My recent travels, to Parsippany, NJ via Baltimore, MD, which involved three talks in two days, followed by multiple games of chess, bookended by two long drives, came to a dramatic conlcusion yesterday when I had to drive home in the snow. Not fun! There was so much snow on the road that you frequently couldn't see the lane markings, making it difficult to orient yourself. And let me tell you, the truckers who use that particular road weren't going to let a little snow slow them down. Then, after creeping along at thirty miles per hour for long stretches (on what is usually a seventy mile an hour road), I finally got to my exit. And wouldn't you know it, there was a car stuck at the top of the exit ramp, spinning its tires helplessly and blocking everyone else. I got there just as the situation was clearing itself up. A couple of good samaritans in line ahead of me were able to give him enough of a push to get him free.

Which is my long-winded way of saying that the new Problem of the Week has now been posted. I've also posted an official solution to last week's problem. So go have a look and let me know what you think.

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Wow, does that sound exhausting ( I assume you played 2 full games of chess before the, what, 6 hour drive?). Not to mention a bit dangerous. Hope you recover quickly.

Let the circle have circumference = c

The circumscribed square can have side length = a

The diagonal of the square, which is also the diameter of the circle, has length = d

It will be seen that the square is divided into 2 right angled triangles, each of them a 1, 1, √2 triangle

Therefore, c = a x π√2 (by Pythag and ratio of c to d).

For ease of expression, let π√2 = n, so c = an

We know that c + 4a = 40

Therefore, 40 – 4a = an, or an + 4a = 40, or a (n + 4) = 40

(We can then also solve for c, such that c = an = 40 – 4a)

Substituting the numbers, this resolves as a = 4.738 inches, therefore 4a = 18.95 inches. This is the perimeter of the circumscribed square – one of the cut wire portions. The other cut wire portion is therefore 40 – 18.95 = 21.05 inches (the circumference of the circle).

The shorter portion is therefore the 18.95 inch section, the perimeter of the square.

(I have expressed all numerical calculations approximated to 4 sig figs, given the irrational numbers involved in the calculation)

Hi Bill. Actually, we decided to withdraw before the last round on Monday. Ned and Brian were a bit under the weather, and I wanted to get on the road in an ultimately unsuccessful attempt to beat the snow. So we bullied Curt into going along with the plan!

The same as Phil B's, but with more explicit algebra:

The length of the wire is known, call it W.

The two pieces of the wire are the circumference of the circle, call it c, and the perimeter of the square, call it p. So

[1] W = c + p

By construction, the diameter of the circle is equal to the diagonal of the square. So

[2] c = π*d

Call the side of the square s; thus the diagonal of the square is

[3] d = √2*s

Substituting for this d in [2]

c = π*√2*s

Since s = p/4, we get

c = (π*√2/4)*p

The expression π*√2/4 is a constant, equal to the ratio of the circumference of the circle to the perimeter of the square. For simplicity, call that z. So

[4] c = z*p

Substituting [4] into [1]:

W = z*p + p, so

[5] p = W/(1+z)

and by [5] and [4]

[6] c = z*W/(1+z)

Since W = 40 and z = 1.1107, all irrationals to to four places:

p = 18.9508

c = 21.0491

So the smaller piece is p, with length 18.9508.

Nothing really to add to Rick and Phil B's solutions, except that as a matter of preference, I preferred not to express the solution in decimal approximation. Since this is a mathematics problem, I left my solution in exact form, namely that the shorter piece has a length of 160/(4+pi*sqrt(2)).

I might agree with Rick that his extra algebra helps a little with the explanation, and also with Sean T in that substituting the actual numbers introduces an element of inelegance which might well be omitted – indeed I considered this approach myself, but in the end I put the ugly numbers in because I thought that Jason might expect of his students (at least his first year students) that they should do a little arithmetic - and I was putting myself in the place of such a student.

However, before Jason jumps down all our throats, it needs to be pointed out that we have all made 3 assumptions in arriving at our solutions, viz:-

1)That Pythagoras is correct – for several elegant proofs, just see this post over the last couple of weeks or so.

2)That there exists a relationship between the circumference (c) and the diameter (d) of a circle, such that the two are related by a constant of proportionality (which, by convention, we usually call π) such that c = πd

3)From Euclid's “The Elements, Proposition 9 of Book IV”, - ”About a given square to circumscribe a circle”, that the theorem described is axiomatic.

Just in case some joker wants to talk about the problem not necessarily being set within a Euclidian space, I refer you to Jason’s earlier objections in a previous “problem of the week” post.

Jason – please tell me, are there any other assumptions that I/we have made, justifiably or otherwise, that perhaps I’ve not picked up on?

Or if you want not to have a radical in the denominator, I guess you can do (hoping the latex gods smile on me):

$latex 80\cdot({\pi\sqrt{2}-4}\over{\pi^2-8})$

Nope, the latex gods hate me.

$latex {80}\cdot{ ( {{\pi\sqrt{2}-4}\over{\pi^2-8}} ) }$

Thanks for the better formatting, Another Matt. What would you call that procedure, though. Normally, eliminating the radical in the denominator is referred to as "rationalizing the denominator". I don't think I would be comfortable with that term in this case, though, given that pi^2 term.

jrosenhouse wrote (February 17, 2015):

>

the new Problem of the Week has now been posted.>

A piece of wire 40 inches long is cut into two pieces. [...]By a straight (plane) cut?

Through a cylindrical (straight) wire? …

I'm mostly just joking about the pedantry of eliminating radical terms from denominators at all costs. This is a case where it's clearly not simpler to do so, although it's a good thing to teach algebra students this kind of manipulation. I might also say that this form

wouldbe simpler if the sqrt(2) term werei(a kind of radical term, yes?) instead, since division by complex values requires conjugate multiplication anyway.Hope to see your annual summary of the event. It will partially compensate for my having to miss it this year :).

Hi Brian. I expect to be posting my annual summary sometime over the weekend. I thought I played pretty well, until I let the computer have a look at my games.

Frank Wappler wrote (#10, February 18, 2015):

>

By a straight (plane) cut?>

Through a cylindrical (straight) wire? …Under these conditions, the given 40 inches long piece of wire may be thought as having been cut into one piece of (still) 40 inches length, and another piece which is at most 40 inches long , or shorter.

In order to satisfy the requirement that the ratio of the lengths of these two pieces has the value $latex \frac{\sqrt{8}}{\pi}~\approx~0.9 $ the shorter piece must be about 36 inches long.

(Don't let it drop into a snow bank! &)

Hi Jason.. I saw this just now

http://priceonomics.com/the-time-everyone-corrected-the-worlds-smartest/

Nd wondered if you had come.across this researching your Monty Hall book...