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So I heard that if one inscribes the largest circle that can fit into a equilateral triangle, then divides the perimeter of the triangle by the diameter of the inscribed circle, it gives a value which can be called "triangle $\pi$", and that value ($\sqrt{27}$) can be used in the place of regular $\pi$ to derive volumes of the other platonic solids. Is that true? Is there a different $\pi$ for triangles? What is that value? Is it close to $\sqrt{27}$? Can it be used to find volumes of platonic solids, especially the icosahedron and the one that looks like a pyramid flipped and stacked on its twin? 4 part question. Thanks we have been arguing about it at work for weeks

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So, what is the question? – Gerry Myerson Jun 14 '12 at 23:48
$\pi$ is a constant, it has only one value... – Alex Becker Jun 15 '12 at 0:03
The word you are looking for is octahedron. – Gerry Myerson Jun 15 '12 at 1:56
If you want to know the volumes of the Platonic solids, try…. – Gerry Myerson Jun 15 '12 at 1:59
The relationship of that number to triangles is somewhat analogous to the relationship of $\pi$ to circles, but calling it "triangle pi" is problematic in two respects: (1) It is not standard terminology, and (2) There are all sorts of special properties of the number $\pi$ that would not apply to that number. $\pi$, for example is a transcendental number. I can imagine it being called "triangle pi$ within the context of a particular article about it, but I'd be a bit surprised if the author proposed adopting that language as standard terminology. – Michael Hardy Jun 15 '12 at 3:11

Yes, the ratio of the perimeter of an equilateral triangle to the diameter of its incircle is indeed $3\sqrt3$, or $\sqrt{27}$ as you wrote. I guess you can call it "triangle pi" if you want, but I don't know of anyone else who does, and people might not take you very seriously if you did.

Actually, I do find it rather interesting that the volume of a regular octahedron of inradius $r$ is $\frac43\overset{\scriptsize\triangle}\pi r^3$, just like the volume of a sphere except with $\overset{\scriptsize\triangle}\pi = 3\sqrt3$. But there don't seem to be any similar relationships with the other Platonic solids.

To address just one of your edited questions: "Is there a different $\pi$ for triangles?" Not really. $\pi$ is the name of a specific number whose value is about $3.14159\ldots$ You can always find a different number which is related to, say, equilateral triangles in an way analogous to how $\pi$ is related to circles, and give it a name of your choosing, such as "triangle $\pi$". But that's just an analogy. It doesn't make that number "a different $\pi$ for triangles", any more than Stephen Harper is "a different Barack Obama for Canada".

On further reflection, whoever chose the diameter of the incircle to take the ratio against, as opposed to that of the circumcircle, the nine point circle, or any other circle associated with the triangle, definitely knew what they were doing. If the inradius is $r$ and you choose $\overset{\scriptsize\triangle}\pi$ so that the triangle's perimeter is $2\overset{\scriptsize\triangle}\pi r$, then it is also the case that the area of the triangle is $\overset{\scriptsize\triangle}\pi r^2$! This actually holds for arbitrary triangles, not just equilateral ones — and it doesn't work for any other value of $r$ different from the inradius.

One can start to demystify this by observing that the incenter is the unique point whose perpendicular distance from all three edges of the triangle is the same, and equal to the inradius. Why this matters is easiest to see when generalized to $n$-sided polygons:

If there exists a point whose perpendicular distance from all sides of a simple polygon is equal to the same value $r$, then the area of the polygon is $\frac12 r$ times the perimeter of the polygon.

This now starts to look like an obvious fact: simply join all the vertices of the polygon to the inner point, and add up the areas of the triangles formed. And it explains the above property of the inradius, which always exists for any triangle (though typically such a point does not exist at all for an arbitrary $n$-sided polygon with $n>3$).

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