Volume 8, Number 3 of the Bulletin kicks off with the first part of a 2-part interview with Hamilton College Assistant Professor of Mathematics Courtney Gibbons. In this first half, she discusses how she got interested in math, tells us about how she learns, studies, and creates math, and explains the notions of “fixing set” and “fixing number.” She also tells us about some of her goals as a mathematician.

The best way to learn math is to do lots of problems: after all, that’s what math is!

– Courtney Gibbons

Next, Emily and Jasmine continue their exploration of *n*-pointed stars, or, more accurately, convex (*n*, *k*)-stars. In this installment, they succeed in finding and proving a formula for the sum of the tip angles of such stars. The cover of this issue shows examples of (17, *k*)-stars. If you can find a good friend to explore mathematics together with, it can be a lot of fun and quite rewarding. What do you notice about the (11, 5)-star at right?

Next up, our second installment of Meditate to the Math which features math related to cutting a chocolate bar in half along one of its diagonals.

Anna continues her investigation of cross sections of the graph of *z* = *xy* and finds a cool family of hyperbolas.

Konstanze Rietsch of King’s College London contributes a curious number puzzle borrowing characters from Lewis Carroll to state it.

In our second installment of our mini-series on the derivative, we explore basic properties of the derivative and then apply them to deduce the derivatives of any polynomial.

This issue’s Learn by Doing explains two different ways to find all primitive Pythagorean triples and then relates the two methods.

What we hope is that the Bulletin induces you to *do math*. That’s why this issue is filled with math questions and content that shows how others do math.

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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Both problems involve a snowball in the shape of a perfect sphere. Mark a point on the surface of the snowball. The snowball rolls without slipping. Also, there is no unnecessary rotation. (In other words, the snowball rotates only around an axis perpendicular to the plane of the great circle that is tangent to the path that the snowball travels.) In each problem, the snowball starts with the marked point touching the surface it is rolling on.

**Snowball Problem 1**. If the snowball rolls about on a plane, what is the locus of all the points in space that the marked point can visit?

**Snowball Problem 2**. If the snowball rolls about on another sphere of the same radius, what is the locus of all the points in space that the marked point can visit?

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After reading *Stained Glass Angles* in Volume 7, Number 4, Marion Walter happened upon an amazing stained glass window by John Rose at the Eugene Public Library in Oregon. That inspired her to suggest a *Math Buffet* column featuring mathematically inspired stained glass windows. One thing led to another, and in this issue, we feature stained glass windows from 7 designers. The cover shows a detail of a stained glass window designed by Millie Wert, a graduate of Harpeth Hall. (Added January 27, 2015: For more math inspired stained-glass windows by students of Thaddeus Wert at Harpet Hall, please visit his blog.)

Next, comes the second half of Margo Dawes’ interview of Cathleen Morawetz, Professor Emerita at the Courant Institute of Mathematical Science at New York University.

Emily and Jasmine continue discussing *n*-pointed stars. Last issue, they showed that the sum of the tip angles of any 5-pointed star is 180 degrees. This time, they try to generalize to *n*-pointed stars, like the 7-pointed star at right. They end up in a discussion about definitions: just what is an *n*-pointed star anyway?

Pull out your compass and straightedge for this issue’s *Learn by Doing*. All of our *Learn by Doing* installments offer the opportunity to explore math through problem solving. The problems start out with the aim of gaining familiarity with the topic at hand and gradually probe deeper and become more challenging. Can you make it all the way through? If not and you’re a subscriber, don’t hesitate to email us for help.

Anna had so much fun exploring cross sections of a paraboloid of revolution, she decided to investigate cross sections of the graph of *z* = *xy*.

If you want to tell us anything about your mathematical explorations, we encourage you to email us about it.

In this issue, we’re beginning a mini-series on the derivative. There are many fine books and textbooks devoted to the subject of calculus. For example, there’s Courant and John’s *Introduction to Calculus and Analysis, Volume 1,* Rudin’s book *Principles of Mathematical Analysis*, and many, many more. Instead of attempting to produce our own textbook, we’re going to give a non-rigorous, but hopefully illuminating explanation. If one grasps local linearity, many results in calculus can be intuited and we hope to show how.

There’s more, including: a card game we call “Full Deck” that addresses the critical technical skill needed to learn how to multiply and divide with decimal numbers, an introduction to the choose function notation, and Notes from the Club where you can read a summary of Prof. Cornelia Van Cott’s recent visit to the club.

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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At Girls’ Angle, we do love digits!

Solve this crossword alone or with friends and relatives. Scan in or take a digital photo of your completed grid and send it to girlsanglepuzzler “at” gmail.com. We will randomly draw from the correct answers received by midnight on **January 4, 2015** to select a “winner” and send the winner a small prize. Girls’ Angle members will be put into a separate pool for a different prize.

*This crossword puzzle is dedicated to Connie Chow.*

Across

1. Rest

6. Fall mo.

9. Sub

14. Breathing

15. Greek letter

16. Loaded with marinara

17. Hushed

18. Goal

19. Vermeer product

20. Sunrises and hatchings

23. Soft cheese

24. It often gets beat

25. Rights wrongs

27. City in 54D

31. Donor

33. Unlike Smith

34. Acts like a ghost

36. Where cultures are made

39. Für Elise form

41. Type of health plan: Abbr.

42. Rise over run

44. Increases

45. Forest and jungle

48. Divisible by 2

49. Catches up to

50. Most frigid

52. Triangular sail, or, a 15-year-old who lives near the *Endeavour*?

55. Hamburg’s commission

56. Drive the getaway car

57. Equal areas in equal times

64. Burdens

66. Overly

67. Baking bean

68. Bad color combination

69. Trig function

70. Seaweeds

71. Veracity

72. Time unit’s plural form: Abbr.

73. Name with a double consonant

Down

1. Docile

2. Graduate

3. Flying diamond

4. Follows for or what

5. Pentagon property

6. Type of ape

7. Elegant

8. *The Princeton Companion to Mathematics*, for example

9. Hawaii is the last of these: Abbr.

10. Samson’s bane?

11. No longer useful

12. Corrosive

13. Kids

21. They can be bruised

22. Oft used file command

26. Place to eat

27. Brown

28. Continuous image of the circle

29. Writes

30. Total

31. Most exciting match

32. Fe

35. Soften in the mouth

36. Zero to Federer

37. What cosine does to sine

38. Not straight

40. Shrek

43. Hawaiian graduation gift

46. File’s partner

47. Tails

49. Loathe

51. Someone you only drive to work with?

52. Fastener

53. Haliotis

54. State of 27A

55. Clean with string

58. Draw

59. Low quality

60. Shelf with a view

61. Plastic brick

62. Not home

63. Gave more importance

65. Short

Good luck!

(We will not use your contact information for any purpose other than to deliver your prize, should you win. After the winner has been selected, all emails received will be promptly deleted. At the winner’s discretion, we will let you know who won. Anyone who makes more than one submission will be disqualified! Sorry! Also, this offer is only valid in those states in the United States where such things are legal. There is no fee to enter this puzzle contest.)

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This time, we’re leaving it up to readers to guess what our latest cover represents.

This past summer, Girls’ Angle program assistant Margo Dawes traveled to New York City to interview Cathleen Morawetz, Professor Emerita at the Courant Institute of Mathematical Science at New York University.

Last year, Professor Jennifer Roberts, the Elizabeth Cary Agassiz Professor of the Humanities at Harvard University, wrote an essay for Harvard Magazine about “creating opportunities for students to engage in deceleration, patience, and immersive attention.” Inspired by that essay, and recognizing that this could be a valuable exercise in mathematics as well, we attempt to give you such opportunities in mathematics with our new column, *Meditate to the Math*. Our first installment features the 9-point circle. Instead of reading about the 9-point circle, we encourage readers to find a comfortable, quiet place, and contemplate a geometric figure. We hope this will be a way to take part in the process of mathematical discovery.

Next, follow Emily and Jasmine as they contemplate 5-pointed stars. If any of our members or subscribers have an exciting story of mathematical discovery of their own, we welcome you to tell us about it!

This issue’s *Learn by Doing* addresses quadratic residues. Last Summer’s batch of Summer Fun Problem sets included one on quadratic reciprocity by Cailan Li. But before quadratic reciprocity, there are lots of things to say about quadratic residues. We explore some of those neat properties here.

Last issue, Anna made a neat discovery about stereographic projection and paraboloids of revolution. As often happens with mathematical theorems, the first proof is messy and then spiffier proofs are found later. In this issue’s *Anna’s Math Journal*, Anna finds a much nicer proof and then applies the result to describe a few more observations about paraboloids of revolution.

While contemplating paraboloids of revolution, Anna also came upon a way to understand the radical axis of two circles. This observation seemed more convenient to write an article on because she came to this understanding without writing anything down. She explains in *Seeing the Radical Axis*. Lightning Factorial supplements her article by briefly defining the radical axis for readers not yet familiar with the concept.

A few weeks ago at the Girls’ Angle club, some members helped to simplify Lunga Lee’s excessively long descriptions of various functions. You can try your hand at this in *Function Madness*.

Also inside are another installment of *Math In Your World*, some exercises about real algebraic varieties (to follow-up on Dr. Zamaere’s introduction of them in her interview in the previous issue), and some notes from the club, which include a summary of Emily Pittore’s recent visit. Emily is a robotic vision engineer from iRobot, the maker of the Roomba vacuum cleaner.

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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Take a multiplication table and build a column of cubes over each entry. For each column, use as many cubes as the product it sits upon. The result is a “multiplication sculpture” or “multiplication tower.” The picture shows a 15 by 15 multiplication sculpture built by Jane Kostick in 2008. For more examples, check out Maria Droujkova’s photo collection at moebius noodles.

Last spring, club members at Girls’ Angle built a 4 by 4 version out of cubes that were 3 inches on a side. They thought about its properties, such as how many cubes make up an *N* by *N* multiplication sculpture.

We’re going to address this last question and end up with a nice, concrete proof of a famous algebraic identity.

Over the *x*, *y* entry, there are *xy* cubes, so we have to add up all products *xy* where *x* and *y* range over the values from 1 to *N*, and this totals

.

Wait! Where else does that expression occur?

It is also the sum of the first *N* (positive) perfect cubes! In math notation,

.

The implication is that the number of cubes sitting over the last row and column of the multiplication table must be . If we can show this directly, we’d have a nice, concrete proof of the sum of cubes formula.

The columns of cubes over the last row of the multiplication table form a staircase with steps of height *N*. Saw these *N* columns off and get a flat, staircase-shaped plank, *N* cubes long and cubes high. Next, saw off the columns of cubes over the last column (of the multiplication table). You’ll get an almost identical staircase-shaped plank. The only difference is that the last step of total height is gone because it was removed when the columns over the last row (of the multiplication table) were sawed off.

Turn one of these planks over, and the two planks will fit together perfectly to form an by rectangle, and an by rectangle has cubes in it!

Thus, a secret key to the identity

is hidden in plain sight in the multiplication table that many of us learned in elementary school!

To read about how Jane made this 15 by 15 multiplication sculpture, and see hints about more of its properties, check out the December, 2008 issue of the Girls’ Angle Bulletin, pages 12-14 and 25-28.

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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!

*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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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!

*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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