Month: April 2024

Easy Lemmas

Olly / / Math Posts

Today I worked on exercises in Analysis with an Introduction to Proof and did a little mathematical puttering.

I’m also ready today to start sharing my work on the proposition about the inscribed equilateral triangle. I’ll start by proving that a triangle is equilateral if and only if its three angles are equal.

First, recall from my post Revenge of the Squares, Part 1 that the angles opposite (i.e. subtended by) the equal sides of an isosceles triangle are also equal. The proof of the following related lemma is very similar.

Lemma: If two angles of a triangle are equal, then the sides that subtend them are also equal.

Triangle Before
Triangle After 1

Consider the triangle $\triangle ABC$ with $\angle ACB\cong\angle ABC$.

Draw a line bisecting $\angle BAC$ and extend it to intersect $\overline{BC}$. Let the point of intersection be called $D$.

Now, by the angle-angle-side property, $\triangle ACD\cong\triangle ABD$, since $\angle ACB\cong\angle ABC$, $\angle CAD\cong\angle BAD$, and $\overline{AD}$ is shared.

Therefore, $AC=AB$. $\overline {AC}$ and $\overline {AB}$ are the sides that subtend the equal angles $\angle ACB$ and $\angle ABC$, so the lemma is proven.

(Note that I have added AAS to my list of triangle congruence properties to prove.1)

Proposition: A triangle is equilateral if and only if its three angles are equal.

Equilateral Triangle

Consider a triangle $\triangle ABC$.

Assume that $\triangle ABC$ is equilateral.

Then, since $AB=AC$, $\angle ACB\cong\angle ABC$, by the theorem cited above. Similarly, since $AC=BC$, $\angle ABC\cong\angle BAC$.

It follows that $\angle ACB\cong\angle ABC\cong\angle BAC$.

Thus, if $\triangle ABC$ is equilateral, then its three angles are equal.

Assume instead that the three angles of $\triangle ABC$ are equal.

Then, by the lemma, since $\angle ACB\cong\angle ABC$, $AB=AC$. Similarly, since $\angle ABC\cong\angle BAC$, $AC=BC$.

It follows that $AB=AC=BC$.

Thus, if the three angles of $\triangle ABC$ are equal, then it is equilateral.

This proves the equivalence.

Stay tuned tomorrow for some circle geometry.

  1. To do. ↩︎

Easy Living

Olly / / Math Posts

My mental state remained good today, but I also slept a lot. I did some housekeeping on the blog to make it easier for me to return to questions I’ve raised but not answered. My study time was devoted to work on the proposition about inscribed angles subtended by the same arc and to considering whether and how my other recent proofs depend on the parallel postulate. The proof of the inscribed triangle proposition does depend on it, I believe. It is not necessary to the proof that the sum of the angles of a triangle is $\pi$ radians, but it is necessary that it is the same for all triangles, which is not the case in some systems of geometry.

Before and After

Olly / / Math Posts

Today was devoted to fun with geometry (as was some time late last night). I came up with a simpler proof of the proposition about the inscribed equilateral triangle, which I’m pleased about; the first one was a bit gnarly. I also worked on the lemmas to accompany it. It needs most of the same ones, but not the one about parallelograms.

Here are before and after shots showing the working figure for my first proof and that for my second. One proof required four new lines and two new named points. The other involved just one one of each.

Before 1
After 1

(I’m looking forward to finishing this notebook so that I can move to an unruled one.)

Groundwork

Olly / / Math Posts

I spent quite a bit of time today working on the inscribed equilateral triangle problem. I experimented both by hand and using the Desmos geometry tool, which allows you to vary parameters in your drawings. At this point, I’ve come up with a proof that convinces me. There is one step that needs to be formalized and a couple that depend on facts that I know to be true but would like to prove. These include that the sides of a parallelogram are pairwise equal1 and that a triangle is equilateral if and only if all of its angles are equal. The argument also depends heavily on a fact I just learned today, conjecturing it based on experimentation and then confirming it with a Google search: inscribed angles subtended by the same arc of a circle are equal (as shown below). I’m not sure whether I will be able to prove that or not. My education in geometry focused almost exclusively on lines and triangles, and I know little of circle geometry.

Two Subtended Angles

(As readers may have guessed, I’m feeling much better today.)

  1. To do. ↩︎

Triangle, Triangle

Olly / / Math Posts

Today I worked for as long as I could on the proposition from yesterday about the inscribed equilateral triangle. I’m feeling a little better, but my mind is still sluggish, so I didn’t make much progress.

I’m wondering whether proving the proposition for a special case first would be helpful. The edge cases where $D$ is the same point as $A$ or $C$ are trivial. The case where $AD = CD$ might provide some insight. On the other hand, using any of the unique characteristics of that case to prove it could leave me as far from a general proof as before.

Inscribed Equilateral Triangle

Ptolemy’s Theorem and Others

Olly / / Math Posts

Things continued difficult today, but I felt a bit better after dinner. I watched four videos from the Numberphile channel, notably one about proving a geometric relationship called Ptolemy’s Theorem by a technique called inversion of the plane. It was a great video, but quite involved. I would not recommend it to those without a strong interest in math.

When I am feeling better, I would like to have a go at proving Ptolemy’s Theorem by another method.1 The lecturer in the video said that it can be done using plane geometry, provided you are clever, and can also be done by crunching trig ratios, something I have some experience in. I’d also like to prove, by some other method, a fact that can be proven using Ptolemy’s Theorem: that in the figure shown below, the distance from $D$ to $B$ is the sum of the distances from $D$ to $A$ and $C$.

Inscribed Equilateral Triangle

A cool thing about Ptolemy’s Theorem is that the Pythagorean Theorem falls right out of it. In fact, you could say that the Pythagorean Theorem is a special case of Ptolemy’s. This means that Ptolemy’s Theorem must only be true if you assume the Parallel Postulate. I wonder if that is the case for the proposition above.

  1. To do. ↩︎

Later

Olly / / Updates

Today I watched part of an interesting video by 3blue1brown about the surface area of a sphere. I had a lot of trouble concentrating on it, though. Near the end it began to require more participation from the viewer, and I decided to save the rest for later. There’s some hope underlying that, I guess.