E/NE 4: Right Down the Line

[stoutfellow]

In Euclidean geometry, it's not too hard to verify the existence of rectangles. The discussion in the previous post leads up to the converse, which is the climax of this series of posts: if rectangles exist, then Euclid's parallel postulate holds. (Or, turning things around, there are no rectangles in hyperbolic geometry.) The proof is under the cut.

Suppose ABCD is a rectangle. For the sake of visualization, let's orient things so that C lies below A and B is off to the right. Now, the line AB certainly can't intersect CD. Suppose, then, that m is some line through A other than AB; I claim that m must intersect the line CD.

So, let's suppose that m doesn't intersect the line CD. There are three conceivable situations: m might enter ABCD - i.e., it might run from the upper left to the lower right; it might run along AD (but in that case it certainly intersects CD); or it might run from the upper right to the lower left. If the third case holds, reflect ABCD in AD, to get a new rectangle ABC'D' which m does enter; since the line C'D' is the same as the line CD, we're back to the first case. In other words, we can safely assume that m enters ABCD.

By the same sort of argument as we've used before, m must exit through another side of ABCD - either BC or CD. Since we're assuming m doesn't intersect the line CD, we can rule out the second possibility: m must exit through BC.

Here's the idea of the proof. Create a series of rectangles, side by side, congruent to ABCD - BEFC, EGHF, and so on. m is going to enter each of those rectangles from the left side and exit from the right, and each time it exits from the right it will be further away from the line AB. Specifically, if it crosses BC x units below B, it will cross EF 2x units below E, GH 3x units below G, and so on. Sooner or later, we'll get a multiple of x which exceeds the length of AD (which is the length of BC, of EF, and so on), and as soon as that happens we have a contradiction - the intersection point can't lie on that line segment.

So, let's fill in the details. Let m intersect BC at X. ABX is a right triangle. Furthermore, the angle-sum of ABX is 180 degrees. (m divides ABCD into this triangle and the quadrilateral AXCD; the rectangle has angle-sum 360 degrees, and the angle-sum of the triangle plus the angle-sum of the quadrilateral is the angle-sum of the rectangle, plus 180 degrees for the two angles at X, for a total of 540. Since the angle-sums of the triangle and quadrilateral are at most 180 and 360 respectively by Saccheri-Lagrange, they must actually equal 180 and 360.) In particular, the angle AXB and the angle XAB sum to 90.

Let m strike EF at Y; let the perpendicular to EF through X strike EF at Z. I claim that the triangles ABX and XZY are congruent. First, they have right angles at B and Z respectively. The lengths AB and XZ are equal (AB and BE are congruent by design; BEZX is a rectangle, by one of the results last time; and BE and XZ are equal in length, by another of the last set of results). Now, the angles AXB, BXZ, and ZXY add up to 180, and BXZ is 90 degrees, so AXB and ZXY sum to 90; but then it follows from the last sentence of the last paragraph that ZXY and XAB are congruent. Now we can invoke angle-side-angle to show that the triangles are congruent. Therefore ZY has the same length as BX. So does EZ, again by a result from last time. Thus, if BX has length x, EY has length 2x. Clearly we can repeat this argument in the next rectangle, and so on. The result follows.

Let me restate the result: If l is a line and P a point not on l, and if there is a rectangle with P at one vertex and one side lying along l, then the edge of the rectangle opposite to l is the only line through P not intersecting l.

So, suppose there is at least one rectangle. Let Q be the foot of the perpendicular to l through P, and let R be a point of l different from Q. There exists a rectangle whose sides have lengths equal to PQ and QR, and we can place a congruent rectangle with P, Q, and R as three of its vertices. The assertion in the last paragraph now shows that there's only one line through P not intersecting l - i.e., Euclid's parallel postulate holds.

It's possible to push this a little further: if there's even one triangle whose angle-sum is 180, then there is a right triangle with angle-sum 180 (drop the perpendicular from one vertex to the opposite side), and if there's a right triangle with angle-sum 180, then there's a rectangle (rotate the triangle 180 degrees around the midpoint of the hypotenuse).

One final thought: one might wonder whether there's an intermediate geometry between Euclidean and hyperbolic - in other words, whether a geometry can be constructed, satisfying the axioms of neutral geometry, and in which, for some choices of P and l, there's only one line through P not intersecting l, while for others there's more than one. The answer is no; it's possible to show that, if there's even one point of that first type, then there is a rectangle. But I've gone on long enough, so I won't go into the details of that proof.

Comments

[countrycousin.livejournal.com]

Thank you. Illuminating.