Tag: geometry

Parabolic Reflections

Olly / / Math Posts

Today I spent all of my study time investigating the parabola problem from last week. There are still details to be worked out, but I’m fairly sure the approach I’m using will allow me to prove that the angles I conjectured might be equal actually are. I thought today I’d give a better explanation of what my conjecture was.

A well known principle in physics, called the Law of Reflection, is that when light reflects from a flat surface, the angle of incidence (that is, the angle at which the light hits the surface) is equal to the angle of reflection (that is, the angle at which the reflected light leaves the surface). This is shown in the diagram below.

Angles Of Incidence And Reflection 1

A well known property of parabolas is that, if light is emitted from the focus of the parabola ($F$ in the diagram below), then it will reflect from the surface of the parabola in a direction parallel to the parabola’s orientation (vertical, in the diagram). This is why parabolic mirrors are used in car headlights to reflect all light outward from a bulb at the focus.1

Reflecting Parabola

My question is basically how this property of parabolas relates to the Law of Reflection. It is not obvious how one would measure the angle at which light hits or reflects from a curved surface. What would that mean? One possibility is to measure the angle between the path of the light and the line tangent to the curve at the point of reflection. (In the diagram, the point of reflection is called $P$ and the blue line is tangent to the parabola at $P$.2)

My conjecture is that the a generalization of the Law of Reflection holds true for light emitted from the focus of the parabola and reflected at $P$, if one interprets the angles of incidence and reflection as the angles formed with the line tangent to the parabola at $P$. That is, I conjectured that the angle between the blue tangent line in the diagram and the vertical line through point $P$ is the same as the angle between the blue tangent line and the segment $\overline{PF}$.

  1. Moving in the other direction, if light hits the surface of the parabola from a direction parallel to the parabola’s orientation, it will reflect along a line that passes through the focus of the parabola. This is why parabolic mirrors are used in power generation to concentrate all incoming light on a collector a the focus. ↩︎
  2. The line tangent to a parabola at a point intersects the parabola at that point and no other, and its slope is sometimes described as the slope of the parabola at that point. ↩︎

Unfocus Meets Focus

Olly / / Math Posts

Today was rocky, though I am feeling a little better now. My math activity for the day was more reading in The Joy of X. The section I read included discussion of an imagined elliptical pool table on which a ball shot from one focus will always fall into a hole at the other focus. Since I know that pool balls (at least ideal ones) bounce off the cushion at the same angle at which they hit it, that had me wondering whether line segments like those shown below in red must always form equal angles with the line tangent to the ellipse at the point where they intersect. (The tangent line is shown in blue.) This seems likely, but I’m not sure how I would go about investigating it.1

Ellipse With Tangent

Naturally, I then wondered a similar thing about parabolas, so useful for reflecting and focusing light, which behaves in the same way as ideal pool balls.

Parabola With Tangent Two

(Sorry the explanation of these questions is a bit sketchy. I know it may be difficult for those without a lot of math background to understand, when this is a topic that should be accessible to them. This is the best I can do at the moment, though.)

  1. To do. ↩︎

Extra Extra

Olly / / Math Posts

Sometimes, during my review, my mind will supply facts before they are mentioned or even in contexts where they never are mentioned. Other times, I will read something an think, “Wait, did I ever know that?”

Today I read about trigonometry, both in the appendix to my calculus text and in the OpenStax free textbook Algebra and Trigonometry. It turns out that, when an angle’s vertex is located at the center of a circle, the measure of that angle in radians is the ratio between the length of the arc subtended by the angle and the radius of the circle. That’s not just a property of radians; that’s what a radian is. Surely I must have known that at some point, but it came as total news to me.

Radians 1

(I’m feeling better today, though still not very good. The fun math revelations were helpful. All that I did remember about radians now makes a lot more sense.)

Ellipses

Olly / / Math Posts

Today I did more exercises from the review of conics. Most of them concerned analyzing and graphing ellipses. It was still second year algebra all over again, but I was a little less bored than by the exercises with parabolas yesterday. I think I may have been bucked up by this intriguing 3Blue1Brown video about ellipses that happened to come up on my YouTube homepage last night:

I also did a little extracurricular reading today about the Goldbach conjecture and the related ternary Goldbach conjecture. The latter was recently proven by a mathematician called Harald Helfgott. I have not yet absorbed even the overall method of the proof, but I did understand that the abstract proof only applies to numbers greater than a constant $C$ that is very large in human terms yet small enough that all numbers less than $C$ can be checked by a computer. I thought that was interesting.

Revenge of the Squares, Part 2

Olly / / Math Posts

Today I finished the 46 exercises from the review of analytic geometry, then rounded out my study time with more exercises from Analysis with an Introduction to Proof.

I had expected the proof from Revenge of the Squares, Part 1 to require showing that the figure must be a parallelogram. I managed it without that, but afterward I started thinking about how I would have proved that fact, had I needed it. The results are below.

Lemma: If, when two lines are intersected by another line, a pair of alternate interior angels is equal, then the first two lines are parallel.

Assume that two lines, $A$ and $B$, are not parallel.

Let them both be intersected by a third line, $C$. Since $A$ and $B$ are not parallel, they intersect, and $A$, $B$, and $C$ enclose a triangle.

Three Lines

Let $x$, $y$, and $z$ be the interior angles of the triangle, as shown in the diagram. Let the interior angle alternate to $x$ be called $k$ and the interior angle alternate to $y$ be called $j$.

Since the sum of the measures of the interior angles of a triangle is 180 degrees, $x+y+z=180$. Thus, $x=180-y-z$.

Furthermore, since $x$ and $j$ are complimentary supplementary, $x+j=180$, meaning that $x=180-j$.

Therefore, $$180-j=180-y-z$$ $$-j=-y-z$$ $$j=y+z$$

Since $z\neq 0$, it follows that $j\neq y$.

By similar reasoning, $k\neq x$.

Hence, if two lines are not parallel, then, when they are both intersected by a third line, neither pair of alternate interior angles is equal.

By contrapostition, it follows that if, when two lines are intersected by another line, a pair of alternate interior angels is equal, then the first two lines are parallel.

Proposition: A convex quadrilateral with each pair of opposite sides equal is a parallelogram (i.e. each pair of opposite sides is also parallel).

Consider a convex quadrilateral $ABCD$, where $AB=CD$ and $BC=AD$.

Parallelogram Before
Parellelogram After

Draw $\overline{BD}$.

Since $AB=CD$, $BC=AD$, and $\overline{BD}$ is shared, $\triangle ABD\cong\triangle BCD$ by the side-side-side property.

Therefore, $\angle CBD\cong\angle ADB$ and $\angle BDC\cong\angle ABD$.

It follows, by the lemma, that $\overline{AB}\parallel\overline{CD}$ and $\overline{AD}\parallel\overline{BC}$.

Thus, $ABCD$ is a parallelogram.

SSS again. Proving that remains on the to-do list.

Revenge of the Squares, Part 1

Olly / / Math Posts

I intended to split my study time today between analytic geometry review and logic review, but ended up getting drawn into some exploration as well. The result is that I have two proofs about quadrilaterals ready to post. The first of the two appears below. It’s a proof of my hunch from the post Squares Again that if a convex, four-sided, equilateral polygon has one right angle, then it has four right angles. It turns out that I could prove this without knowing that such a figure must be a parallelogram. (It must, though. That’s the proof for tomorrow.)

Lemma: The angles opposite (i.e. subtended by) the equal sides of an isosceles triangle are also equal.

Triangle Before 1
Triangle After

Consider the isosceles triangle $\triangle ABC$ with $AB=AC$.

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

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

Therefore, $\angle ACD\cong\angle ABD$. These are the angles opposite the equal sides of the original isosceles triangle, so the lemma is proven.

Proposition: If a convex, four-sided, equilateral polygon has one (interior) right angle, then it has four (interior) right angles.

Let $ABCD$ be an convex, four-sided, equilateral polygon, and let $\angle BCD$ be a right angle.

Square Before
Square After

Divide the polygon by drawing $\overline{BD}$.

Note that, since $\overline{BD}$ is shared and all other line segments are equal, $\triangle ABD\cong\triangle BCD$ by the side-side-side property.

Hence, since $\angle BCD$ is a right angle and the angles of congruent triangles are equal, $\angle BAD$ is also a right angle.

Similarly, $\angle ADB\cong\angle CBD$.

Furthermore, by the lemma, since $\triangle ABD$ is isosceles, $\angle ADB\cong\angle ABD$, and since $\triangle ACD$ is isosceles, $\angle BDC\cong\angle CBD$.

Therefore, $\angle ADB\cong\angle ABD\cong\angle BDC\cong\angle CBD$.

Let the measure of those angles be called $x$. Since the sum of the measures of the angles of a triangle is 180 degrees, either triangle gives us the relationship $90+2x=180$, so $x=45$.

The measures of $\angle ADC$ and $\angle ABC$ are both $2x=2(45)=90$. Thus they are also right angles.

Hence, if a convex, four-sided, equilateral polygon has one right angle, then it has four right angles.

This proof uses SSS again—I actually thought of it while trying to work out a proof of SSS—as well as SAS, through the lemma1. It also uses the fact that the measures of the angles of a triangle sum to 180 degrees. I’m pretty sure that theorem is only a couple of steps from the parallel postulate, though.

My original version of this proof didn’t make use of the lemma about isosceles triangles, but instead argued that the two congruent triangles could be mapped onto each other by either rotation or reflection, and that was why all of the acute angles had to be equal. That was kind of a nifty argument, but not easy to formalize.

Stay tuned tomorrow for the proof about parallelograms.

  1. To do. ↩︎

Perpendicular Lines

Olly / / Math Posts

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.

Congruent Triangles

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 $x_F=x_G=\frac{c-b}{m-n}-1$.

Perpendicular Lines

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.

  1. To do. ↩︎

Parallel Lines

Olly / / Math Posts

Today I worked on some exercises in Analysis with an Introduction to Proof, then read the review of analytic geometry I had queued. Near the end of the review, the author quotes the following theorems, indicating that proofs can be found in his precalculus textbook:

1. Two nonvertical lines are parallel if and only if they have the same slope.
2. 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:$$m_2=-\frac{1}{m_1}$$

That got me to thinking about how I would prove those things, and I ended up spending the rest of my study time working on them. I have a complete proof of Theorem 1. It is not that exciting, but it did give me the chance to use some of the logic that I am reviewing in Analysis with an Introduction to Proof.

Proving Theorem 2 seems like the more interesting problem. What does it mean for lines to be perpendicular? If it means that they meet at a right angle, what is that? How do you prove an angle is a right angle? My best idea so far is an argument involving right triangles, but in order for it to work, I need to establish to my satisfaction not just that the Pythagorean theorem holds for all right triangles, but also that only right triangles have sides that relate in the way described in the Pythagorean theorem. It’s pretty easy to convince oneself of the latter, given the former, and I may decide that’s enough. We’ll see.

Anyway, here’s what I did for Theorem 1:

Proposition: Two nonvertical lines are parallel if and only if they have the same slope.

For two lines to be parallel means that either they do not intersect or they are the same line.

Thus, we need to prove two statements:

  1. If two nonvertical lines do not intersect or they are the same line, then they have the same slope.
  2. If two nonvertical lines have the same slope, then either they do not intersect or they are the same line.

To prove statement 2, assume that two nonvertical lines have the same slope, $m$. Their equations can be written as $y=mx+b$ and $y=mx+c$, where $b$ and $c$ are real numbers.

Now assume that the lines intersect, and let $(x_1, y_1)$ be a point that lies on both lines. It follows that $y_1=mx_1+b$ and $y_1=mx_1+c$, and thence that

$$mx_1+b=mx_1+c$$$$b=c$$

When $b=c$, the two lines have the same equation, and are thus the same line.

Thus, when two nonvertical lines have the same slope it follows that if the two lines intersect, then they are the same line.

Hence, if two nonvertical lines have the same slope, then either they do not intersect or they are the same line, which is statement 2.

(The logic in play here is: $[p\implies (q\implies r)]\equiv[p\implies(\lnot q\lor r)]$)

To prove statement 1 by contraposition, assume that two nonvertical lines have different slopes, $m$ and $n$. Their equations can be written as $y=mx+b$ and $y=nx+c$ where $m\neq n$ and $b$ and $c$ are real numbers.

Since their equations are different, they are not the same line.

Furthermore, consider the point $$\left(\frac{c-b}{m-n},\frac{mc-nb}{m-n}\right)$$ which must exist, since $m\neq n$.

Notice that $$m\left(\frac{c-b}{m-n}\right)+b=\frac{(mc-mb)+(mb-nb)}{m-n}=\frac{mc-nb}{m-n}$$ and $$n\left(\frac{c-b}{m-n}\right)+c=\frac{(nc-nb)+(mc-nc)}{m-n}=\frac{mc-nb}{m-n}\text{.}$$

Thus, the point lies on both lines and the lines intersect.

Therefore, if two nonvertical lines have different slopes, then the two lines intersect and they are not the same line.

By contraposition, it follows that if two nonvertical lines do not intersect or they are the same line, then they have the same slope, which is statement 1.

(The logic in play here is: $[\lnot p\implies (q\land\lnot r)]\equiv[\lnot(q\land\lnot r)\implies p]\equiv[(\lnot q\lor r)\implies p]$)

Stay tuned for related proofs about perpendicular lines, probably tomorrow.

Squares Again

Olly / / Math Posts

I’m not sure yet how I’m going to organize my review. So far I’ve been dabbling widely in the materials I have available. Yet I think it would be a good idea for me to mix drills in the calculation-focused subjects I may eventually want to tutor—calculus, analytic geometry, and so on—with review of some more proofs-based math. Otherwise I’m going to feel as if I’m missing half the show.

Today I was reading Chapter Zero by Carol Schumacher, one of my old textbooks, which was written as an introduction to proofs-based math. In its discussion of definitions, it says the following:

[W]e might naively define a square as “a four-sided, equilateral polygon,” but we would quickly see that such a figure need not have right angles. If we are actually only interested in right-angled figures, then we could refine our definition to be “a four-sided, equilateral polygon with four right angles.”

That immediately had me wondering whether, to know that a four-sided, equilateral polygon is a square, one must know the measures of all of its angles. I suspect that one is sufficient. I believe there are no equilateral, four-sided polygons that have one right angle and do not have four right angles. I haven’t sat down to work out a proof yet, though.