Today I watched more of 3Blue1Brown’s Lockdown Math series. (See the new Resources tab for a link.) It turns out that the subseries I was following has four installments, not three, so I may be spending another day on it. I’ve enjoyed the primer on complex numbers 3Blue1Brown provides. Although they cropped up occasionally in my mathematical education, I never learned about them in any depth. I had certainly never understood their connection to trigonometry before.
Now it is time for the proofs of the two remaining cases of the proposition that is the focus of the Parabola-a-Go-Go series. (You will find links to the rest of the series under the new Highlights tab.)
Proposition: Consider the parabola $x^2=4py$, where $0<p$. Let $a$ be a real number. Let $A$ be the vertical line $x=a$. Let $B$ be the line through $F=(0, p)$, the focus of the parabola, and $R=(a,\frac{a^2}{4p})$, the point where $A$ intersects the parabola. Let $C$ be the line tangent to the parabola at $R$. Then, for all $a$, the acute or right angle formed by $A$ and $C$ is equal to the acute or right angle formed by $B$ and $C$.
As noted in Parabola-a-Go-Go, Part 2, the equation of the parabola can be written as $y=\frac{1}{4p}x^2$ and, using calculus, $y’=(\frac{1}{4p})2x=\frac{1}{2p}x$.
Thus, the slope of $C$, the line tangent to the parabola at $R=(a,\frac{a^2}{4p})$, is $\frac{a}{2p}$.
Furthermore, the slope of $B$, the line through $F=(0, p)$ and $R=(a,\frac{a^2}{4p})$ is $$\frac{\frac{a^2}{4p}-p}{a-0}=\frac{a}{4p}-\frac{p}{a}=\frac{a^2-4p^2}{4ap}\text{.}$$
Now suppose that $a>2p$.
In that case, the slopes of $B$ and $C$ are both positive. Note that the slope of $C$ is greater than the slope of $B$, since $\frac{a}{2p}=\frac{2a^2}{4ap}$ and $2a^2>a^2-4p^2$. It follows that the lines intersect one another, line $A$, and the x-axis as shown in the diagram below.
In this diagram, $\alpha$ is the angle formed by $A$ and $C$ and $\beta$ is the angle formed by $B$ and $C$.
Consider the right triangle formed by $A$, $C$, and the x-axis. Note that one of its vertecies is vertical to $\alpha$. Note also that the ratio of its vertical leg to its horizontal leg is equal to the slope of $C$. It follows that the ratio of its horizontal leg to its vertical leg is the reciprocal of the slope of $C$. Yet that ratio is also the tangent of the angle that is vertical to $\alpha$.
Therefore, $$\alpha = \arctan\frac{2p}{a}\text{.}$$
Consider instead the right triangle formed by $A$, $B$, and the x-axis. By similar reasoning we see that the tangent of $\beta+\alpha$ is the reciprocal of the slope of $B$.
Thus, $$\beta+\alpha=\arctan\frac{4ap}{a^2-4p^2}$$$$\beta=\arctan\frac{4ap}{a^2-4p^2}-\alpha=\arctan\frac{4ap}{a^2-4p^2}-\arctan\frac{2p}{a}\text{.}$$
Since $a>2p$, $(\frac{2p}{a})^2<1$. Therefore, by the lemma from Parabola-a-Go-Go, Part 3, $$2\arctan\frac{2p}{a}=\arctan\frac{2(\frac{2p}{a})}{1-(\frac{2p}{a})^2}=\arctan\frac{\frac{4p}{a}}{1-\frac{4p^2}{a^2}}=\arctan\frac{\frac{4ap}{a^2}}{\frac{a^2-4p^2}{a^2}}=\arctan\frac{4ap}{a^2-4p^2}\text{.}$$
It follows that $\arctan\frac{2p}{a}=\arctan\frac{4ap}{a^2-4p^2}-\arctan\frac{2p}{a}$ and thence that $\alpha=\beta$.
Thus the proposition holds when $a>2p$.
Suppose instead that $0<a<2p$.
In that case, the slope of $B$ is negative and the slope of $C$ is positive. It follows that the lines intersect one another, line $A$, and the x-axis as shown in the diagram below.
In this diagram, $\alpha$ is the angle formed by $A$ and $C$ and $\beta$ is the angle formed by $B$ and $C$. An additional angle is labeled $\gamma$ and has the property that $\alpha+\beta+\gamma=\pi$.
Considering the two right triangles formed by the lines $A$, $B$, and $C$ with the x-axis, we see that the tangent of $\alpha$ is the reciprocal of the slope of $C$ and the tangent of $\gamma$ is the absolute value of the reciprocal of the slope of $B$.
Thus, $\alpha = \arctan\frac{2p}{a}$ and $\gamma=\arctan\left|\frac{4ap}{a^2-4p^2}\right|$.
Since the $0<a<2p$, $\frac{4ap}{a^2-4p^2}$ is negative. Thus $\left|\frac{4ap}{a^2-4p^2}\right|=\frac{-4ap}{a^2-4p^2}$.
It is a property of arctangent that $\arctan (-x)=-\arctan x$. (I forgot to write up a proof of this, but the proof is simple. “Left to the reader as an exercise.”) Thus $$\gamma=\arctan\left|\frac{4ap}{a^2-4p^2}\right|=\arctan\frac{-4ap}{a^2-4p^2}=-\arctan\frac{4ap}{a^2-4p^2}\text{.}$$
It follows that $$\beta =\pi – \arctan\frac{2p}{a} -\left(-\arctan\frac{4ap}{a^2-4p^2}\right)=\pi – \arctan\frac{2p}{a} +\arctan\frac{4ap}{a^2-4p^2}\text{.}$$
Since $0<a<2p$, $(\frac{2p}{a})^2>1$. Thus, by the lemma, $$2\arctan\frac{2p}{a}=\pi+\arctan\frac{2(\frac{2p}{a})}{1-(\frac{2p}{a})^2}=\pi+\arctan\frac{4ap}{a^2-4p^2}\text{.}$$
It follows that $\arctan\frac{2p}{a}=\pi-\arctan\frac{2p}{a}+\arctan\frac{4ap}{a^2-4p^2}$ and thence that $\alpha=\beta$.
Therefore the proposition holds for $0<a<2p$ and holds in general.
This gnarly argument was fun to work out, and I got to practice all the things I’ve been reviewing: algebra, analytic geometry, conics, and trig, plus some basic geometry. That said, if you want to see a really good proof of this theorem, check out this page. I looked this up sometime after I finished my own proof, and I’ll admit I was a little discouraged to find there was a way to do the problem that was so much better. (And this definitely is. It uses the definition of the object involved directly, requires less advanced concepts, and doesn’t rely on cases.) Not everyone can be optimally clever all the time, though—one is lucky if one can be optimally clever any of the time—and, as I said, doing it my way was fun and educational. I’m not sure whether either approach can be adapted to ellipses. That is a project for the future.
For those wondering, the diagrams for the Parabola-a-Go-Go series are based on images from the graphing program Desmos. I added additional details in GIMP and, in some cases, annotated the results by hand using a stylus and the photo app of my tablet. Most of my diagrams (the ones with the cream quad-ruled backgrounds) are made on my tablet using the stylus and a freehand note-taking app.
I know this series of posts was not very accessible. I want to explain more of what I do in a way non-specialists can understand; that’s a task that has always appealed to me. After doing my study each day, though, I only have so much energy and time left for the blog, and it tends to be more efficient to write at a higher level. Perhaps some of the competition problems I hope to try will have fairly short solutions that lend themselves to more complete explanations.
It’s beautiful to look at, even if I don’t understand
What Kim said, although I do understand. Olly, I think you have a lovely attitude toward producing gnarly arguments. We learn so much by making facts our own through proving them our own way. I always feel a sense of pleasure and ownership if I can prove something myself, even if there may be slicker arguments that I did not find. And I think it’s worth doing exactly what you did, looking at other arguments once you’ve made something your own. Sometimes one looks at those arguments and finds something that one can add to one’s toolbox for future proofs. Sometimes one looks at those arguments and learns a really powerful new technique or connection. Sometimes one looks at those arguments and says, I would never think of doing it this way, but I know a way to do it that delighted me and found the result. And sometimes the proof that one found oneself is just better in some way or another then the proof one finds in the book. I don’t know if you remember me telling you about the proof that Alberto Calderon found in Antoni Zygmund’s book. I won’t write it down now, but I can tell it to you again sometime when we’re Zooming.
And I hope you appreciate and celebrate all the joy that you’re 51 days have brought your friends!