One point at a time
- You look bored!
- Yeah. I don’t know what to do. Nothing inspires me.
- You are as sent from heaven!
- Yeah?
- Make a circle and colour every point on its circumference red or blue.
- Won’t that take a long time?
- And? You said you had nothing to do.
- OK. But what’s the point?
- Pun intended?
- What pun?
- The old lady in the bakery asserts that regardless of how you colour the points one can always find an isosceles triangle with equally coloured vertices on the circle.
- I didn’t know you shopped at the bakery.
Problem source: Mathematical problems by Andrew Adler.
June 25th, 2010 at 2:42 am
Set up polar coordinates (r, theta) so that the circle is centred at the origin. Let f(x) be the colour of that point
whose theta-coordinate is x. Then f(x) = f(x + 2pi). We are to prove that there are three reals in arithmetic progression, no two of them equal modulo 2 pi, at which f takes equal values.
Pick arbitrary reals a and b (b-a positive and not a multiple of pi or of 2pi/3). If f(a) = f(b), then let c=a and d=b. Otherwise, one of f(a) is red and the other is blue. Let c=(a+b)/2 and let d = whichever of a or b makes f(d)=f(c). Then in any case f(c)=f(d). Now consider 2c-d, (c+d)/2 and 2d-c. If any of them, call it e, is such that f(e)=f(c), then c, d and e are three reals in arithmetic progression as required. If not, then whatever colour f(c) is, f is the other colour at all three of 2c-d, (c+d)/2 and 2d-c, so those three values are three reals in arithmetic progression as required.
June 25th, 2010 at 8:43 am
That beer in the background is a very good beer. Just thought I’d point that out.
I bet you a dime that photo was taken in Texas.
June 25th, 2010 at 10:18 am
The image was taken from http://www.cookingbytheseatofmypants.com/recipes/ale-bread-the-resurrection/ by Jerry Russell. Is he from Texas?
June 25th, 2010 at 4:34 pm
Inspired with Richard proof, I found this one which is more geometrical.
Draw an octagone in the circle. Take three points next to each others.
a) they are all the same color -> ok
b) we have BRB (or similarly RBR)
c) we have BBR (or similarly RBB, RRB or BRR)
b: look two points further and we have x.BRB.y
if x or y = B, we have B.B.B somewhere -> ok
if x = y = R, we have R..R..R -> ok
c) we have either RBBR or BBBR. BBBR -> ok.
if we have RBBR, look at the middle point between the to B’s.
We have either R.BbB.R or R.BrB.R, both ok.
June 26th, 2010 at 12:46 am
Hehe… I was right. Took some digging but I found another recipe on his webpage where he refers to Texans with the pronoun “we”.
June 26th, 2010 at 1:00 am
I did some digging too. Born in California, moved to Texas, now moving
to Alaska. I think.
http://twitter.com/JerryDRussell
July 14th, 2010 at 4:25 am
Since you can color all the points either red OR blue, it wouldn’t take a long time at all. Just color the entire circle either red or blue and you’re done!