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!

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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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:

.

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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In *Bisection Envelopes*, a paper to appear in the journal *Involve*, Noah Fechtor-Pradines proves many interesting properties of the envelope of the set of lines that split a planar shape into two halves of equal area. The cover shows the bisection envelope of the regular pentagon and in *Math Buffet*, you can see more images of bisection envelopes that illustrate just a few of the properties Noah discovered. Try to sketch the bisection envelope of a semicircle, then check out the one in the Bulletin.

In this issue’s interview, meet Associate Professor of Mathematics at Loyola Marymount University, Alissa Crans. She gives a lot of great advice and also tells us how she got into math. She explains some of her goals as the Associate Director of Diversity and Education at the Mathematical Sciences Research Institute and her work at Pathways.

This fall, Akamai mathematician Kate Jenkins visited the Girls’ Angle club and explained Dijkstra’s algorithm for finding optimal paths in graphs. You can try your hand at applying Dijkstra’s’ algorithm in *Mole Map**, USA*. She also described what it is like to work at Akamai, which creates products that make the internet run more efficiently.

Robert Donley, a.k.a. Math Doctor Bob, launches our new column, *Learn by Doing*, with a problem concerning dice. In *Learn by Doing*, we are going to present mathematics by posing problems. We hope you make a point of learning math actively, with pencil and paper in hand. If you particularly enjoy learning by solving problems, check out *Combinatorial Problems and Exercises*, by László Lovász.

Two hurdles to overcome in mathematics are gaining facility with variables and thinking in higher dimensions. For variables, Tim Chow, a mathematician who works at the Center for Communications Research, brings us *It is a Variable!* And for the fourth and higher dimensions, we offer *The Fourth Dimension*. It’s truly remarkable that we are able to think about objects that cannot be realized in the physical world but only exist in our minds. In fact, we define mathematical objects with such precision and detail, that, in some sense, they can even seem more real than physical objects!

In our regular columns, Anna continues looking at products of consecutive integers with respect to whether or not they can be perfect squares and Lightning Factorial discusses Fermat’s Principle of Least Time in *Math In Your World**.*

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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Given a circle of radius , let

Abe a point at a distance from the centerCof the circle. LetBbe the point on the circle nearest to pointA. A line passing through the pointAintersects the circle at pointsKandL. What is the maximum possible area for ?

(The AIME problem, in typical AIME-style, asks the contestant to extract a single integer after putting the answer in a particular form.)

Here’s an illustration of the situation.

We must find the maximal area of the red triangle.

**Solution 1**. If we regard *LK* as the base of the triangle, the height is the length of the perpendicular to *LK* dropped from *B*. If we also drop the perpendicular to *LK* from the center *C*, we get a pair of similar right triangles (triangles *ABB’* and *ACC’* in the figure below).

From the figure, we can see that *BB’* : *CC’* as , and this is also the ratio of the areas of the two red triangles. Hence, maximizing the area of one is the same as maximizing the area of the other.

The area of isosceles triangle *CKL* is , and is maximized when is right. This configuration can be achieved because as the line *AL* rotates down from the tangent, the arc of the circle cut off increases from 0 degrees to 180, and will hit 90 at some point. At this point, the area of triangle *CKL* is 13/2, and so the answer to the problem is .

**Solution 2**. This time, let’s realize the area of *BKL* as the areas of the green and blue isosceles triangles, less the area of isosceles triangle *CBL* in the figure below.

This area can be expressed as . So we have to maximize subject to the constraint that *a* and *b* are related by the fact that *L*, *K*, and *A* are colinear. Using trigonometry, we can readily compute the Cartesian coordinates of points *L*, *K*, and *A*, and use a proportion to express the fact that these 3 points are colinear:

.

This can be rearranged and simplified as follows:

or, using the angle-sum formula for sine,

.

We substitute this into our formula for the area of triangle *BKL* and find that the area is

.

This is maximized when is right and its maximum value equals what we got for the first solution (as it should!).

* * *

Would you recognized essentially this same problem if you are asked to maximize instead?

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The cover shows an example of (quasi) self-similarity in nature: the Romanesco broccoli, and alludes to this issue’s Math Buffet. Flip over the cover and meet Anne Shiu, an L. E. Dickson instructor in the Department of Mathematics at the University of Chicago and a National Science Foundation Postdoctoral Fellow. Dr. Shiu does research in mathematical biology, which is a vast and exciting field.

Would you believe that it is possible to create 3-D images using a camera with a single lens? Antony Orth, a graduate student in the School of Engineering and Applied Sciences at Harvard, explains how in *3-D Movie Technologies*.

UC Berkeley astrophysicist Aaron Lee follows up his introduction to conic sections with *Cosmic Conics **II*, this issue’s installment of Math In Your World. As you read this, Comet ISON hurls toward the sun, where it will venture within one solar diameter of the sun’s surface at the end of November. For more on Comet ISON, visit NASA’s Comet ISON Observing Campaign.

In addition to regular columns *Anna’s Math Journal*, *Coach Barb’s Corner*, and *Notes from the Club* (where you can read about Dr. Anna Frebel’s recent visit to the club), we have our first installment of *Member’s Thoughts* in quite a while. See how one of our members rediscovered some beautiful facts about the Fibonacci numbers by asking and seeking the answer to a very natural math question.

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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A friend sent me the challenge at right that his daughter was given in school.

What would you answer?

(Please decide how you would answer before reading further.)

The reason he sent me this, and the reason I’m blogging about it, is because this is a very good example of how there can be math instruction that ends up being counterproductive.

The teacher sought the specific answer of 10, which can be realized as follows:

O O O K R J O O O O

where the front of the line is to the left and O’s represented unnamed people. All other answers were marked wrong.

But there are many configurations that satisfy all the constraints with different numbers of people in line, such as the following:

O J R K O O,

or

J O R K O,

or

O O O K O O O O O O O R O O O O O O O J O O O O.

Math classes are normally a great way to learn how to be thorough. Countless math problems demand meticulous attention to finding all possibilities. Such problems often press students to eliminate false assumptions, which is a valuable skill applicable in many situations outside of mathematics. This could have been such a problem, but the way it was handled, it became the opposite.

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Sometimes it takes a little work to find the math. In the completed crossword grid, find 8 singular nouns with mathematical meaning. Each of the 8 mathematical terms has at least 5 letters, appears horizontally, and is split by a single black square. Send these 8 mathematical terms along with your contact information to girlsanglepuzzler “at” gmail.com. We will randomly draw from the correct answers received by midnight on October 1, 2013 to select a “winner” and send the winner a small prize. Girls’ Angle members will be put into a separate pool and have a chance to win an Intel web computer.

Across

1. Hawaiian statue

5. MIT’s Tim

11. Grays

15. Kemeny and Kurtz’s language

16. Above the timberline

17. Simpson’s Burns

18. Brick

19. Traveled

20. Z and Ω

21. Reptile

22. Them’s opponent

24. It’s found over a portal

26. Tic-tac-toe side

27. Really understand

28. Feed the kitty

29. Dad

31. Location of the tallest building in the Western hemisphere

35. Less rough

38. Yale alum

39. City where Tao teaches

40. Card game that models a 4 dimensional affine space over

42. Put back in the oven

43. Word after time or washing

46. Edition or release (abbr.)

48. Typesetting software by Knuth

49. Where to find a surgeon

50. Drops on the grass

52. 1,000 and 1,000,000, briefly

54. “Poor baby!”

56. He directed Life of Pi

58. Tennis do-over

60. Entreat

63. 1 + 1 = 3 and 2 + 2 = 5

66. Young woman, sometimes

68. Six

69. Center of gravity

70. Bend light

72. She sees many red eyes

75. Latin and

76. Nest eggs

78. Charged particle

79. Sometimes follows home

80. Burning

82. Longest river in Italy

83. Semiconductor maker

86. Cosmos host

88. sin *s* / cos *s*

90. Scrabble 3-pointer

91. Butter substitute

92. Vest

94. Amorphous things

95. Winged stinger

96. In this place

97. Bird

Down

1. Voilà!

2. Exercise refreshment

3. Block

4. Mr. ChemCake, sometimes (see volume 4, number 4 of the Girls’ Angle Bulletin)

5. A French bread

6. __ Salvador

7. Interest rate for an entire year

8. String instrument

9. The last part of a story

10. Type in again

11. Ending for do, not, and change

12. Objective

13. Type of curve

14. It flew faster than sound

15. Sack

23. Black, Red, and Yellow, for example

25. Belief

27. Can precede up, out, and in

29. Green bullet

30. Work of a Constable

31. Office note

32. Winglike

33. Formerly Clay

34. It borders Ore.

36. Unifying idea

37. Superman’s nemesis

41. The smallest whole number that isn’t a single digit

44. The Leroy P. Steele Prize

45. Shocking creature

47. Steal

51. Betroth

53. Forms a Pythagorean triple with 24 and 25

54. Hotshot fighter pilots

55. Sharpen

56. Snoopy says it

57. Special cup

59. Follows green or iced

61. British title

62. Expulsion

63. Mistaken mistake?

64. Army equivalent of academia’s emer.

65. Remove

67. Plane on radar

71. Meat and peppers in cornmeal

73. After formal tie action

74. Kindly

77. Title for Nadal

80. God of war

81. Fosbury ___

83. Peacekeeper or Minuteman

84. Napoleon stayed here for 300 days

85. Less less?

86. Calf’s mom

87. Pie – mode connector

89. Short suite

93. Ciao

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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