The cover illustrates a neat result that Anna B. discovered and explains in this issue’s *Anna’s Math Journal*. She continued her investigation of paraboloids and discovered that orthogonal projection from a paraboloid coincides with the composition of stereographic projection and a special map M inspired by the optical properties of a paraboloid. For details, check out her column!

We also feature an interview with University of Minnesota assistant professor of mathematics Christine Berkesch Zamaere.

Next, Akamai Technologies computer scientist Kate Jenkins concludes her discussion of algorithms that find the “maximal subsequence” of a sequence. Were you able to figure out an algorithm that determines the maximum subsequence of *N* numbers using *O*(*N*) computations? Kate’s article is just one example of how mathematics applies to problems in industry. In the past decades, so much information has been digitized, including books, pictures, video, weather, architectural plans, music, etc. Where there are numbers, there is the potential for mathematical analysis.

Emily and Jasmine return, this time designing star patterns for different numbers of states. We received positive feedback about their last project where they designed a stained glass window (see Volume 7, Number 4), so we plan to feature them more in the future. The two show how, with a bit of inquisitiveness, there’s mathematics.

We conclude with solutions to this summer’s batch of Summer Fun problem sets. Incidentally, if we had more room, we would have liked to include one more problem in the Summer Fun problem set on permutations. That problem set ended with a result of Zolotarev connecting the signs of certain permutations to the theory of squares modulo *p*, where *p* is a prime number. With more room, we’d have outlined Zolotarev’s proof of quadratic reciprocity using permutations. This proof is “just around the corner” from the material in the permutation problem set and Cailan’s Summer Fun problem set on quadratic reciprocity. As a challenge, you could try to reconstruct Zolotarev’s beautiful proof. Here’s a hint: The idea is to take a deck of *pq* playing cards, where *p* and *q* are distinct odd prime numbers. Consider the following 3 arrangements of the cards into a *p* by *q* rectangle:

Arrangement 1: Deal the cards out row by row, from left to right.

Arrangement 2: Deal the cards out column by column, from top to bottom.

Arrangement 3: Deal the cards out going along a NW-SE diagonal, with wraparound.

Consider the permutations defined in going from arrangement 1 to 2, from 2 to 3, and from 3 to 1.

Prof. Jerry Shurman of Reed College has written up a beautiful presentation of Zolotarev’s proof.

We hope you enjoy our latest issue!

Finally, a reminder: when you subscribe to the Girls’ Angle Bulletin, you’re not just getting a subscription to a magazine. You are also gaining access to the Girls’ Angle mentors. We urge all subscribers and members to write us with your math questions or anything else in the Bulletin or having to do with mathematics in general. We will respond. We want you to get active and *do* mathematics. Parts of the Bulletin are written to induce you to wonder and respond with more questions. Don’t let those questions fade away and become forgotten. Send them to us!

We continue to encourage people to subscribe to our print version, so we have removed some content from the electronic version. Subscriptions are a great way to support Girls’ Angle while getting something concrete back in return. We hope you subscribe!

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Marion Walter’s Theorem: In a triangle, draw line segments from each vertex to the trisection points on the opposite side. The six drawn line segments will form the edges of a central hexagon. The ratio of the area of the hexagon to that of the whole triangle is 1/10.

An efficient way to prove Marion Walter’s theorem is to use mass points.

In this post, I’ll give details because the proof is a model example of the mass points technique. If you’re having difficulty learning the technique, I hope this post will help it all come together for you. As always, try to use mass points to prove the theorem yourself, and, only after you have tried, read on. If you’ve never heard of mass points before, google it or check out Volume 7, Number 3 of the Girls’ Angle Bulletin.

Spoiler Alert! Proof Below!

It all begins with Archimedes’ Law of the Lever, which we will apply over and over.

Two point masses connected by a massless rod will balance at their center of mass. Archimedes tells us that the distances of each mass from the center of mass satisfy the equation .

The center of mass will lie along the line segment connecting the two masses and sit nearer the heavier one. If the masses are equal, then the center of mass will be exactly halfway between. If one mass is twice the other, then the center of mass will be 2/3 of the way from the lighter to the heavier. (In general, the center of mass is the weighted average of the positions of each object weighted by the mass of that object.)

For a proof of this “piecemeal” way to compute the center of mass and for more information about the center of mass and other applications, see, for instance, Section 19.1 of the Feynman Lectures on Physics, Volume 1, or checkout Volume 7, Number 3 of the Girls’ Angle Bulletin, particularly page 20.

In addition to the Law of the Lever, a beautiful property of the center of mass that we will use repeatedly is that the center of mass can be computed piecemeal. That is, we can split the objects into different sets, compute the center of mass of the objects in each set separately and pretend that each set is replaced by a single point mass equal to the total mass of the objects in that set and located at that set’s center of mass. Then we can compute the center of mass of these pretend point masses to learn the location of the center of mass of the original collection of objects.

For point masses, this “piecemeal” fact enables us to find the center of mass of any number of point masses with repeated application of Archimedes’ Law of the Lever. All we have to do is compute 2 masses at a time, as illustrated in the following figures.

**Proof of Marion Walter’s Theorem**

Here’s a figure illustrating Marion Walter’s theorem with the 3 medians of the triangle added in.

We’ll focus on the green triangle and compute how much of the tip of that triangle is inside the orange hexagon:

Let be the measure of .

The area of triangle *XOY* is .

The area of triangle *AOB* is .

Therefore, the ratio of the area of triangle *XOY* to the area of triangle *AOB* is:

.

To compute these two ratios of lengths, we will use the technique of mass points to compute each length as a fraction of the length of the cevian that it lies on.

Back to the larger figure:

Let’s start with the indicated lengths (which correspond to *OY* and *OB* in the labeling of the prior diagram). We’ll use mass points to figure out their lengths as a fraction of the median they’re both on. Let’s start with the longer length.

If we knew what masses to place at the endpoints of the median so that their center of mass sat at the intersection of the medians, then we could use Archimedes’ Law of the Lever to compute the ratio of the lengths that the median is split into. But the problem is, we don’t know that ratio. That’s essentially the ratio we’re trying to find.

What we do know is the ratios of key lengths along the sides of the triangle, because the whole problem is set up by trisecting the sides. This suggests using 3 masses, one at each vertex, as shown below:

Can we figure out how to assign masses to the 3 vertices so that the center of mass of all 3 masses will be where the medians intersect?

Notice that if the two leftmost masses are equal, then, by Archimedes’ Law of the Lever, their center of mass will be at one end of the red median, since the median bisects the side. The importance of assigning the 2 leftmost masses so that their center of mass is at one end of the red median is that it guarantees that the center of mass of all 3 masses will lie on the red median (because of the “piecemeal” property for computing the center of mass).

When the unknown mass (indicated by “?”) is zero, the center of mass of all 3 masses will be midway between the two leftmost masses. As the unknown mass increases, the center of mass of all 3 masses will move along the red median toward the unknown mass. What must the unknown mass be so that the center of mass of all 3 masses will be right where the medians intersect?

We know we can compute the center of mass of the 3 masses piecemeal, and we used this fact to see that by making the 2 leftmost masses equal, the center of mass of all 3 masses will be somewhere on the red median. And now comes a key idea: *if we cleverly arrange it so that the center of mass of the 2 lower masses is exactly halfway between them, then the center of mass of all 3 masses will also lie along a second median, and, therefore, it will be at the intersection of both medians!*

In other words, to make the center of mass of all 3 masses be at the intersection of the medians, we now only have to concentrate on making the center of mass of the lower 2 masses be at their midpoint. Using Archimedes’ Law of the Lever, we know this happens when the unknown mass is also *m*.

Thus, the center of mass of 3 equal masses placed at the vertices of a triangle is located at the intersection of the triangle’s medians.

Now, we can compute the ratio of the lengths that the red median is split into by the other medians. We go back to computing the center of mass piecemeal by first computing the center of mass of the 2 leftmost masses and imagining a single point mass of mass 2*m* placed there. We’ve purposefully assigned the masses so that the center of mass of this 2*m*-mass and the *m*-mass in the lower right corner will be at the intersection of the medians, hence the ratio of the lengths that the red median is split into, by Archimedes’ Law of the Lever, must be 1 : 2. We deduce that the length we are looking for (indicated by the double-headed arrow) is 1/3 the length of the red median.

Now let’s turn our attention to the other length:

This time, we try to assign masses to the vertices so that their center of mass will be where the red and orange cevians intersect in the figure below.

Since the red cevian is a median, we know to place equal masses at the 2 leftmost vertices, just as before. This time, however, we want to assign a mass to the lower right vertex in such a way that the center of mass of the 2 lower masses will be at the foot of the orange cevian, which is 1/3 of the way from left to right. Using Archimedes’ Law of the Lever, we compute that the lower right mass must be *m*/2.

We now compute the center of mass piecemeal, starting with the 2 leftmost masses. Their center of mass is at the foot of the red median (by design), so we imagine a point mass of mass 2*m* placed there. The red median is split into two pieces by the orange cevian. Archimedes’ Law of the Lever tells us that the ratio of the lengths of these two pieces is equal to the ratio *m*/2 to 2*m*, which is 1/4. Hence, the center of mass of these 3 point masses is 1/5 of the way along the red median, measured from its foot.

Therefore, the distance we’re interested in (indicated by the double-headed arrow in the figure above) is 1/3 – 1/5 = 2/15 of the length of the red median.

Next, we turn our attention to the other 2 important lengths indicated below:

What fraction of the purple median are they?

The longer length stretches from the vertex to the intersection of the medians, and we already know how to balance the triangle there. That’s achieved with equal point masses at each vertex and we conclude that the longer length is 2/3 of the length of the purple median.

For the shorter length, we try to assign point masses to the vertices so that their center of mass is where the purple median and orange cevian intersect. The figure above shows what the masses need to be to achieve that. Computing this center of mass location piecemeal, starting with the 2 rightmost masses and then applying Archimedes’ Law of the Lever, we find that the orange cevian actually bisects the purple median! Hence, the shorter length is 2/3 – 1/2 = 1/6 of the length of the purple median.

Putting it all together, we find that:

.

(Note that the numbers are fractions of lengths, not absolute lengths. That is *OX* is not 1/6 nor is *OA* equal to 2/3, but since we are taking ratios, *OX*/*OA* is equal to (1/6)/(2/3).)

To find the fraction of the triangle’s area that the hexagon occupies, we note that the tips of the 6 green triangular sections that dip into the hexagon are all 1/10 of their respective green triangles. That’s because the computations for each green triangle will look just like the computation we just went through by symmetry.

Therefore, the area of the hexagon, which consists of the 6 tips of the green triangles, is exactly 1/10 the area of the whole, and the theorem is proven.

To show how efficient this technique is, I’ve scanned in my actual scratch work. When you’ve gotten the hang of the mass point technique, you can compute the necessary masses “as you go.”

One set of mass assignments is uncircled and the other set is circled. After the fact, I tried to indicate which parts of the computation correspond to which line segments in the figure.

For more practice with mass points, there is a Summer Fun problem set devoted to the technique by Johnny Tang in the latest issue of the Girls’ Angle Bulletin (Volume 7, Number 5). Also, check out these lecture notes by Tom Rike for a talk he gave at the Berkeley Math Circle. At MathWorld, there’s a slightly different approach to proving Marion Walter’s theorem that also essentially uses mass points. There are many other proofs that use different techniques, such as the one outlined at the end of this handout by James King.

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A paraboloid of revolution adorns the cover. Anna investigates cross sections of paraboloids in this issue’s Anna’s Math Journal. It sure feels like Anna is embarking on an interesting mathematical journey with this new topic. We hope you’ll be inspired to follow-up on her work.

However, first up is the concluding half of our interview with University of Oregon Professor Emerita Marie Vitulli. Read some of the ways she thinks gender bias in mathematics can be countered.

Next, Akamai Technologies computer scientist Kate Jenkins discusses algorithms that find the “maximal subsequence” of a sequence. Her first part closes with an interesting challenge. Can you find a solution to her challenge before she gives it in the next issue?

Brit Valeria Golosov presents a fictional account of how she imagines that Brahmagupta derived his famous formula for the area of a cyclic quadrilateral. Valeria is entering her final year of secondary school in London.

This issue’s Math In Your World was specifically requested by Vida John. We love receiving content requests from members and subscribers. This Bulletin is written for members and subscribers and members and subscribers are welcome to control Bulletin content by emailing us comments and suggestions. Please don’t be shy about emailing us about anything to do with math! We also welcome and encourage all members and subscribers to send in solutions to the Summer Fun problem sets. We might even publish your solution in the Bulletin.

This summer’s batch of Summer Fun problem sets address magic squares, mass points, quadratic reciprocity, and permutations. Contributors include Johnny Tang and Cailan Li, both recent high school graduates who will be heading to college this coming fall. The central theme of Volume 7, Number 3 of the Bulletin was the concept of center of mass which underlies the technique of mass points. In that issue, we didn’t have enough room to include many problems to practice the technique. So that’s one reason why we included a problem set on mass points. The problems range from introductory level to some that will hopefully entertain those experienced in the technique. Cailan’s problem set takes readers from the rudiments of modular arithmetic all the way through a proof of Gauss’s Law of Quadratic Reciprocity, following a proof by D. H. Lehmer. The set on permutations culminates with a result of Zolotarev that links signs of certain permutations to the Legendre symbol introduced in Cailan’s problem set.

To whet your appetite, suppose *A*, *B*, and *C* are the angle of a triangle. Can you prove that

with equality if and only if the triangle is equilateral? For a spiffy way to prove this, check out the Summer Fun problem sets!

We conclude with a brief account of a wonderful field trip we took to MIT’s Department of Aeronautics and Astronautics, which was generously organized by Professor Karen Willcox,

We hope you enjoy it!

Finally, a reminder: when you subscribe to the Girls’ Angle Bulletin, you’re not just getting a subscription to a magazine. You are also gaining access to the Girls’ Angle mentors. We urge all subscribers and members to write us with your math questions or anything else in the Bulletin or having to do with mathematics in general. We will respond. We want you to get active and *do* mathematics. Parts of the Bulletin are written to induce you to wonder and respond with more questions. Don’t let those questions fade away and become forgotten. Send them to us!

We continue to encourage people to subscribe to our print version, so we have removed some content from the electronic version. Subscriptions are a great way to support Girls’ Angle while getting something concrete back in return. We hope you subscribe!

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Would anyone know how to construct a cube in one point perspective?

This is an excellent question because it isolates an important, simplified situation that enables one to study key aspects of perspective drawing.

(If you are completely new to perspective drawing, I’d suggest working through the Summer Fun problem set on pages 21-22 of the June, 2013 issue of the Girls’ Angle Bulletin.)

**What is the “point” in a one-point perspective drawing of a cube?**

The “point” in “one-point perspective” refers to a vanishing point. So before we draw cubes, let’s review the concept of vanishing point.

Vanishing points correspond to sets of parallel lines in the following way. Imagine a bunch of parallel lines in space. Assume that these lines are not parallel to the drawing canvas. A perspective drawing of these lines will look like a bunch of spokes emanating from a common point. This common point on the canvas is the vanishing point associated to the parallel lines in space. Each parallel line is drawn as a ray emanating from the vanishing point.

**Where should the vanishing point be on the canvas?**

If we fix a set of parallel lines, the exact location of their vanishing point depends on from where exactly we are meant to view the drawing. Perspective drawings are designed to be viewed from a specific location. The further the observer moves away from this ideal vantage point, the more distorted the drawing will seem. When you make a perspective drawing, you must decide where exactly you want an observer to place her eye when viewing your drawing.

The famous painting *The Ambassadors* by Hans Holbein features a highly distorted skull. It appears distorted because that skull is a perspective drawing designed for viewers off to the right of the painting, whereas the rest of the painting is designed to be viewed from a standard location directly in front.

Only after you have decided where you want the observer’s eye to be can you determine the location of vanishing points. To find the vanishing point associated to a set of parallel lines, place your eye at the specific viewing spot you want your drawing to be viewed from. Then, look at the canvas in the direction of the parallel lines. The point on the canvas directly in front of your eye is the vanishing point. In other words, the vanishing point for a set of parallel lines is where the one (of these lines) that passes through your observing eye intersects the drawing canvas. If you imagine following a point travelling along another of these parallel lines into the distance with your eye, your line of sight will converge to the special line that passes through your eye. In this way, we can see that drawings of such lines are rays emanating from the vanishing point.

Parallel lines that are parallel to the plane of the drawing canvas have no vanishing point on the canvas. When you look in their direction, your line of sight won’t even intersect the drawing canvas, even if the drawing canvas were an infinite plane. (I’m assuming that we’re drawing on a flat canvas.)

**Back to the cube**

A cube has 12 edges. These can be arranged into 3 groups of 4 mutually parallel edges. These 3 groups potentially define 3 vanishing points. If the edges in 2 of these groups are parallel to the canvas, then only the 3rd group will yield a vanishing point. This, then, is the setup for a one-point perspective drawing of a cube. A one-point perspective drawing of a cube arises when you make a perspective drawing of a cube that has 8 edges parallel to the canvas (or, equivalently, two faces parallel to the canvas).

In a cube, edges meet at right angles. When 8 of the edges of the cube are parallel to the drawing canvas, the other 4 will be perpendicular to it. The vanishing point for these 4 edges is therefore located right where the viewer of the drawing looks when looking straight onto the canvas. I’ll refer to this point as *V* for the rest of this blog post.

Let’s begin drawing.

Let’s start by placing a dot on the canvas directly in front of the observer’s eye to mark the point *V*.

Next consider the front face of the cube. This face forms the base of a square pyramid whose apex is the observer’s eye. The drawing canvas intersects this pyramid in a plane parallel to the front face. Using properties of similar figures, we can deduce that the front face gets projected onto a square. Let’s now draw in this square. (If the front face of the cube is in front of the drawing canvas, then the sides of the square pyramid must be extended to meet the canvas.)

The 4 edges of the cube perpendicular to the canvas are drawn as line segments that extend from each corner of the square (that represents the front face) toward their vanishing point, which is *V*. This brings up the next problem: How long should these line segments be?

The answer is that it depends on how far away from the canvas the observer’s eye is supposed to be. If the observer is to view the drawing from far away, then the 4 perpendicular edges will be represented by short line segments. But if the observer is supposed to view the drawing from up close, then the 4 perpendicular edges will be represented by line segments that extend almost all the way to the vanishing point. This fact is illustrated below.

The black square represents the front face of a cube. Close one eye and situate your other eye directly over the vanishing point. When you are very far away from the image, both the green and the blue squares will appear to be too far from the black square in depth to depict the back face of a proper cube. As you move your open eye closer to the canvas (but always staying directly over the vanishing point), there will be a place where the black and green squares do look like the front and back of a perfect cube, while the blue square appears far in the distance. Move your open eye even closer to the drawing, and the apparent distance between the black and green squares will appear to shrink, and at some point, the black and blue squares will appear to form the front and back faces of a perfect cube.

Suppose you have decided that the viewer should view the drawing from a distance *D* away from the canvas. Then the line segments representing the 4 edges perpendicular to the drawing canvas must be a very precise length to properly depict a cube. How do we figure out how long to draw these line segments?

I’ll first explain one method to compute this in the case of a cube that is upright and not rotated relative to the ground. With such a cube, we can refer to its bottom face. Consider one of the diagonals of the bottom face. For definiteness, let’s pick the diagonal that points into the distance to the right. In the drawing, the diagonal will extend toward some vanishing point *V*’ different from *V*. Where is *V*’? To find *V*’, we put our eye at the observation point, which is exactly a distance *D* directly in front of *V*, and we look in the direction of the diagonal. The diagonal bisects the right angles in the base of the cube, so to turn from the direction perpendicular to the canvas to the direction of the diagonal, one must turn right by 45 degrees. We will then be staring directly at *V*’.

The observer’s eye, the point *V*’, and the point *V* form the vertices of a 45-90-45 isosceles right triangle with *V* the right-angled vertex. Therefore *V*’ is located exactly a distance *D* to the right of *V*, as shown in the diagram below.

When we draw in the diagonal, it intersects the line connecting *V* to the lower right corner of the square depicting the front face precisely where we must draw the lower right corner of the square that depicts the back face. This intersection is indicated by the red arrow in the diagram above. That’s how we can determine the exact lengths of the 4 line segments representing the 4 edges of the cube perpendicular to the drawing canvas. The rest of the cube can now be drawn in, as shown below. If you close one eye and place your open eye exactly the distance *D* directly over point *V*, you’ll be rewarded with the illusion of a perfect cube.

To handle cubes that are not upright (but still in one-point perspective), you can rotate everything about the axis that passes through the observer’s eye and *V* until the cube becomes upright, draw that, and then rotate everything (including the drawing canvas) back to the original position.

There are other ways to construct one-point perspective drawings of cubes and we encourage you to invent your own. In fact, we have avoided giving a step-by-step procedure or lots of diagrams in the hope that you’ll create your own.

**Conclusion**

Even though the one-point perspective represents the simplest kind of perspective drawing of a cube, it already reveals the critical importance of designing the drawing for a specific observer location and the value of the vanishing point concept. Elaborate cityscapes can be constructed in one-point perspective. Joel Babb‘s drawing for the cover of Volume 6, Number 6 of the Girls’ Angle Bulletin is essentially a 1-point perspective drawing. For an example of constructing a 2-point perspective drawing of a perfect cube, check out **Rowena**‘s article in the April, 2010 issue of the Girls’ Angle Bulletin.

If you make a one-point perspective drawing of several cubes of different sizes, we’d love to see it. Feel free to scan it in and send it to us at girlsangle@gmail.com. When a drawing shows more than one cube, it becomes crucial to consistently design the drawing for a specific observation point. If you don’t, you’ll end up with a drawing of bricks of different shapes instead of a drawing of cubes.

Thank you for your question, Jamie!

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The cover features a planar configuration by Leah Berman, associate professor of mathematics at the University of Alaska Fairbanks. There are 240 lines and 240 points arranged so that each line contains 6 of the 240 points and each point sits on 6 of the 240 lines. More images of planar configurations by Leah and Nadine Alise can be found in *Mathematical Buffet**.*

This issue’s interview is with University of Oregon Professor Emerita Marie Vitulli. In this first part of a 2 part interview, we learn about Prof. Vitulli’s field and how she got into mathematics.

Special thanks to Professor Emeritus Thomas Moore for contributing a problem about Pythagorean triples. If you haven’t heard of Pythagorean triples, Prof. Moore gives a brief introduction and more challenges in *Pythagorean Triples Challenge*.

Angles pervade much of this issue. In *Learn By Doing*, Addie Summer covers the basic of angle measure. Then, Lightning Factorial follows Emily and Jasmine as they use angles to design a stained glass window. Finally, in this issue’s *Math In Your World*, I write about one of John and Jane Kostick’s latest creations, which they dub the Quintetra Assembly. To compute the necessary angles needed to create a wood sculpture for this amazing polyhedron, several angles must be computed. In this article, I sketch how to determine these angles and include a net, courtesy of the Kosticks, for Jane’s Quintetra block, 30 of which can be used to build a model of the Quintetra Assembly.

Finally, Anna continues her investigation of *x* to the *x*.

We hope you enjoy it!

Finally, a reminder: when you subscribe to the Girls’ Angle Bulletin, you’re not just getting a subscription to a magazine. You are also gaining access to the Girls’ Angle mentors. We urge all subscribers and members to write us with your math questions or anything else in the Bulletin or having to do with mathematics in general. We will respond. We want you to get active and *do* mathematics. Parts of the Bulletin are written to induce you to wonder and respond with more questions. Don’t let those questions fade away and become forgotten. Send them to us!

We continue to encourage people to subscribe to our print version, so we have removed some content from the electronic version. Subscriptions are a great way to support Girls’ Angle while getting something concrete back in return. We hope you subscribe!

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Much of this issue is about the center of mass, including Julia Zimmerman’s cover drawing which features a mobile fantasy. In *Math In Your World*, we cover the basics involved in ensuring that a mobile will balance properly. You could use this knowledge to get your pet hamster to lift your car. In *Center of Mass*, we explain the “piecemeal” property of the center of mass: that you can compute the center of mass part by part. This property forms the basis of the problem solving technique known as “mass points.”

To understand the center of mass and torque, it helps to learn about vectors. So Robert Donley, a.k.a. Math Doctor Bob, introduces vectors in his second installment of *Learn by Doing*. If you work through the problems in his column, you’ll learn about vector addition, scalar multiplication, and the dot and cross products. Because Bob wanted to lay the groundwork for understanding physical vectors such as force and torque, he confined himself to discussing vectors over the real numbers.

In our interview, meet University of Michigan Professor of Mathematics Karen Smith. Prof. Smith discusses a wide range of topics, including math that interests her, how she goes about solving math problems, and how she got into math. Her route to math is rather exceptional in that after graduating from Princeton with a Bachelor’s degree, she taught in the school system before going to math graduate school. She is a recent Clay Scholar and she contributed an autobiographical essay to the book *Complexities: Women In Mathematics*, edited by Bettye Anne Case & Anne M. Leggett. In this interview, she even describes an unsolved problem which middle school students could begin to explore.

In *Anna’s Math Journal*, Anna launches into an exploration of the function after being asked the following question: For what number *x* other than 1/2 is ?

“You” helps 3/7 find a nifty solution to a curious dilemma in Coach Barb’s Corner…involving fractions, of course!

Finally, if you like dissection problems, see if you can find a nice way to tile a regular hexagon into pieces that can be rearranged to form an equilateral triangle. **Henri**, a student at the Buckingham, Browne, and Nichols middle school in Cambridge, was presented with just this challenge, and we exhibit his beautiful solution inside. This problem is an instance of the Wallace-Bolyai-Gerwien theorem which says that any polygon can be dissected into parts that can be rearranged to form any other polygon of the same area. Even though there is a general construction to create such tilings, there’s still the challenge of finding nice ones, and **Henri**’s is quite elegant.

We hope you enjoy it!

*do* mathematics. Parts of the Bulletin are written to induce you to wonder and respond with more questions. Don’t let those questions fade away and become forgotten. Send them to us!

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Recall that the chain rule tells us how to compute the derivative of a composite function. Specifically, suppose and are differentiable functions. Let be the derivative of with respect to and let be the derivative of with respect to .

Then .

One way to gain an intuitive feel for the chain rule is to think of an example of a composite function that you understand well and then see that what you think about the example is consistent with what the chain rule says. Here, I’ll describe a scenario that I often use to explain the chain rule and seems to help.

Imaging filming a car racing along a long straight road. The film begins just as the car takes off from the starting line. The car accelerates rapidly. Maybe it slows down to avoid a passing chicken. Whatever. Let represent the distance the driver has traveled from the starting line at time .

The driver’s speed at time is given by the derivative .

Now let’s watch this film in a movie theater. If we operate the projector normally, you’ll see the race car traveling along the road just as if you were watching the actual race car, and you would judge the speed of the race car to be just as you saw it when you filmed it. (There’s a possible ambiguity here that I want to make sure does not become a point of confusion. One calculates the speed of the car by paying attention to the distances the car travels along the road and *not* to the distance traversed by the image of the car along the theater screen, which could even be no distance at all if the camera is following the car.)

But what if we feed the film through the projector twice as fast as normal?

Well, then, like those sped up comedy skits, you’d see the race car zipping along seemingly twice as fast as when you filmed it. It would finish the race in half the time.

And what if we feed the film through the projector half as fast as normal?

Then you’ve entered slowmo. The car creeps along at half the speed it was actually going.

What if the reel gets stuck in the projector?

Then you see the race car get stuck in place.

And what if we feed the reel through the projector in reverse?

You get the idea.

The situation is modeled by a composite function , where is the function that tells the time along the film as a function of the actual time that passes as you sit in the theater watching. Under normal theater operations, . That is, film time and actual time are in sync. In slowmo, we might have .

The derivative tells us what we’d judge the speed of the race car to be by watching the movie.

The common chain rule error is to write . But if you write this, you’re saying that the speed of the race car on the screen does not depend on how we feed the reel through the projector. But we know that the apparent speed of the race car is scaled exactly by . That is, we intuit that , just as the chain rule dictates!

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The Girls’ Angle WIM Video series is the brainchild of Girls’ Angle director Elisenda Grigsby. WIM Videos are often shot on the spur of the moment and feature women in mathematics explaining some piece of math that excited them when they were students. We hope to create a large gallery that will showcase the diverse group of women working in mathematics today. If you support this concept, please consider making a charitable donation to Girls’ Angle. Girls’ Angle is a 501(c)3 and every amount helps!

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I’ll use the vertex labels to also denote the corresponding angle measures.

Take the sines of all 3 angles in the triangle: , , and .

Would you believe it if I told you that these three sines satisfy the triangle inequality?

Even more, would you believe it if I told you that a triangle with sides of length , , and is similar to the original triangle?

Please think about this before reading further.

SPOILER ALERT!

Do you recognize the statement as implied by the law of sines?

In fact, the law of sines even tells us that the similarity ratio is equal to the diameter of the circle circumscribed about the original triangle.

We can use this understanding to prove the law of sines. First, note that in a circle of diameter 1, an inscribed angle measuring degrees subtends a chord of length (see figure at left). To see this, note that the blue chord has length because the central angle of the arc subtended by the inscribed angle with measure degrees has measure degrees.

That means that if we scale a triangle so that its circumscribed circle has unit diameter, the lengths of the 3 sides of the scaled triangle must be the sines of its 3 angles, and since we know what the corresponding sides are, we get the law of sines, complete with similarity ratio being the diameter of the circumscribed circle.

Just for kicks, we can use the law of sines to give the following proof of the law of cosines.

In a triangle, the angles add up to 180 degrees, so . Using this, we can compute using standard trigonometric identities:

And that’s the law of cosines. (Because the equation is homogeneous in the sines, the law of sines informs us that we can replace with , with , and with , to get an equivalent equation, which is how the law of cosines is usually written.)

In general, if you have an equation that is homogeneous in the lengths of the sides of a triangle, you automatically have an equation in terms of the sines of the angles of the triangle, via the law of sines. For example, by dropping the altitude from vertex in the yellow triangle, we see that . The lengths all appear with exponent 1, so, via the law of sines, we can replace with , with , and with to get , and since , we also have , and, voila!, out pops the trigonometric identity for the sine of a sum. (Here, we needed angles and to be positive and have a sum less than 180 degrees, so one has to think a little more to get the trig identity without restriction.)

Ptolemy’s theorem relates the lengths of the sides of a cyclic quadrilateral and the lengths of its diagonals. The equation is homogeneous in these lengths, so we can immediately convert to a trigonometric identity. We label the vertices and angles as in the red circle. Notice that because they subtend the same arc. So the 4 unlabeled angles in the corners are equal to the labeled ones in some order.

Ptolemy’s theorem says that . Using the law of sines (which can be envisioned as scaling the figure so that the circle has diameter 1), this identity involving lengths is converted into a trigonometric identity:

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Turning this around, Ptolemy’s theorem can be proven by proving this trig identity, and to prove it, one only needs a few basic trig identities and the fact that . Try it and see for yourself!

Now, check out problem #15 from the 2013 AIME II.

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How many dots are in the base?

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