Solving Problems Submitted to MAA Journals (Part 6b)

The following problem appeared in Volume 97, Issue 3 (2024) of Mathematics Magazine.

Two points $P$ and $Q$ are chosen at random (uniformly) from the interior of a unit circle. What is the probability that the circle whose diameter is segment $\overline{PQ}$ lies entirely in the interior of the unit circle?

As discussed in the previous post, I guessed from simulation that the answer is $2/3$ . Naturally, simulation is not a proof, and so I started thinking about how to prove this.

My first thought was to make the problem simpler by letting only one point be chosen at random instead of two. Suppose that the point $P$ is fixed at a distance $t$ from the origin. What is the probability that the point $Q$ , chosen at random, uniformly, from the interior of the unit circle, has the desired property?

My second thought is that, by radial symmetry, I could rotate the figure so that the point $P$ is located at $(t,0)$ . In this way, the probability in question is ultimately going to be a function of $t$ .

There is a very nice way to compute such probabilities since $Q$ is chosen at uniformly from the unit circle. Let $A_t$ be the probability that the point $Q$ has the desired property. Since the area of the unit circle is $\pi(1)^2 = \pi$ , the probability of desired property happening is

$\displaystyle \frac{\hbox{area}(A_t)}{\pi}$ .

So, if I could figure out the shape of $A_t$ , I could compute this conditional probability given the location of the point $P$ .

But, once again, I initially had no idea of what this shape would look like. So, once again, I turned to simulation with Mathematica.

First, a technical detail that I ignored in the previous post. To generate points $(x,y)$ at random inside the unit circle, one might think to let $x = r \cos \theta$ and $y = r \sin \theta$ , where the distance from the origin $r$ is chosen at random between 0 and 1 and the angle $\theta$ is chosen at random from $0 to 2\pi$ . Unfortunately, this simple simulation generates too many points that are close to the origin and not enough that are close to the circle:

To see why this happened, let $R$ denote the distance of a randomly chosen point from the origin. Then the event $R < r$ is the same as saying that the point lies inside the circle centered at the origin with radius $r$ , so that the probability of this event should be

$F(r) = P(R < r) = \displaystyle \frac{\pi r^2}{\pi (1)^2} = r^2$ .

However, in the above simulation, $R$ was chosen uniformly from $[0,1]$ , so that $P(R < r) = r$ . All this to say, the above simulation did not produce points uniformly chosen from the unit circle.

To remedy this, we employ the standard technique of using the inverse of the above function $F(r)$ , which is clearly $F^{-1}(r) = \sqrt{r}$ . In other words, we will chose randomly chosen radius to have the form $R= \sqrt{U}$ , where $U$ is chosen uniformly on $[0,1]$ . In this way,

$P(R < r) = P( \sqrt{U} < r) = P(U < r^2) = r^2$ ,

as required. Making this modification (highlighted in yellow) produces points that are more evenly distributed in the unit circle; any bunching of points or empty spaces are simply due to the luck of the draw.

In the next post, I’ll turn to the simulation of $A_t$ .

Lagrange Points and Polynomial Equations: Part 3

From Wikipedia, Lagrange points are points of equilibrium for small-mass objects under the gravitational influence of two massive orbiting bodies. There are five such points in the Sun-Earth system, called $L_1$ , $L_2$ , $L_3$ , $L_4$ , and $L_5$ .

The stable equilibrium points $L_4$ and $L_5$ are easiest to explain: they are the corners of equilateral triangles in the plane of Earth’s orbit. The points $L_1$ and $L_2$ are also equilibrium points, but they are unstable. Nevertheless, they have practical applications for spaceflight.

The position of $L_2$ can be found by numerically solving the fifth-order polynomial equation

$(m_1+m_2)t^5+(3m_1+2m_2)t^4+(3m_1+m_2)t^3$

$-(m_2+3m_3)t^2-(2m_2+3m_3)t-(m_2+m_3)=0$ .

In this equation, $m_1$ is the mass of the Sun, $m_2$ is the mass of Earth, $m_3$ is the mass of the spacecraft, and $t$ is the distance from the Earth to $L_2$ measured as a proportion of the distance from the Sun to Earth. In other words, if the distance from the Sun to Earth is 1 unit, then the distance from the Earth to $L_2$ is $t$ units. The above equation is derived using principles from physics which are not elaborated upon here.

We notice that the coefficients of $t^5$ , $t^4$ , and $t^3$ are all positive, while the coefficients of $t^2$ , $t$ , and the constant term are all negative. Therefore, since there is only one change in sign, this equation has only one positive real root by Descartes’ Rule of Signs.

Since $m_3$ is orders of magnitude smaller than both $m_1$ and $m_2$ , this may safely approximated by

$(m_1+m_2)t^5+(3m_1+2m_2)t^4+(3m_1+m_2)t^3 - m_2 t^2- 2m_2 t-m_2=0$ .

This new equation can be rewritten as

$t^5 + \displaystyle \frac{3m_1 + 2m_2}{m_1 + m_2} t^4 + \frac{3m_1+m_2}{m_1+m_2} t^3 - \frac{m_2}{m_1+m_2} t^2 - \frac{2m_2}{m_1+m_2} t- \frac{m_2}{m_1+m_2} = 0$ .

If we define

$\mu = \displaystyle \frac{m_2}{m_1+m_2}$ ,

we see that

$\displaystyle \frac{3m_1 + 2m_2}{m_1 + m_2} = \frac{3m_1 + 3m_2}{m_1 + m_2} - \frac{m_2}{m_1 + m_2} = 3 - \mu$

and

$\displaystyle \frac{3m_1 + m_2}{m_1 + m_2} = \frac{3m_1 + 3m_2}{m_1 + m_2} - \frac{2m_2}{m_1 + m_2} = 3 - 2\mu$ ,

so that the equation may be written as

$t^5 + (3-\mu) t^4 + (3-2\mu) - \mu t^2 - 2\mu t - \mu = 0$ ,

matching the equation found at Wikipedia.

For the Sun and Earth, $m_1 \approx 1.9885 \times 10^{30} ~ \hbox{kg}$ and $m_2 \approx 5.9724 \times 10^{24} ~ \hbox{kg}$ , so that

$mu = \displaystyle \frac{5.9724 \times 10^{24}}{1.9885 \times 10^{30} + 5.9724 \times 10^{24}} \approx 3.00346 \times 10^{-6}$ .

This yields a quintic equation that is hopeless to solve using standard techniques from Precalculus, but the root can be found graphically by seeing where the function crosses the $x-$ axis (or, in this case, the $t-$ axis):

As it turns out, the root is $t \approx 0.01004$ , so that $L_2$ is located $1.004\%$ of the distance from the Earth to the Sun in the direction away from the Sun.

Lagrange Points and Polynomial Equations: Part 2

We begin with $L_1$ , whose position can be found by numerically solving the fifth-order polynomial equation

$(m_1+m_3)x^5+(3m_1+2m_3)x^4+(3m_1+m_3)x^3$

$-(3m_2+m_3)x^2-(3m_2+m_3)x-(m_2+m_3)=0$ .

In this equation, $m_1$ is the mass of the Sun, $m_2$ is the mass of Earth, $m_3$ is the mass of the spacecraft, and $x$ is the distance from the Earth to $L_1$ measured as a proportion of the distance from the Sun to $L_1$ . In other words, if the distance from the Sun to $L_1$ is 1 unit, then the distance from the Earth to $L_1$ is $x$ units. The above equation is derived using principles from physics which are not elaborated upon here.

We notice that the coefficients of $x^5$ , $x^4$ , and $x^3$ are all positive, while the coefficients of $x^2$ , $x$ , and the constant term are all negative. Therefore, since there is only one change in sign, this equation has only one positive real root by Descartes’ Rule of Signs.

Since $m_3$ is orders of magnitude smaller than both $m_1$ and $m_2$ , this may safely approximated by

$m_1 x^5 + 3m_1 x^4 + 3m_1 x^3 - 3m_2 x^2 - 3m_2x - m_2=0$ .

Unfortunately, the unit $x$ is not as natural for Earth-bound observers as $t$ , the proportion of the distance of $L_1$ to Earth as a proportion of the distance from the Sun to Earth. Since $L_1$ is between the Sun and Earth, the distance from the Sun to Earth is $x+1$ units, so that $t = x/(x+1)$ . We then solve for $x$ in terms of $t$ (just like finding an inverse function):

$t = \displaystyle \frac{x}{x+1}$

$t(x+1) = x$

$tx + t = x$

$t = x - tx$

$t= x(1-t)$

$\displaystyle \frac{t}{1-t} = x$ .

Substituting into the above equation, we find an equation for $t$ :

$\displaystyle \frac{m_1t^5}{(1-t)^5} + \frac{3m_1t^4}{(1-t)^4} + \frac{3m_1t^3}{(1-t)^3} - \frac{3m_2t^2}{(1-t)^2} - \frac{3m_2t}{1-t} - m_2=0$

$m_1t^5 + 3m_1t^4(1-t) + 3m_1t^3(1-t)^2 - 3m_2t^2(1-t)^3 - 3m_2t(1-t)^4 - m_2(1-t)^5=0$

Expanding, we find

$m_1 t^5 + 3m_1 (t^4 - t^5) + 3m_1 (t^3-2t^4+t^5) - 3m_2 (t^2-3t^3+3t^4-t^5)$

$-3m_2(t - 4t^2 + 6t^3 - 4t^4 + t^5) - m_2(1 - 5t + 10t^2 - 10 t^3 + 5t^4 + t^5) = 0$

Collecting like terms, we find

$(m_1 - 3m_1 + 3m_1 + 3m_2 - 3m_2 + m_2)t^5 + (3m_1-6m_1-9m_2+12m_2-5m_2)t^4$

$+ (3m_1+9m_2-18m_2+10m_2)t^3 + (-3m_2+12m_2-10m_2) t^2$

$+ (-3m_2+5m_2)t - m_2 = 0$ ,

$(m_1+m_2) t^5 - (3m_1 +2m_2) t^4 + (3m_1 + m_2) t^3 - m_2 t^2 + 2m_2 t- m_2 = 0$ .

Again, this equation has only one positive real root since the original quintic in $x$ only had one positive real root. This new equation can be rewritten as

$t^5 - \displaystyle \frac{3m_1 + 2m_2}{m_1 + m_2} t^4 + \frac{3m_1+m_2}{m_1+m_2} t^3 - \frac{m_2}{m_1+m_2} t^2 + \frac{2m_2}{m_1+m_2} t- \frac{m_2}{m_1+m_2} = 0$ .

If we define

$\mu = \displaystyle \frac{m_2}{m_1+m_2}$ ,

we see that

$\displaystyle \frac{3m_1 + 2m_2}{m_1 + m_2} = \frac{3m_1 + 3m_2}{m_1 + m_2} - \frac{m_2}{m_1 + m_2} = 3 - \mu$

and

$\displaystyle \frac{3m_1 + m_2}{m_1 + m_2} = \frac{3m_1 + 3m_2}{m_1 + m_2} - \frac{2m_2}{m_1 + m_2} = 3 - 2\mu$ ,

so that the equation may be written as

$t^5 + (\mu-3) t^4 + (3-2\mu) - \mu t^2 + 2\mu t - \mu = 0$ ,

matching the equation found at Wikipedia.

For the Sun and Earth, $m_1 \approx 1.9885 \times 10^{30} ~ \hbox{kg}$ and $m_2 \approx 5.9724 \times 10^{24} ~ \hbox{kg}$ , so that

$\mu = \displaystyle \frac{5.9724 \times 10^{24}}{1.9885 \times 10^{30} + 5.9724 \times 10^{24}} \approx 3.00346 \times 10^{-6}$ .

As it turns out, the root is $t \approx 0.00997$ , so that $L_1$ is located $0.997\%$ of the distance from the Earth to the Sun in the direction of the Sun.

Inverse Function

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My Favorite One-Liners: Part 63

In this series, I’m compiling some of the quips and one-liners that I’ll use with my students to hopefully make my lessons more memorable for them.

I’ll use today’s one-liner to explain why mathematicians settled on a particular convention that could have been chosen differently. For example, let’s consider the definition of $y = \sin^{-1} x$ by first looking at the graph of $f(x) = \sin x$ .

Of course, we can’t find an inverse for this function; colloquially, the graph of $f$ fails the horizontal line test. More precisely, there exist two numbers $x_1$ and $x_2$ so that $x_1 \ne x_2$ but $f(x_1) = f(x_2)$ . Indeed, there are infinitely many such pairs.

So how will we find the inverse of $f$ ? Well, we can’t. But we can do something almost as good: we can define a new function $g$ that’s going look an awful lot like $f$ . We will restrict the domain of this new function $g$ so that $g$ satisfies the horizontal line test.

For the sine function, there are plenty of good options from which to choose. Indeed, here are four legitimate options just using the two periods of the sine function shown above. The fourth option is unorthodox, but it nevertheless satisfies the horizontal line test (as long as we’re careful with $\pm 2\pi$ .

So which of these options should we choose? Historically, mathematicians have settled for the interval $[-\pi/2, \pi/2]$ .

So, I’ll ask my students, why have mathematicians chosen this interval? That I can answer with one word: tradition.

For further reading, see my series on inverse functions.

My Favorite One-Liners: Part 16

In this series, I’m compiling some of the quips and one-liners that I’ll use with my students to hopefully make my lessons more memorable for them.

One of the basic notions of functions that’s taught in Precalculus and in Discrete Mathematics is the notion of an inverse function: if $f: A \to B$ is a one-to-one and onto function, then there is an inverse function $f^{-1}: B \to A$ so that

$f^{-1}(f(a)) = a$ for all $a \in A$ and

$f(f^{-1}(b)) = b$ for all $b \in B$ .

If $A = B = \mathbb{R}$ , this is commonly taught in high school as a function that satisfies the horizontal line test.

In other words, if the function $f$ is applied to $a$ , the result is $f(a)$ . When the inverse function is applied to that, the answer is the original number $a$ . Therefore, I’ll tell my class, “By applying the function $f^{-1}$ , we uh-uh-uh-uh-uh-uh-uh-undo it.”

If I have a few country music fans in the class, this always generates a bit of a laugh.

See also the amazing duet with Carrie Underwood and Steven Tyler at the 2011 ACM awards:

My Favorite One-Liners: Part 9

In this series, I’m compiling some of the quips and one-liners that I’ll use with my students to hopefully make my lessons more memorable for them.

Today, I’d like to discuss a common mistake students make in trigonometry… as well as the one-liner that I use to (hopefully) help students not make this mistake in the future.

Question. Find all solutions (rounded to the nearest tenth of a degree) of $\sin x = 0.8$ .

Erroneous Solution. Plugging into a calculator, we find that $x \approx 53.1^o$ .

arcsine1

The student correctly found the unique angle $x$ between $-90^o$ and $90^o$ so that $\sin x = 0.8$ . That’s the definition of the arcsine function. However, there are plenty of other angles whose sine is equal to $0.7$ . This can happen in two ways.

First, if $\sin x > 0$, then the angle $x$ could be in either the first quadrant or the second quadrant (thanks to the mnemonic All Scholars Take Calculus). So $x$ could be (accurate to one decimal place) equal to either $53.1^o$ or else $180^o - 53.1^o = 126.9^o$ . Students can visualize this by drawing a picture, talking through each step of its construction (first black, then red, then brown, then green, then blue).

However, most students don’t really believe that there’s a second angle that works until they see the results of a calculator.

Second, any angle that’s coterminal with either of these two angles also works. This can be drawn into the above picture and, as before, confirmed with a calculator.

So the complete answer (again, approximate to one decimal place) should be $53.1^{\circ} + 360n^o$ and $126.9 + 360n^{\circ}$ , where $n$ is an integer. Since integers can be negative, there’s no need to write $\pm$ in the solution.

Therefore, the student who simply answers $53.1^o$ has missed infinitely many solutions. The student has missed every nontrivial angle that’s coterminal with $53.1^o$ and also every angle in the second quadrant that also works.

Here’s my one-liner — which never fails to get an embarrassed laugh — that hopefully helps students remember that merely using the arcsine function is not enough for solving problems such as this one.

You’ve forgotten infinitely many solutions. So how many points should I take off?

For further reading, here’s my series on inverse functions.

Engaging students: Inverse trigonometric functions

In my capstone class for future secondary math teachers, I ask my students to come up with ideas for engaging their students with different topics in the secondary mathematics curriculum. In other words, the point of the assignment was not to devise a full-blown lesson plan on this topic. Instead, I asked my students to think about three different ways of getting their students interested in the topic in the first place.

I plan to share some of the best of these ideas on this blog (after asking my students’ permission, of course).

This student submission comes from my former student Joe Wood. His topic, from Precalculus: inverse trigonometric functions.

What are the contributions of various cultures to this topic?

Trig functions have a very long history spanning many countries and cultures. Greek astronomers such as Aristarchus, Claudius, and Ptolemy first used trigonometry; however, according to the University of Connecticut, these Greek astronomers were primarily concerned with “the length of the chord of a circle as a function of the circular arc joining its endpoints.” Many of these astronomers, Ptolemy especially, were concerned with planetary and celestial body’s rotations, so this made sense.

While the Greeks first studied trigonometric concepts, it was the Indian people who really studied sine and cosine functions with the angle as a variable. The information was then brought to the Arabic and Persian cultures. One significant figure, a Persian by the name Abu Rayhan Biruni, used trig to accurately estimate the circumference of Earth and its radius before the end of the 11^th century.

Fast-forward about 700 years, a Swiss mathematician, Daniel Bernoulli, used the “A.sin” notation to represent the inverse of sine. Shortly after, another Swiss mathematician used “A t” to represent the inverse of tangent. That man was none other than Leonhard Euler. It was not until 1813 that the notation sin^-1 and tan^-1 were introduced by Sir John Fredrick William Herschel, an English mathematician.

As we can see, the development of inverse trigonometric functions took quite the cultural rollercoaster ride before stopping some place we see being familiar. It took many cultures, and even more years to develop this sophisticated branch of mathematics.

How could you as a teacher create an activity or project that involves your topic?

Last Semester I taught a lesson on the trigonometric identities. I found this cool cut and paste activity for the students that allowed them to warm up to the trig identities by not having to do the process themselves, but still having to see every step of converting one trig function into another with the identities. Below, you will find the activity, then the instructions, and finally how to modify the activity to fit inverse trig identities specifically.

Directions:
1.) Begin by cutting out all the pieces.
2.) Students will take any of the four puzzle pieces with the black squiggly line.
3.) Find an equivalent puzzle piece by using some trig identity.
4.) Repeat step 3 until there are no more equivalent pieces.
5.) Grab the next puzzle pieces with the black squiggly line.
6.) Repeat steps 3-5 until all puzzle pieces have been used.
Ex.) Begin with cscx-sinx. Lay next to that piece, the piece that reads =1/sinx – sinx, then the piece that reads =1/sinx – sin²x/sinx. Contiue the trend until you reach =cotx * cosx. Then move to the next squiggly lined piece.

Modify:
This game can be modified using inverse trig functions. Start with pieces such as sin^-1(sin(30⁰)) in squiggles. Have a piece showing sin^-1(sin(30⁰)) with a line through the sines. Then a piece that just shows 30⁰. Next a piece in a squiggly line that is sin^-1(sqrt(2)/2) that connects to a piece of 45⁰, but make them write why this works.

How does this topic extend what your students should have learned in previous courses?

Obviously, by this time students should know what trigonometric functions are and how to use this. Students should also know from previous classes what inverse functions are. Studying inverse trig functions then is a continuation of these topics. As I teacher I would begin relating inverse trig functions by refreshing the students on what inverse functions are. The class would then move into the concept that the trig expression of an angle returns a ratio of two sides of a triangle. We would slowly move into what happens then if you know the sides of a triangle but need the angle. From there we would discuss trigonometric expressions using the angles as variables. Finally, we would make the connection that that is a function, and on the proper interval should have an inverse function. That is when the extension into the new topic of inverse trigonometric functions would seriously begin.

Undo It

$f^{-1}(f(a)) = a$ for all $a \in A$ and

$f(f^{-1}(b)) = b$ for all $b \in B$ .

If $A = B = \mathbb{R}$ , this is commonly taught in high school as a function that satisfies the horizontal line test.

If I have a few country music fans in the class, this always generates a bit of a laugh.

See also the amazing duet with Carrie Underwood and Steven Tyler at the 2011 ACM awards:

Engaging students: Computing inverse functions

I plan to share some of the best of these ideas on this blog (after asking my students’ permission, of course).

This student submission again comes from my former student Carissa Birdsong. Her topic, from Algebra: computing inverse functions.

How can this topic be used in your students’ future courses in mathematics or science?

When students are learning any algorithm in math, it helps keep their interest if they know what this can be possibly used for in the future. In pre-calculus, students need to find the inverse of cosine, sine, tangent, etc. to find certain angles. In order to grasp the students’ attention, the teacher can show videos of bottle rockets being shot off at different angles. Then the teacher will explain that in order to find most of these angles, one must use the inverse property. Then the teacher can go into depth of how to find the inverse of a function. But, the students must understand that using inverse to find angle measurements will not happen in this curriculum, but in future classes such as pre-calculus, trigonometry and physics.

How could you as a teacher create an activity or project that involves your topic?

Human Representation of Inverse Function

Move the desks to the sides of the room, making a big open space in the middle.
Assign each student a partner.
Have a strip of tape down the middle of the room prior to class. Have the students line up facing their partner with the strip of tape in between them.
Have the side on the “right” be side A and the side on the “left” be side B. (The teacher will choose which side is the right or left, depending on where the front of the classroom is)
Side A will pick a position to stand in (the teacher must monitor to make sure the students are being appropriate). The students are encouraged to change their face, arms, head, etc. to pick the most creative position possible.
Now side B will mimic their specific partner on side A.
Once the students have locked in their position, the teacher will point out that side B is reflective of side A. Therefore, side B is the inverse function of A.

*Make sure that the students understand that side B is not doing the exact same thing that side A is doing, but the opposite, the reflection. The inverse of a function “undoes” the function itself. If someone were to take away side A, and bring in a new crop of people to reflect side B, it should be EXACTLY what side A had done. The inverse of the inverse of a function must take you back to the original function.

*After the teacher teaches how to find the inverse of a function, and can elaborate on the graphing of each function, he or she can refer back to this activity and show that there is an invisible line between the function and the inverse function, making clear that they reflect each other, just as the students did.

How has this topic appeared in pop culture (movies, TV, current music, video games, etc.)?

Even though most students probably haven’t seen Top Secret, they will probably appreciate watching any sort of movie or television during class. In the making of Top Secret, the actors film a scene walking backwards and saying lines in reverse order. In the movie, this scene is played in reverse, so they look like they were just speaking gibberish and walking forward. They did this so Val Kilmer can do cool tricks like throw a book on the top shelve and slide up a pole.

The teacher could show his or her class the original scene, straight from the movie.

Then ask, “How do you think the actors did this?” “What language are they speaking?” Hopefully a student will catch on fast and say that they just filmed it backwards. Then the teacher can show the scene played forwards.

These two scenes are inverse each other. Going from the beginning to the end of one takes you to the beginning of the other. And going from the beginning to the end of the other, takes you to the beginning of one. Most functions have an inverse function. This means there is a function that is reverse of its inverse. This does NOT mean that the inverse of a function is just the original backwards (i.e. y=3+x and x+3=y). The function of f has the input x and the output y, whereas the inverse of the function f has the input y and the output x.

Resources:

https://www.youtube.com/watch?v=jXlucE4iUDE

https://www.youtube.com/watch?v=2Mr_XAM8CMw