by We finally published the Winter Spring SUMMER(ish) edition of our department newsletter!  This issue is named the Taniyama Times after Yutaka Taniyama (谷山豊 ) of the Taniyama-Shimura conjecture (proved by Andrew Wiles, and giving Fermat’s Last theorem as a wee little corollary).

This issue will admittedly hold less interest for non-alumni than most of our issues, since it’s primarily about where the Class of 2009 has been spending the past year.  Still, it does contain the following fun problems to work on!

Problem 4.2.1: (2006 AMC10) If xy =x³—y, what is h◊(h◊h)?

Problem 4.2.2: Which fits better: a square peg in a round hole or a round peg in a square hole?

Problem 4.2.3: The figure below shows the first three circles in an infinite sequence. What is the total area of the circles? What is the total circumference? Answers are welcome in the comments, and you might just be acknowledged in the next newsletter!  [If we remember, which is sort of a risk since I’m right now remembering that we forgot to check that when we put together this issue.]

### 7 Responses to “It’s a New Newsletter!”

1. Joshua Says:

#1: It’s hard to read since my fonts here don’t support all those symbols, I think, but the answer seems to be h.

#2: round peg in square hole is easy, pi/4. square peg in round hole is 2/pi since the inscribed square is half the area of the circumscribed. Then pi/4 is more than 3/4 and 2/pi is less than 3/4, so the round peg fits better.

#3: I think I’ll come back to this one sometime when I have scratch paper. I don’t see any beautiful shortcut here since there’s only one square, but maybe I’m missing something nice.

2. Confusio Says:

4.2.3: I think I have it:

If the first radius R is 1, then the sum of areas tends to 3,2368… (in other words: pi*R2*(3+2*SQRT(2))/(4*SQRT(2)) and the sum of the circles tends to infinity.

Its a finite area contained in an infinite limit.

Nice problem and nice page!

3. Consumer Reviews Says:

Thank you for sharing the problems! I am going to share them with my class, and see where they go with them.
Kind regards.

4. Roberta Wedge Says:

I tried to leave a comment on your entry re the Polygon buildings in Somers Town, but the software wouldn’t let me. If this gets through, please could you email me? firstname dot lastname at gmail dot com

5. Ξ Says:

Roberta, I still don’t know what is going on with that (as we’ve talked about over email) but it looks like the software has relented! Your comment just went through in this post.

6. Mark Says:

Well, 10 minutes and a post-it note suggest the total area of all the circles tends towards (pi/4)/(12*sqrt(2)-16) times the area of the square.

My approach was to suppose the big circle has radius 1, find the radius of the 2nd circle, and square it. That gives you the ratio for an infinite geometric series, with the first term being pi/4 for the area of the first circle relative to the square. The figure quoted above it its sum. That 2nd radius (call it r) can be found by observing that the distance along the y-axis between the circles’ tangent point and the top of the square (call it h) is 1- sqrt(2)/2. h is also equal to r*(1+sqrt(2)/2), and then it’s trivial to solve for r.

7. Raúl Says:

What I get in problem 4.2.3 is:
a) The following recurrence between one radius and the next one: r sub (k + 1) = lambda times r sub k, where lambda = (sqrt 2 – 1)/(sqrt 2 + 1) < 1.
b) The total length of the circles is 2.pi.sum of r sub k = 2.pi.r sub 1/(1 – lambda) = 1,06 times length of greatest circle. Here r sub 1 = a/2, and a = side of the first square.
c) The total area is pi times a squared times (9/16)(3sqrt 2 + 2)/14).
In any case, both the total length and the total area converge. each of the sum is the sum of an infinite geometric series; the ratios are lambda and lambda squared, respectively.
The total area can also be obtained approximately by using the above recurrence—squared— and adding up. After multiplication by pi—and neglecting the term [r sub (n + 1)]*2—, the above result follows. ¡Remarkable!