endgroup$ – dxiv Jun 10 '18 at 19:22
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[ EDIT ] Thanks @Blue for catching the original typo. Corrected and reposted as CW.


According to WA the sum of the center angles equals $\,2 \pi\,$, which proves that the six triangles with the given sides can indeed be arranged as pictured around a common vertex in the plane:

$$ \arctan \frac{403}{396} + \arctan \frac{1053}{396} + \arctan \frac{675}{900} + \arctan \frac{1925}{900} + \arctan \frac{2107}{276} + \arctan \frac{493}{276} = 2 \pi $$

This is an exact calculation which can be verified by hand using the $\tan$ of sum of angles formula.

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  • $\begingroup$ Yes, it looks like the peaks won't quit meet at a point. (We'll have a Mercedes star in the middle.) I've generated many of these programmatically. Still, if I find a set for which WA says the internal arctangents add up to 2$\pi$, wouldn't it still be just a better approximation? Thanks. $\endgroup$ – poetasis Jun 8 '18 at 22:26
  • $\begingroup$ @poetasis The above is an exact calculation, not an approximation. It can be verified by hand using the sum of angles $\tan$ formula. $\endgroup$ – dxiv Jun 8 '18 at 22:31

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