I haven’t seen this proof before, but there are so many proofs of the Pythagorean theorem that it wouldn’t surprise me if this has already been discovered. If you’ve seen this proof published elsewhere, please send email to Girls’ Angle and let me know.

I am aware of other “origami” proofs of the Pythagorean theorem, although the ones I’ve seen don’t particularly use origami so much as that they take some diagram used to illustrate a proof and then render the lines in the diagram as creases in paper.

This particular proof is not elegant, but it does show how one might discover the Pythagorean theorem through a natural line of exploration starting with an origami square. Also, it really uses the notion of an origami fold as its key ingredient.

All you have to do is think about what happens when you fold a corner of the origami square to a point along one of the further edges, as illustrated below.

That single fold produces 3 triangles and a quadrilateral.

Notice that the 3 triangles are all similar right triangles. (Here, we must use the fact that a square has 4 right angles.)

Since the entire model is determined by the parameter , a natural question to ask is: What are the lengths of all the other line segments in terms of ?

To get the answer, we will repeatedly apply two facts. First, that the 3 triangles are similar to each other. And second, that the edge lengths along a side of the square add up to the side length of the square, which we’ll go ahead and take to be one unit for now.

We’ll start by letting be the length of the other leg of . We’ll then express the other lengths in terms of and and find an equation that relates the two. By solving for in terms of , we will effectively have succeeded in expressing the lengths in terms of .

Here goes!

We add the label to the diagram:

The hypotenuse of and the leg of length make up what was the right side of our square, so the hypotenuse of has length :

One of the legs of extends to the full side of the square:

Using the fact that and are similar, we can find the other side lengths of :

One leg of and the hypotenuse of form what was the upper side of the origami square:

Next, we use similarity of with to see that its other sides have length:

and .

So far, we’ve used the right, bottom, and top sides of the original square. So now we use the left side. Studying the figure, we see that the original left side of the square has become one leg of and the last two sides we computed of . Thus,

.

If you solve for in terms of using this equation, you will find that .

Thus, the legs of are and and its hypotenuse has length .

Notice that as varies from 0 to 1, the ratio of the leg lengths of covers the entire range from 0 to infinity (i.e. the range of is when is restricted to the interval ). This means that a representative of every similarity class of right triangle can be folded by making a suitable choice of .

Therefore, by scaling the origami square if necessary, we can make congruent to any given right triangle.

The form of the expressions for the lengths of the sides of in terms of suggests the possibility of finding an explicit relationship between these 3 side lengths, especially since one of the side lengths is just , and if we can find such a relationship that is also homogeneous in these lengths, then the relationship will hold for *all* right triangles since a homogeneous relationship is preserved by scaling. And, indeed, the desired algebraic identity is , hence, the Pythagorean theorem. (Making note of this algebraic identity without knowing the Pythagorean theorem might be a bit of a trick, but it doesn’t seem unreasonable to think it possible.)

(Note that as well, and this corresponds to the equation , where and are the legs of a right triangle with hypotenuse of length . However, this identity is *not* homogeneous in the leg lengths, so it is not true that for all right triangles. The equation holds only for right triangles where , i.e. those folded from a unit square.)

In an earlier post, we described David Gale’s method for constructing Pythagorean triples. There, our aim was to find Pythagorean triples, not prove the Pythagorean theorem, so when we spoke of finding the lengths of , we had in mind a method that made use of the Pythagorean theorem. One could say that the content of this blog post is that you don’t have to assume the Pythagorean theorem there because you can deduce it instead. In other words, one-fold origami can deliver the Pythagorean theorem from scratch as well as all Pythagorean triples.

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Volume 8, Number 5 of the Bulletin kicks off with an interview of Ivana Alexandrova. Ivana is an Assistant Professor of Mathematics at the State University of New York, Albany. Among other things, she maintains a webpage of weekly problems for high school students. Check it out!

The topic of induction came up quite a few times this spring at the Girls’ Angle club, so next comes an article on this widely used proof technique.

This issue’s Learn by Doing features irrational numbers and culminates in a series of problems that let you reconstruct a proof of the irrationality of due to Charles Hermite.

Anna tackles one of Prof. Alexandrova’s weekly problems for high school students in Anna’s Math Journal, finding 3 different ways to solve the problem, which is to compute . Can you find your own solution?

Next comes our 4th installment on the derivative where we find the derivatives of the basic trigonometric functions. The way we deduce the derivative of sine is similar in spirit to the way we showed that the area under one hump of a sine curve is exactly 2.

Since this is our June issue, we include the 2015 Summer Fun problem sets. This batch contains problems pertaining to telescoping series (by Fan Wei), induction, the symmetric group (by Noah Fechtor-Pradines), and derivatives.

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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First, the problem:

Let be a diameter of circle . Extend through to . Point lies on so that line is tangent to . Point is the foot of the perpendicular from to line . Suppose , and let denote the maximum possible length of the segment . Find .

Here’s an illustration of the situation:

We must maximize the length of the red line segment .

As the diameter rotates, both and move. It’s typically easier to maximize the length of a line segment if one of its endpoints is stationary, so let’s instead maximize the length of the mid-segment of triangle , specifically, the mid-segment that joins the center of and the midpoint of (which are labeled and , respectively, in the following figure):

As diameter rotates, what path does trace? Since is the midpoint of , the point traces out the curve obtained by squashing the circle by a factor of 1/2 in the vertical direction:

reaches its maximum length precisely when is perpendicular to the tangent to the squashed circle at . The slope of the tangent to the squashed circle at is half the slope of the tangent to at point . Therefore, we want the slope of to be twice the slope of (since *is* perpendicular to the tangent to at ), which is the same thing as saying that we want to be twice .

Since is the midpoint of , we see that is maximized exactly when , i.e. when and trisect the segment .

Thus, is maximized when and . Hence, the maximal value of is .

Remembering that is half that of , we conclude that the answer to the AIME problem is .

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Volume 8, Number 4 of the Bulletin kicks off with the concluding half of our 2-part interview with Hamilton College Assistant Professor of Mathematics Courtney Gibbons. In this second half, one of the objects she describes are the Cayley graphs of groups. This inspired the creation of several Cayley graphs by members of Girls’ Angle, which are featured in this issue’s Math Buffet.

The cover itself also shows a Cayley graph of .

Next, Emily and Jasmine continue their exploration of *n*-pointed stars. This time, they follow-up on an observation that Emily made about the (17, 7)-star on the cover of Volume 8, Number 3. She noticed that the (17, 7)-star contains (17, *k*)-stars for *k* = 1, 2, 3, 4, 5, and 6. Together, they not only show that this is an instance of a general phenomenon, but they also find the relationship between the sizes of such embedded stars within stars.

Anna wraps up her study of cross sections of the surface *z* = *xy*.

In Part 3 of The Derivative, we explain both the Chain Rule and the Quotient Rule along the line emphasized in this series: local linearity. We hope that the Chain Rule, especially, appears “obvious” from this point-of-view. For another take on the Chain Rule, here’s an earlier post that uses movies to explain it.

Next, Stuart Sidney, Emeritus Professor at the University of Connecticut, entertains us with curious facts about palindromic numbers.

We introduce a mathematical variant of the classic Hot Potato game that can be used as a vehicle to explore quite a few math concepts and in quite a bit of depth.

Finally, in Notes From the Club, we give a brief account of Bathsheba Grossman’s recent Support Network visit to Girls’ Angle. If you haven’t seen her mathematically inspired sculptures, you’re in for a treat. She mesmerized us for over an hour with her inspiring creations.

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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We claim that every point in space on the side of the plane with the rolling snowball and within a snowball diameter of the plane can be visited by the marked point.

Let *C* be the circumference of the snowball. Any positive distance less than 2*C* can serve as the base of an isosceles triangle with equal sides of length *C*. Draw this triangle in the plane and put the snowball at one endpoint of the base with its marked point touching the plane. Roll the snowball along the two sides of length *C*. It ends up at the other endpoint of the base with its marked point touching the plane. This shows that the marked point can visit any point within a circle of radius 2*C* from its starting point. By taking several such journeys, we see that every point in the plane can be visited. For any point *P* on the same side of the plane as the snowball and within a snowball’s diameter of the plane, position the snowball so that its marked point is at *P*. We can roll this snowball until the marked point is touching the plane and since all points in the plane are visitable, we conclude that *P* can also be visited.

We claim that the locus of visitable points is a cardioid of revolution.

Let’s mark the point on the sphere where the marked point on the snowball touches it at the start with a golden dot. Look at the situation from the point-of-view of the common tangent plane. The two balls roll as mirror images of each other. The marked points are reflections of each other in the common tangent plane.

In other words, no matter how the snowball rolls about the sphere, the marked point will be located at the reflection of the golden dot in the common tangent.

We conclude that the locus of visitable points is a surface with rotational symmetry (about the axis one of whose poles is the golden dot). A vertical cross section by a plane containing the pole of the stationary sphere is the curve traced out by a dot on a circle as it rolls about a circle of the same diameter, and this is known to be a cardioid.

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Last Sunday, Girls’ Angle hosted a booth at the Girl Scouts STEM Expo in Framingham, Massachusetts. Visitors were challenged to accomplish 3 tasks, one of which was solving this crossword puzzle. They succeeded, so now we’re opening this puzzle to the general public. 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 **April 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. (If you worked on this puzzle at the Girl Scouts STEM Expo, you can’t enter this raffle. Sorry!)

Across

1. Prefix meaning

5. Soften with warmth

9. Laboring train sound

13. Tip

14. Something to follow

15. Pool division

16. Even

19. Temporary cut protector

20. Like mathematical concepts

21. Dr. Seuss’s ___-I-Am

24. Teach

27. Steal

29. Make collinear

31. What a ghost might say

32. 2012 Ben Affleck film

33. A fact only you know

35. Opposite of SSW

37. They used shells, dots, and bars for numbers

38. Slope

41. A failure

43. Something you hack with

44. Like a proper subset of a set

47. Darkens

48. Say “pi is equal to 3.14.”

50. Turns

51. Mimic

52. An example of something by Bach

55. Erase (abbr.)

56. Jack climbed one

58. Baby’s first utterance?

60. Line

66. Circle, square, or triangle

67. Fencing tool

68. Ready

69. Rugged cliff

70. Men with children

71. Major tennis tournament

Down

1. A little bit

2. Precedes graph, cycle, or gram

3. Depress the accelerator

4. It’s often labeled with *x*, *y*, or *z*

5. A group of three

6. Noisy mob

7. Can be found at the end of some scores

8. Tiny

9. Bonnie’s partner

10. Dislike intensely

11. Not justified

12. Earth science

17. Parts of an act

18. Smallest amount of computer memory

21. Name for a congruence theorem

22. Pub drink

23. Very small alien rock

25. Change a jpeg to a gif

26. Top quality

28. Big strangler

30. Helps you hold on to things

32. Entertain

34. It can be green or black

36. Kind of eagle

39. They’re often yoked

40. Make new and improve

41. They recommend how much to eat

42. Mouth along to music

45. Poetic before

46. Broadband choice

49. Smashed into

52. Held on tightly

53. Abbreviation often found on police rap sheets

54. Drains

57. Blue green

59. The Matterhorn is one of them

61. Type of light (abbr.)

62. Place to unwind

63. Address on the net

64. Precedes horse, cucumber, and lion

65. Unit for pulse

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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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!

*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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