Today I spent all my study time on exploration and proofs, as opposed review exercises. My first goal was to prove Theorem 2 from yesterday:
Two lines with slopes $m_1$ and $m_2$ are perpendicular if and only if $m_1m_2=-1$; that is, their slopes are negative reciprocals:$$m_2=-\frac{1}{m_1}$$
I succeeded in that, and also came up with an argument for the converse of the Pythagorean Theorem, which is needed for the proof.
Argument for the converse of the Pythagorean Theorem
Proposition: Given a triangle with side lengths $a$, $b$, and $c$, if $a^2+b^2=c^2$, then the triangle is a right triangle.
Let $\triangle ABC$ be a triangle such that $CA=\sqrt{AB^2+BC^2}$.
Construct a right triangle $\triangle EFG$ such that $EF=AB$ and $FG=BC$.
By the Pythegorean Theorem, $GE=\sqrt{EF^2+FG^2}$. Furthermore, $\sqrt{EF^2+FG^2}=\sqrt{AB^2+BC^2}=AC$.
Thus, $EF=AB$, $FG=BC$, and $GE=AC$, meaning that by the side-side-side property, $\triangle EFG$ is congruent to $\triangle ABC$.
The angles of congruent triangles are equal. Therefore, since $\triangle EFG$ is a right triangle, $\triangle ABC$ is also a right triangle.
The use of the SSS property here arose from a hint I found on the Internet. I would like to be able to prove that property, but have had no luck so far.1 It isn’t obvious to me why it should be true that having all sides equal guarantees congruence for triangles when it does not for other polygons.
An abbreviated proof of Theorem 2
Proposition: Two lines with slopes $m$ and $n$ are perpendicular if and only if $mn=-1$; that is, their slopes are negative reciprocals:$$n=-\frac{1}{m}$$
Let $y=mx+b$ and $y=nx+c$ be two lines such that $m\neq n$.
By Theorem 1 (from yesterday), since $m\neq n$, the two lines are not parallel. Thus, they intersect.
Let $E$ be the point at which the lines intersect.
We know from the proof of Theorem 1 that the coordinates of $E$ are $\left(\frac{c-b}{m-n},\frac{mc-nb}{m-n}\right)$.
Let $F\neq E$ be point on $y=mx+b$ with coordinates $(x_F, y_F)$ and let $G\neq E$ be point on $y=nx+c$ with coordinates $(x_G, y_G)$.
Furthermore, let $x_F=x_G=\frac{c-b}{m-n}-1$.
By plugging $\frac{c-b}{m-n}-1$ into both equations, we can find the values of $y_F$ and $y_G$ in terms of $m$, $n$, $b$, and $c$.
Using the formula below, we can then find the distance between each pair of points:
$$P_1P_2=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$$ where $P_1=(x_1,y_1)$ and $P_2=(x_2,y_2)$.
The algebra yields the following: $FG=|m-n|$, $EF=\sqrt{1+m^2}$, and $EG=\sqrt{1+n^2}$.
Now, assume that the lines $y=mx+b$ and $y=nx+c$ are perpendicular. This means that they meet at a right angle. Thus $\triangle EFG$ is a right triangle, and by the Pythagorean Theorem
$$\left(\sqrt{1+m^2}\right)^2+\left(\sqrt{1+n^2}\right)^2=\left(|m-n|\right)^2$$$$m^2+n^2+2=m^2-2mn+n^2$$$$2=-2mn$$$$mn=-1$$
Therefore, if the two lines, $y=mx+b$ and $y=nx+c$, are perpendicular, then $mn=-1$.
Instead, assume that $mn=-1$. In that case, $$\left(\sqrt{1+m^2}\right)^2+\left(\sqrt{1+n^2}\right)^2=m^2+n^2-2mn=(m-n)^2\text{.}$$
It follows, by the converse of the Pythagorean Theorem, that $\triangle EFG$ is a right triangle. Since the right angle of a right triangle is opposite the hypotenuse, $\angle FEG$ is a right angle. Hence the lines $y=mx+b$ and $y=nx+c$ are perpendicular.
Therefore, if the slopes of two lines, $y=mx+b$ and $y=nx+c$, have a product of $-1$, then the lines are perpendicular.
I had some other ideas today, as well, including regarding the question about squares from my post Squares Again. Expect those in the next few days.
- To do. ↩︎
Doesn’t it have to do with the degrees of freedom for the angles? All triangles with the same length sides have the same angles (I think you could prove that with some geometry and Pythagoras’ theorem), while in higher order polygons have more degrees of freedom for their angles, so defining the sides alone does not define the polygon.
I feel like saying that the angles of higher order polygons have more degrees of freedom than the angles of triangles is what I want to prove, so it can’t be a means of proving it. Maybe I don’t understand what you mean by degrees of freedom, though.
(1) Very polished proofs. I hope they’re making you happy.
(2) What’s different about triangles? This is really the converse of your question, so I hope it isn’t a spoiler. You’re right that triangles aren’t like other polygons. One way to rephrase SSS is that triangles are rigid. Imagine you make a polygon where the sides are steel beams riveted together at the vertices. If the polygon isn’t a triangle, then you can change its shape without bending the beams: you pivot the beams at those rivets. It’s easy to distort. SSS says you can’t distort a triangle by pivoting it at the rivets. The only way to distort it is either to change the length of the beams (or bend them) or to tear the rivets. Triangles are therefore rigid in a way other polygons aren’t. That’s why engineers build bridges using triangles, not other polygons. Rather than have beams making a quadrilateral, they’ll add a beam across the quadrilateral to make two triangles. The pair of triangles are rigid. The quadrilateral isn’t. I had never noticed that about bridges (and other engineering structures) until I was learning geometry and my dad, a mechanical engineer, pointed it out to me. Initially it confused me, but then the penny dropped, and I started seeing triangles everywhere.
That doesn’t address your question. You sort of ask WHY triangles are rigid and other shapes are not. I say that IF triangles were rigid and other shapes were not, then triangles would be really useful in building things. I think that’s cool, and that it doesn’t take away your right to think more about SSS yourself.
(3) Still not a spoiler: If you’re interested in SSS historically, then Book I, Prop. 4 of Euclid’s Elements is SAS. He states it as a theorem, but he uses a proof technique he never uses again in the Elements. From a modern perspective, SAS is usually an axiom, not a theorem. Book I, Prop. 7 is pretty much SSS. So Euclid gets from SAS to SSS pretty quickly. There’s no reason for you to do things the way Euclid does. For instance, you might want to use algebra and Pythagoras and coordinate systems to prove SSS. I have no idea how and whether this would work. The questions in your blog are yours to think about, not mine. If you do want to follow Euclid, though, you could ask yourself how you might get from SAS and very little else to SSS. For a more useful hint, you could look at Props 4-7 in Book I and just see what they say and try to figure out what the proofs might be. Or you could just look at the pictures and try to figure out the theorems and their proofs.
Or none of these things. It’s your study / review / research time.
(4) As always, encourage me to get excited on my own but shut up if that’s the best way to be a fan. I very much am a fan, and I’m so thrilled you’re having fun, and having it so elegantly.
– Tim
(1) The proofs are indeed making me happy, although I may have overdone it a bit today. I am very tired.
(2) Heh. I originally wrote something more wordy saying that I do have practical experience of the fact that triangles are rigid and mentioning that that is why they are used in building. So I am aware that triangles as physical objects have that property, and it had occurred to me that it is related to the mathematical SSS property. I’m still not sure how to understand why SSS is true for triangles as mathematical objects, however.
(3) I will consider how SAS and SSS relate, and possibly I will look at Euclid. I’ve tried to read Euclid in the past, and the problem I always have is that I want to prove every proposition myself, which inevitably leads to my getting stuck.
(4) I value your comments very much.
Tim, I’m so happy that you and Olly have this way of interacting; I know you mean a lot to her. Thank you