Tag: to-do

The Proof at Last

Apr 25, 2024
More low-level depression and concentration problems today, but I did do a bit of watching and reading. I also finished writing up the proof below, which I know you’ve all be waiting for. As I explained yesterday, I’ve tried to make it a bit more readable than other proofs I’ve posted recently.

Proposition: If $\triangle ABC$ is an equilateral triangle inscribed in a circle and $D$ is a point on the circle between $A$ and $C$, then $AD+CD=BD$.

Place a point $E$ on $\overline{BD}$ so that the distance from $D$ to $E$ is the same as the distance from $D$ to $C$, as shown below.

Draw the segment $\overline {CE}$.

Since the original triangle, $\triangle ABC$, is equilateral, its three angles must be equal by the proposition from the post Easy Lemmas.

The sum of the measures of the three angles of a triangle is $\pi$ radians. Thus, when the three angles are equal, each one must measure $\frac{\pi}{3}$. It follows that the angles of $\triangle ABC$ have this measure: $\angle BAC=\angle ABC=\angle ACB=\frac{\pi}{3}$.

Because they are both subtended by arc $BC$, by the second proposition from the post Angles and Arcs, $\angle BAC$, the leftmost of the angles of the original equilateral triangle, is equal to $\angle BDC$, the rightmost of the angles with their verticies at $D$.

Similarly, because they are both subtended by arc $AB$, $\angle ACB$, the rightmost of the angles of the original equilateral triangle, is equal to $\angle ADB$, the leftmost of the angles with their verticies at $D$.

It follows that the two angles with their verticies at $D$, $\angle BDC$ and $\angle ADB$, both measure $\frac{\pi}{3}$.

Consider the newly formed triangle $\triangle CDE$. Two of its sides, $\overline{CD}$ and $\overline{DE}$, are equal by construction (i.e. because of the way point $E$ was chosen). Thus, the two angles subtended by those sides, $\angle CED$ and $\angle DCE$, are equal to one another by the lemma from the post Revenge of the Squares, Part 1.

Furthermore, since $\angle CDE$, the other angle of $\triangle CDE$, is one of the angles that measures $\frac{\pi}{3}$, it follows that $\angle CED$ and $\angle DCE$ together must measure $\pi-\frac{\pi}{3}=\frac{2\pi}{3}$ and that each one individually must measure $\frac{\pi}{3}$.

Hence the three angles of $\triangle CDE$ are all equal, from which it follows, by the proposition in the post Easy Lemmas, that $\triangle CDE$ is equilateral.

Now consider $\angle BEC$, the angle supplementary to $\angle CED$, the uppermost of the angles of $\triangle CDE$. Since the angles of $\triangle CDE$ all measure $\frac{\pi}{3}$, the supplementary angle $\angle BEC$, must measure $\frac{2\pi}{3}$.

Note that, since each one measures $\frac{\pi}{3}$, the sum of $\angle ADB$ and $\angle BDC$, the two equal angles with their verticies at $D$, is also $\frac{2\pi}{3}$. Thus, the measure of $\angle ADC$, the large angle they form, is also $\frac{2\pi}{3}$.

Yet that means that the two triangles highlighted above in yellow, $\triangle BCE$ and $\triangle ACD$, are congruent by the side-side-angle property, which applies when the angle in question is greater than or equal to $\frac{\pi}{2}$. The triangles’ large angles, $\angle BEC$ and $\angle ACD$, are equal, as just shown, and measure $\frac{2\pi}{3}$; their long sides, $\overline {BC}$ and $\overline {AC}$, are equal, since both are sides of the original equilateral triangle $\triangle ABC$; and another pair of their sides, $\overline {CE}$ and $\overline {CD}$, are equal because both are sides of the new, smaller equilateral triangle $\triangle CDE$.

It follows that the remaining sides, $\overline {BE}$ and $\overline{AD}$, are equal, as well. Hence $BE=AD$.

As above, $DE=CD$ by construction.

Thus $AD+CD=BE+DE$. Yet $\overline {BE}$ and $\overline {DE}$ together make up the larger segment $\overline {BD}$. Therefore $AD+CD=BD$ and the theorem of is proven.

With the addition of SSA for right and obtuse angles, I think I might have a complete set of triangle congruence properties to prove.¹

I could almost have done this proof without assigning measures to the angles. Equality between $\angle BEC$ and $\angle ADC$ could be established using the property of triangles illustrated below, and I think I could have shown that they are obtuse as well. Without angle measures, though, I couldn’t think of a way to show that, with two sides of $\triangle CDE$ equal by construction, not only must the angles they subtended be equal to one another, but both must be equal to the third angle.

Let me know what you thought of this verbose-mode proof. I found it a little difficult to write without an audience clearly in mind.
1. To do. ↩︎
Math Posts

geometry, proof, to-do
Angles and Arcs

Apr 23, 2024
My back is somewhat better today. I spent an hour on exercises in Analysis with an Introduction to Proof and I plan to read some more of Infinite Powers this evening, as well.

I’m also ready to prove the proposition about inscribed angles from my post Groundwork. It turns out that proposition falls right out of another that reader Tim McL suggested starting with when I said I knew little about the geometry of circles.

Proposition: If a central angle and an inscribed angle are subtended by the same arc of a circle, then the measure of the central angle is twice that of the inscribed angle.

Let $A$ and $B$ be points on a circle with center $D$.

Draw the central angle subtended by arc $AB$.

Draw an inscribed angle also subtended by arc $AB$ and let its vertex be called $C$.

Join $\overline {CD}$.

Now suppose that the sides of the two angles intersect only at $A$ and $B$ (as in the figure above).

Since $\overline{BD}$ and $\overline{CD}$ are both radii of the circle, $BD=CD$, from which it follows that $\angle CBD = \angle BCD$ (as I’ve proven previously).

Similarly, $\overline{AD}$ and $\overline{CD}$ are both radii of the circle, so $AD=CD$ and $\angle CAD = \angle ACD$.

Futhermore, since the sum of the angles of a triangle is $\pi$ radians, $2\angle BCD + \angle BDC = \pi$ and $2\angle ACD + \angle ADC = \pi$.

It is also true that $\angle BCD + \angle ADC + \angle ADB = 2\pi$.¹

Combining these equations will yield $\angle ADB = 2(\angle BCD + \angle ACD)$.

But $\angle ACB = \angle BCD + \angle ACD$, so $\angle ADB = 2\angle ACB$, and the theorem is proven for this case.

Suppose instead that sides of the angles intersect other than at $A$ and $B$, as shown above. (Note that this argument does apply in the edge case where $\overline{BC}$ intersects $\overline{AD}$ at $D$.)

Then, $\angle ACB =\angle ACD-\angle BCD$ and $\angle ADB = \angle BDC – \angle ADC$.

As above, $2\angle BCD + \angle BDC = \pi$ and $2\angle ACD + \angle ADC = \pi$.

Combining these equations will yield $\angle ADB = 2(\angle ACD – \angle BCD)$.

Hence $\angle ADB = 2\angle ACB$, and the theorem is proven for this case, as well.

After proving this proposition, I looked up how Euclid did it. His solution was largely the same. It was slightly more elegant, using the property of triangles illustrated below to cut out some of the algebra, but not so elegant as to dispense with the cases.

Proposition: If two inscribed angles are subtended by the same arc of a circle, then they are equal.

I’m just going to sketch a proof of this. By the proposition above, the blue and green inscribed angles subtended by arc $AB$ both have half the measure of the red central angle subtended by the same arc. Therefore, they must be equal to each other.

Footnotes
1. Do I need to prove this? I have an idea of how I would do it, but it seem so fundamental. ↩︎
Math Posts

geometry, proof, to-do
Easy Lemmas

Apr 20, 2024
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.

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.

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. ↩︎
Math Posts

geometry, proof, to-do
Groundwork

Apr 17, 2024
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.

(As readers may have guessed, I’m feeling much better today.)
1. To do. ↩︎
Math Posts

geometry, to-do
Ptolemy’s Theorem and Others

Apr 15, 2024
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 $C$.

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. ↩︎
Math Posts

geometry, to-do, video
Fun with Tilings

Apr 11, 2024
Today I’ve been playing with tilings to try to understand the relationship between radial symmetry and periodicity. I’ve come up with the following to show that each can exist independently of the other.

I’m still wondering, though, if a tiling can be radially symmetrical about more than one point and be non-periodic. I suspect not.¹
1. To do. ↩︎
Math Posts

geometry, to-do
Shiny New Week

Apr 1, 2024
Today I spent two hours reading a section of Analysis with an Introduction to Proof and doing the associated exercises. I’m feeling better than I was for most of last week, though the mental weather is still not entirely clear.

Part of the section I read explained the inadequacy of examples, even quite a few examples, in proving general statements. As an illustration, it offered the statement that $n^2+n+17$ is prime for every natural number $n$. This statement is true for values of $n$ up to 15, but is not true for 16 or 17. The latter, in particular, could be predicted without trying each value of $n$, since $17^2$, $17$, and $17$ are clearly all divisible by a common factor (meaning that their sum must also be divisible by that factor).

Later, an exercise asked me to find a counterexample to the statement that $3^n+2$ is prime for every natural number $n$. This had me wondering whether there was a way to predict a counterexample, as there had been in the earlier case.¹ I could not think of one, however, so I answered the question by trying each $n$. It turns out that I didn’t have to go far, as $n=5$ breaks the pattern. I still wonder if there is a better way, though.

Thank you to all of you who have commented on my posts recently. It really helps me stay committed to my project. I will try to reply soon.
1. To do. ↩︎
Math Posts

number theory, to-do
Not Doing Much

Mar 24, 2024
My mental health was better today, but I haven’t yet bounced back physically. I took a long nap after church, and I was only up for some light math reading after. The section of The Mathematical Tourist that I read was about tests for primality, which is appropriate given my current interest in primes. It made me wonder if SageMath tests numbers directly or looks them up in a table. There might be a forum where I could ask.¹
1. To do. ↩︎
Math Posts

number theory, to-do
Functioning and Functions

Mar 16, 2024
Today I read the first section of the first chapter of my calculus book and worked on the associated exercises. (Yippee!) My concentration, which was such a problem on Thursday, was fine today, and I’m feeling encouraged.

The section I read was about functions and discussed the vertical line test. According to the test, a graph represents a function of $x$ if and only if no vertical line intersects the graph more than once. The way I remember first being taught the test, though, the “of $x$” condition was not included. It always bothered me that $y=x^2$, an upward facing parabola, should be a function, while $x=y^2$, a rightward facing parabola, should not. It is, of course. It’s just a function of $y$ rather than a function of $x$. This has me wondering whether, say, the graph of a diagonally facing parabola could be interpreted as representing a function, and what it would be a function of.¹
1. To do. ↩︎
Math Posts

algebra (basic), to-do
Repairs are Ongoing

Mar 6, 2024
The depressive episode I started yesterday persists, but has been much less severe today. This afternoon I finished the last few chapters of The Joy of X without too much struggle. Now I will have to find a new book for my light math reading. I have several options in my library already, but I would welcome recommendations, as well.

The final chapter of The Joy of X was about infinity and included a treatment of Cantor’s diagonal argument that the set of real numbers is uncountable. (I’m sorry, non-specialist readers, but I’m not up to explaining what that is right now, though it is fairly accessible.) The Joy of X did not account in any way for the fact that some numbers have more than one decimal representation (e.g. that $0.1\overline{0}=0.0\overline{9}$). I’m sure a more formal presentation of the argument would have. I intend to think for a while about how I would do it before looking up the accepted method. Not today, though.¹
1. To do. ↩︎
Math Posts

number theory, to-do