Can we compute the surface area and the volume of the Penrose triangle as an impossible object, not as a visual illusion?
I'm interested in this question because I'm trying to find a proof that mathematics cannot distinguish between possible and impossible objects. That is, an object does not necessarily exist because we can compute its mathematical properties.
Thanks.
tl; dr: There are at least two questions here: One concerning "impossible objects" and what it means for some mathematical object to "exist," the other about geometry.
I'm interested in this question because I'm trying to find a proof that mathematics cannot distinguish between possible and impossible objects. That is, an object does not necessarily exist because we can compute its mathematical properties.
Consider the questions
To answer any of these, we need a definition of triangle, but we also need to agree what is meant by exist. For the second we need in addition a definition of right angle.
Suppose we define a triangle to be a Euclidean plane figure comprising three line segments (the sides) such that each pair of sides meet at precisely one common endpoint (a vertex of the triangle), and each side meets the other two sides.
In this setting there does not exist a triangle with four sides in the sense that a triangle with four sides is logically self-contradictory. The same argument can be used in any setting where a triangle is a figure with three sides.
In the same setting there does not exist a triangle with three interior right angles: In Euclidean geometry, we can prove the sum of the interior angles of a triangle is a straight angle, i.e., two right angles.
Finally, we might agree that triangles exist because there are constructs within Euclidean geometry satisfying the definition.
Alternatively, we might agree that triangles do not exist because triangles, and Euclidean geometry itself, and arguably all of mathematics, are mere idealizations of the physical world around us.
These considerations illustrate why we need to exercise care to speak of impossibility.
Let's revisit the second question. Suppose we define a triangle to be a spherical figure comprising three arcs of great circles (the sides) such that each pair of sides meet at precisely one common endpoint (a vertex of the triangle), and each side meets the other two sides.
In this setting, there exists a triangle having three interior right angles. For example, take two longitudes meeting perpendicularly and extending from one pole to the equator, and close the figure with the quarter-equator joining the "free" endpoints.
If the sphere is assumed to have Euclidean radius $R$ (or spherical diameter $\pi R$ if we want to refer only to intrinsic geometry), a "triple-right" triangle has perimeter $\frac{3}{2}\pi R$ and area $\frac{1}{2}\pi R^{2}$.
To a Euclidean plane mathematician, such a figure is impossible in the sense of logically incompatible with theorems of Euclidean plane geometry. To a spherical mathematician, a triple-right triangle is one-eighth of a globe.
Logically, the considerations about Euclidean versus spherical triangles parallel the question about the Penrose triangle. The Penrose triangle is not logically self-contradictory. Nonetheless, with a suitable definition involving three equal-length sides of square cross section meeting perpendicularly, a Penrose triangle is incompatible with three-dimensional Euclidean geometry.
On the other hand, as noted by Stinking Bishop in the comments, there exists a three-dimensional universe with locally-Euclidean geometry, topologically the product of the Euclidean plane and a circle, in which the Penrose triangle does exist as a mathematical object, analogously to a triple-right triangle on a sphere.
In Euclidean three-space, consider the infinite object made by concatenating successive $1 \times 1 \times \ell$ blocks parallel to the Cartesian coordinate axes in cyclic succession. More precisely, place a copy with opposite corners at $(0, 0, 0)$ and $(\ell, 1, 1)$; place a second copy with corners at $(\ell+1, 0, 0)$ and $(\ell, \ell, 1)$; place a third copy with corners at $(\ell + 1, \ell + 1, 0)$ and $(\ell, \ell, \ell)$. Call the union of these three pieces a fundamental unit.
The chain is now extended by translating the fundamental unit by integer multiples of $(\ell, \ell, \ell)$. With $\ell = 4$ as in the diagram, the union is a doubly-infinite ordered list of unit cubes, each with two attached faces and four unattached faces. Whether or not $\ell$ is an integer, the fundamental unit has volume $3\ell$ and exposed surface area $12\ell$ in the infinite chain.
Introduce a Cartesian coordinate system $(x, y, z)$ whose first two axes are perpendicular to $(1, 1, 1)$ and whose third axis is parallel to $(1, 1, 1)$. Let the additive group $(\mathbf{Z}, +)$ act as translation by multiples of $(\ell, \ell, \ell)$, i.e., $$ n \cdot (x, y, z) = (x, y, z) + n\ell(1, 1, 1). $$ The "slab" $D = \{(x, y, z) : 0 \leq z \leq \ell\sqrt{3}\}$ is a fundamental domain for the group action, and the infinite chain of fundamental units is invariant under the group action. Since translations are isometries, all the geometry descends to the quotient. The ambient space is the product of a Euclidean plane and an orthogonal circle of circumference $\ell\sqrt{3}$. The image of the fundamental unit is a topological loop with its "initial" cube abutting the "final" cube, i.e., a Penrose triangle.
While a Penrose does exist in this locally-Euclidean quotient space, it does not exist (embed isometrically as a subset) in Euclidean three-space: If we start at the corner $(0, 0, 0)$ of the fundamental unit and travel successively to $(\ell, 0, 0)$, $(\ell, \ell, 0)$, and $(\ell, \ell, \ell)$, we return to our initial location in the Penrose triangle (as indicated by any of the blue arrows), but we do not return to our initial location in Euclidean space.
