Yes, this is an actual question.

And yes, in some geometries, rectangles do not exist. (To be clear, a rectangle is defined as a quadrilateral with four right angles.)

Before proceeding with this line of thought, let me clarify what a geometry is. Most folks encounter geometry in high school, learning how to calculate perimeter, area, surface area, and volume for various shapes and how to prove triangle congruence with such criterion as side-angle-side (SAS), angle-side-angle (ASA), and hypotenuse leg (H-L). Geometry in this setting is just a step in the high school mathematics ladder.

That high school geometry, however, is not the be-all-end-all of geometry. Rather, it is a type of geometry, called Euclidean geometry, one of the three that exist in two dimensions. The other two are spherical geometry and hyperbolic geometry. A major difference between these three geometries is their treatment of parallel lines. Specifically, in spherical geometry parallel lines do not exist; in Euclidean geometry, for every line *l *and every point *P *not on *l*, there exists a unique parallel line to *l *through *P*; and in hyperbolic geometry, there exists an infinite number of lines through *P *that are parallel to *l*.

In Euclidean geometry, parallel lines are understood as being two distinct lines that are equidistant from one another – that is, every perpendicular joining a point *P *on *l *to a point *Q *on *m* has the same length. This is the familiar variation. In geometry broadly, a more accurate definition of parallel lines is: two distinct lines that do not intersect at any point. In the world of Euclidean geometry, this non-incidence applies for equidistant lines.

In the world of spherical geometry, the lines of spherical geometry are great circles; imagine a sphere being “straight” cut at various angles so that each cut intersects the center of the sphere and you’ll have a collection of great circles. Graphically (see above), we can see that any great circle will intersect at two points with any other great circle. Therefore, parallel lines do not exist.

What separates hyperbolic geometry from Euclidean geometry is Hilbert’s Parallel Postulate, or HPP. Hilbert’s Parallel Postulate, which is added to neutral geometry^{1} to create Euclidean geometry, states that for any line *l *and any point on on *l*, there exists a unique line *m *through *P *such that *l *is parallel to *m*. In hyperbolic geometry, HPP is thrown out, replaced by its negation: There exists a line *l *and a point *P *not on *l* such that at least 2 distinct lines that pass through *P *are parallel to *l*.

On a flat plane, this business of multiple parallel lines through a point can be visualized in the following picture. The picture looks questionable because it is attempting to project lines on a curved surface, the setting of hyperbolic geometry, to a flat plane, but I assure you that it is legitimate.

What does all this have to do with rectangles? It turns out that the negation of Hilbert’s Parallel Postulate in hyperbolic geometry and the absence of parallel lines in spherical mandates and implies rectangle non-existence.

We focus here on hyperbolic geometry. In this geometry, rectangles cannot exist because adopting a *reductio ad absurdum *hypothesis that rectangles exist leads to an assertion of HPP, which invalidates the defining feature of hyperbolic geometry (i.e. HPP is false and its negation if true). How is this?

This is where the discussion devolves to more technical math talk. If you don’t like technical math talk, I advise that you flee now. Maybe go to YouTube and watch the music video for this old Katy Perry song that has been stuck in my head for the past week.

For those who have decided to stay, let us continue.

To prove^{2} that rectangles cannot exist in hyperbolic geometry, we begin with the RAA hypothesis that rectangles do exist. Another theorem^{3} tells us that the existence of rectangles implies that all Saccheri and Lambert quadrilaterals are rectangles.

(Saccheri quadrilaterals are quadrilaterals with right base angles and congruent sides and Saccheri quadrilaterals are quadrilaterals with at least three right angles.)

Suppose that there is a line *l,* a point *P*, and a line *m* through *P *such that *l *is parallel to *m*. (My geometry textbook refers to this as the standard configuration *PQlm*. Very catchy.) Drop a perpendicular from *P *on *m *to a point, called the foot, *Q *on *l.* Now suppose that there exists a “competing” parallel *n* through *P *such that *l *is also parallel to *n*. These steps give us the following picture:

(Side note: I admit that *n* does not look parallel to *l* according to the “Is the lines parallel? Are the lines perpendicular?” exercises of youth, but this is college geometry. It’s how it goes…)

Now, for any *Y* on the ray of *n* that is on the same side of *m *as *Q* (that is, we don’t want a *Y *“above” *m*), drop a perpendicular with foot *X* to *m* and a perpendicular with foot *S *to line *PQ*. Note that *S* and *Y *are on the same side of *m* and *SY *is parallel to *m*, having the common perpendicular line *PQ*.

As a Lambert quadrilateral (three right angles), []*PSYX* is a rectangle according to the RAA hypothesis that rectangles exist. Hence, segment *PS* is congruent to segment *XY*. Since *n* is not equal to *m* and *n* is not perpendicular to *m*, the rays of *m* and *n* must form an acute angle.

There is in an axiom in geometry, called Aristotle’s Angle Unboundedness Axiom, from which we infer that the angle formed by rays of *m* and *n *has a perpendicular segment to *m* from a point *Y* on *n *to a foot *X *on *m* such that *XY* > *PQ*. That is, somewhere on the angle the perpendicular segment “connecting” *m* and *n* is “longer” than the segment *PQ*.

Note that previously we allowed “any *Y*.” Here, we’re using a specific *Y*. Previous statements on the relationship between *XY *and other segments apply. Since segment *XY* is congruent to segment *PS *(as they are the opposite sides of a rectangle), it follows that if *XY* > *PQ*, then *PS *> *PQ* and *Q *is between *P *and *S*.

*Q* is on *l*, so *Q *being between *P *and *S *implies that *P* and *S* are on opposite sides of *l*. Moreover, as *SY* is parallel to *l* (common perpendicular, as with *PS *and *m*), *S *and *Y *must be on the same side of *l*. If *P* and *S* are on opposite sides of *l *and *S* and *Y *are on the same side of *l*, then *P *and *Y* must be on opposite sides of *l*. Thus, *n* (i.e. the line *PY*) must meet *l*, which means that it is, in fact, not parallel to *l*.

As *P* and *l* are arbitrary, HPP (the posulate that there exists one unique line that is through *P* and parallel to *l*) holds. This contradicts the rejection and negation of HPP that is the defining point of hyperbolic geometry. **QED.**^{4}

1. Also known as absolute geometry, neutral geometry admits all axioms of incidence, betweenness, and congruence but does not account for a parallel postulate. These axioms are: [list axioms]

2. While scientific proof is a myth, math and logic are all about proofs.

3. That would be the Saccheri Angle Theorem, part (b). The other two parts concern triangles with angle sums of less than 180 degrees and greater than 180 degrees. Assuming that rectangles exist assumes that all triangles have angle sum of 180 degrees, implying that all Saccheri and Legendre quadrilaterals are rectangles.

4. This proof was adapted from one that my mathematics professor completed with the class during a lecture. Credit to Dr. Wantz.

I couldn’t have said it better!

LikeLiked by 1 person

Are you sure you don’t want to give it a shot? 😉

LikeLike