E/NE 3: They're All Right
Oct. 4th, 2005 11:54 am![[personal profile]](https://www.dreamwidth.org/img/silk/identity/user.png){kind=small}
stoutfellowThe proof of Saccheri-Legendre that I gave last time is, as I said, pretty involved. I need that theorem, though, for what comes next; if you're willing to accept it as true, most of what's left will be quite a bit easier. Give it a try, OK?
The Saccheri-Legendre Theorem told us that the sum of the angles of a triangle can't be more than 180 degrees. Now, think of a quadrilateral ABCD. If we draw the line AC, we divide the quadrilateral into two triangles, ABC and ACD (draw the picture!) and the sum of the angles of those two triangles is the sum of the angles of ABCD. (BCD = BCA+ACD, and DAB = DAC+CAB, so ABC + BCD + CDA + DAB = ABC + (BCA + ACD) + CDA + (DAC + CAB) = (ABC + BCA + CAB) + (ACD + CDA + DAC).) This gives us two results. First, the sum of the angles of ABCD is at most 360 degrees; second, if the sum of the angles of ABCD is exactly 360, then the sum of the angles of ABC is exactly 180, and likewise for the sum of the angles of ACD - if you have two numbers, neither one greater than 180, and their sum is 360, they must both be 180.
Now let's talk about rectangles. Suppose that ABCD is a quadrilateral, and all four of its angles are right. (Note that the angle-sum is then 360 degrees.) Then ABC and ACD are right triangles, and each has angle-sum 180. Since the angle ABC is 90 degrees, the sum of the angles ACB and CBA is also 90, and similarly for the sum of the angles ACD and CAD. But angles ACB and ACD together make up the right angle BCD, so they sum to 90 degrees also; therefore angle CBA and angle ACD are equal. Likewise angles ACB and CAD are equal. Since the triangles ABC and CDA share the side AC, and the angles flanking it are congruent, the angle-side-angle theorem tells us that the triangles are congruent. In particular, the sides AB and CD have the same length, and so do the sides BC and DA. (This is certainly what you'd expect in Euclidean geometry, but remember, we're doing neutral geometry at the moment!)
Next, take the rectangle ABCD and reflect it in the side BC, producing a new, congruent rectangle BCGH. Since ABC and CBH are right angles, A, B, and H lie on a straight line, and likewise for C, D, and G. This means that AGHD is another rectangle, just as tall as ABCD and twice as wide. Repeating this process, I can create rectangles as large as I want - think of a checkerboard, but with rectangles instead of squares. So, for instance, if I start with a 1x1 rectangle, I can make a 1x2, a 1x3, ..., and then a 2x1, 2x2, 2x3, and so on - I can make an mxn rectangle, for any positive whole numbers m and n. But is that all? Can I make one with fractional sides? Yes! I need the following result.
Let E be a point on the side AB of the rectangle ABCD, in between A and B. Let l be the line through E perpendicular to AB. Now, l enters the triangle ABC through side AB, so it must exit through one of the other sides. It can't exit through BC, because then AB, BC, and l would form a triangle with two right angles, and that violates Saccheri-Legendre. So l exits ABC through AC. But then it enters ACD through AC, and must exit through another side. It can't exit through AD - Saccheri-Legendre again - and so it exits through CD, say at F. Now the quadrilateral AEFD has three right angles (at A, E, and D), so the angle EFD can't be more than 90 degrees. But the quadrilateral on the other side, EBCF, also has three right angles, so EFC can't be more than 90 degrees. But EFD and EFC add up to 180 degrees, so they must both be exactly 90 degrees. In other words, AEFD is a rectangle. (So is EBCF.)
So what does this say? Suppose I want to construct a 2.5 x 3.3 rectangle. First, I construct a 3x4 rectangle RSTU by making repeated copies of my 1x1 rectangle. Along the 3-unit side RS, let V be the point 2.5 units from R. As in the last paragraph, construct a rectangle RVWU. This one's 2.5 x 4. Now, along the side RU, let X be the point 3.3 units from R, and construct the 2.5 x 3.3 rectangle RVYX.
What all this amounts to is this: if there's one rectangle, of whatever width and length, then there are rectangles of all possible widths and lengths. This turns out to be critically important; but that's the topic of the next post.
The Saccheri-Legendre Theorem told us that the sum of the angles of a triangle can't be more than 180 degrees. Now, think of a quadrilateral ABCD. If we draw the line AC, we divide the quadrilateral into two triangles, ABC and ACD (draw the picture!) and the sum of the angles of those two triangles is the sum of the angles of ABCD. (BCD = BCA+ACD, and DAB = DAC+CAB, so ABC + BCD + CDA + DAB = ABC + (BCA + ACD) + CDA + (DAC + CAB) = (ABC + BCA + CAB) + (ACD + CDA + DAC).) This gives us two results. First, the sum of the angles of ABCD is at most 360 degrees; second, if the sum of the angles of ABCD is exactly 360, then the sum of the angles of ABC is exactly 180, and likewise for the sum of the angles of ACD - if you have two numbers, neither one greater than 180, and their sum is 360, they must both be 180.
Now let's talk about rectangles. Suppose that ABCD is a quadrilateral, and all four of its angles are right. (Note that the angle-sum is then 360 degrees.) Then ABC and ACD are right triangles, and each has angle-sum 180. Since the angle ABC is 90 degrees, the sum of the angles ACB and CBA is also 90, and similarly for the sum of the angles ACD and CAD. But angles ACB and ACD together make up the right angle BCD, so they sum to 90 degrees also; therefore angle CBA and angle ACD are equal. Likewise angles ACB and CAD are equal. Since the triangles ABC and CDA share the side AC, and the angles flanking it are congruent, the angle-side-angle theorem tells us that the triangles are congruent. In particular, the sides AB and CD have the same length, and so do the sides BC and DA. (This is certainly what you'd expect in Euclidean geometry, but remember, we're doing neutral geometry at the moment!)
Next, take the rectangle ABCD and reflect it in the side BC, producing a new, congruent rectangle BCGH. Since ABC and CBH are right angles, A, B, and H lie on a straight line, and likewise for C, D, and G. This means that AGHD is another rectangle, just as tall as ABCD and twice as wide. Repeating this process, I can create rectangles as large as I want - think of a checkerboard, but with rectangles instead of squares. So, for instance, if I start with a 1x1 rectangle, I can make a 1x2, a 1x3, ..., and then a 2x1, 2x2, 2x3, and so on - I can make an mxn rectangle, for any positive whole numbers m and n. But is that all? Can I make one with fractional sides? Yes! I need the following result.
Let E be a point on the side AB of the rectangle ABCD, in between A and B. Let l be the line through E perpendicular to AB. Now, l enters the triangle ABC through side AB, so it must exit through one of the other sides. It can't exit through BC, because then AB, BC, and l would form a triangle with two right angles, and that violates Saccheri-Legendre. So l exits ABC through AC. But then it enters ACD through AC, and must exit through another side. It can't exit through AD - Saccheri-Legendre again - and so it exits through CD, say at F. Now the quadrilateral AEFD has three right angles (at A, E, and D), so the angle EFD can't be more than 90 degrees. But the quadrilateral on the other side, EBCF, also has three right angles, so EFC can't be more than 90 degrees. But EFD and EFC add up to 180 degrees, so they must both be exactly 90 degrees. In other words, AEFD is a rectangle. (So is EBCF.)
So what does this say? Suppose I want to construct a 2.5 x 3.3 rectangle. First, I construct a 3x4 rectangle RSTU by making repeated copies of my 1x1 rectangle. Along the 3-unit side RS, let V be the point 2.5 units from R. As in the last paragraph, construct a rectangle RVWU. This one's 2.5 x 4. Now, along the side RU, let X be the point 3.3 units from R, and construct the 2.5 x 3.3 rectangle RVYX.
What all this amounts to is this: if there's one rectangle, of whatever width and length, then there are rectangles of all possible widths and lengths. This turns out to be critically important; but that's the topic of the next post.
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no subject
Date: 2005-10-04 05:52 pm (UTC)In particular, Suppose that ABCD is a quadrilateral, and all four of its angles are right. How do we know we can do that without the parallel postulate? We can take a straight line, construct a bisector at a point and do that again at two other points to get a quadrilateral for which three angles are right. How do we know the fourth? Or what other method do we have for constructing a rectangle? I agree, if we have one, the rest follows. But from what you've shown so far, I don't see that we know that we can get one. But see first sentence.
But your last sentence seems to reinforce my question, so perhaps I am premature.
no subject
Date: 2005-10-04 06:01 pm (UTC)Exactly the right question. Stay tuned.
OT: symbol fonts?
Date: 2005-10-06 02:52 am (UTC)But many of the symbols show up correctly.
Is there some particular font one has to install to get the rest of the story?
Thanks / Richard
Re: OT: symbol fonts?
Date: 2005-10-06 09:43 am (UTC)Re: OT: symbol fonts?
Date: 2005-10-06 10:57 am (UTC)