# The Saccheri Quadrilateral

We’ve been exploring non-Euclidean geometries lately, and the Saccheri Quadrilateral plays a pivotal role in this particular mathematical history.

The Saccheri quadrilateral is a biperpendicular quadrilateral with two congruent legs. It’s an object that is “obviously” a rectangle in Euclidean geometry, but proving that without the aid of the parallel postulate turns out to be rather tricky.

In fact, just proving that the measure of the green angle is less than the measure of the orange angle is pretty tough!

## 6 Comments

## Justin Lanier · March 25, 2012 at 10:13 pm

You mentioned on Twitter that you’re looking for help with a proof of this. Unfortunately, there isn’t one—because it isn’t true! (At least if I’m understanding you correctly) The “weak” exterior angle theorem—that an exterior angle of a triangle is greater than either of its remote interior angles (Elements I.16)—isn’t valid on a sphere. Consider a triangle with a 100º interior angle at the north pole and two right interior angles at the equator. The exterior angles at the equator are also right, and so less than the 100º interior angle at the pole.

The same thing happens in your Saccheri quadrilateral. Make a Saccheri quadrilateral with a 100º pole angle, then lower its “top” side closer and closer to the equator. Its summit angles then get as close to 90º as you please (since the “top” side acts more and more like the equator; or if you prefer, its area goes to 0, and so its angle sum goes to a Euclidean 360º). And presto—an example of a Saccheri quadrilateral on a sphere whose exterior green angle is greater than its summit orange angles.

A final note: Euclid’s proof of the weak exterior angle theorem fails on a sphere because he uses his second postulate—that any straight line can be extend indefinitely—to determine a point outside of the given triangle. But on a sphere, extending this line in this fashion doesn’t necessarily give a point outside of the triangle, because the line could wrap all of the way around the sphere and back into the triangle!

I hope that’s useful. Happy non-Euclideaning!

## Justin Lanier · March 25, 2012 at 10:31 pm

So I didn’t catch onto your question quite right, but the thrust of my argument still holds: if you make the pole angle obtuse, the green angle can get as close to equalling that obtuse angle as you please, and the orange summit angles can get as close to right as you please by pushing the quadrilateral’s top toward the equator.

## MrHonner · March 25, 2012 at 10:31 pm

Justin-

Thanks for the well-articulated (and cited!) reply. I think I understand your argument, but I don’t think it speaks exactly to what I’m trying to show (probably because there is some information and context that hasn’t been provided in this particular problem).

I’m not looking to prove the exterior angle theorem here. This is part of the proof that, in a Saccheri quadrilateral with unequal legs, the longer side is opposite the larger angle.

The green angle is part of the original biperpendicular quadrilateral with distinct legs, and the leg to the green angle is the longer leg. We’re trying to prove that the green angle is the smaller of the two original summit angles. This seems to rely on the fact that the green angle is indeed less than the orange angle, which is the summit angle of a Saccheri quadrilateral constructed inside the original biperpendicular quadrilateral.

Hopefully that clears things up. But I’m still not sure (a) how to prove it, or (b) if it’s true!

## Mike Lawler · April 10, 2013 at 9:15 pm

For the special case that the angle at the north pole is 90 degrees, I think the fact that the green angle is larger than the orange angle is a consequence of (i) the spherical law of sines and (ii) that the distance from the left hand orange angle to the green angle increases as the green angle “moves” down from the north pole to the right hand side orange angle.

I know that this isn’t the general case you were looking for, but it seemed like a simple case to start with. I’m interested to take a look at the case that the angle at the pole is acute tomorrow.

## MrHonner · April 10, 2013 at 10:21 pm

Thanks for chiming in, Mike–I’m glad the problem has caught your interest.

I’m afraid to start thinking too deeply about it right now, as delving into non-Euclidean geometry tends to make everything else I’m working on make less sense!

## Mike Lawler · April 11, 2013 at 6:26 am

I thought it was a really interesting problem for two reasons.

First, I only recently learned about an unsolved geometry problem named “the perfect cuboid.” It seems innocent enough – find a box (all 90 degree angles) with integer side lengths such that all of the face diagonals are also integers and the internal diagonal is also an integer. So, you need three integers a, b, and c, that produce 3 Pythagorean triples and also have the property that a^2 + b2 + c^2 is a perfect square. I was amazed to learn that this problem was unsolved. There are known solutions, btw, if you removed the condition about the interior diagonal. These shapes are called “Euler Bricks.”

Having seen this seemingly simply unsolved problem, it made me wonder if your problem – which is equally easy to state – would also prove difficult. I now believe that what you are trying to prove is indeed true if the pole angle is 90. Will try to think about the acute case as it seems pretty interesting.

The other reason it was interesting is that it reminded me of a problem I encountered in my work about 10 years ago. I left academia in the late 90s and now work as an underwriter for a large reinsurance company. The company had a data base containing the lat. and long. every 6 hours for every Atlantic hurricane going back at least 50 years (maybe longer, I don’t remember the detail).

The problem that some of the property insurance underwriters were trying to solve was this – I want to put in a known (lat., long.) into the database and know every hurricane that came within 50 miles of that position.

To solve this problem exactly requires some detailed spherical geometry, but to solve it roughly right (since a 50 mile radius circle is close enough to flat) you still need to understand a distance formula on a sphere that involved lat. and long. You also need to understand how to find the distance from a point to a line since just checking the known coordiantes of the hurricanes isn’t good enough as they will move more than 50 miles in 6 hours.

The property group was struggling with this problem and asked me to help out. It was quite a surprise to see spherical geometry coming up in the context of an insurance problem!

In any case, your problem reminded me a little bit about this old project which made it fun to work it.