Circle set of points

MaThS

Need some help..

We define the outermost edge (or the circumference) of a circle to be a set of points. Let each of these points be colored black or white. Prove that regardless of choice, it is always possible to inscribe in this circle an isosceles triangle whose three vertices are of the same color. Is this possible for any other type of triangle?

**Posted:** Thu Jan 06, 2005 7:56 am

grobber

I think it's enough to consider a regular heptagon inscribed in the circle. I didn't check it throroughly, but it looks like we can find three of its vertices which are colored the same and form an isosceles triangle.

Am I wrong? Confused

**Posted:** Thu Jan 06, 2005 8:31 am

Myth

What other kinds of triangles should we consider?
It is clear, we can't guarantee that there is a regular triangle or right triangle.

**Posted:** Thu Jan 06, 2005 8:39 am

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Mario_KaRt

I think he is asking if it is possible to inscribe any other triangle in the circle with the given conditions, but it is not part of the original problem statement...

**Posted:** Thu Jan 06, 2005 8:44 am

grobber

Just another thing about the problem: a regular nonagon works for sure, because at least $5$ of the points will have the same color, and it is well-known that if you choose $5$ numbers among $1,2,\ldots,9$ , there will be an arithmetic progression with $3$ terms among those (I'm sure the connection between the two problems is clear Smile

).

**Posted:** Thu Jan 06, 2005 8:45 am

Myth

I can reveal great secret for you: regular pentagon works too Mr. Green

**Posted:** Thu Jan 06, 2005 8:50 am

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grobber

It's my turn to reveal a great secret Smile

: we can find an isosceles triangle, no matter what finite number of colors we use (not just the very small number $2$ ). We just use Van der Waerden's Theorem and that idea with the arithmetic progression.

**Posted:** Thu Jan 06, 2005 9:04 am

Mario_KaRt

In that case, can anyone prove it only for an isoceles triangle? I would like to see how a proof would look like for this problem...

**Posted:** Thu Jan 06, 2005 9:13 am

grobber

You're refering to the case with $r$ colors?

In that case, basicly, the proof has already been posted Smile

. One of the versions of Van der Waerden's Theorem states that there is an $n(3,r)$ s.t. whenever $n\ge n(3,r)$ , any coloring of the numbers $1,2,\ldots,n$ with $r$ colors yields an arithmetic progression with $3$ elements, all colored the same.

Now consider a regular polygon with $n\ge n(3,r)$ vertices on the circle, and number its vertices $1,2,\ldots,n$ in a clockwise direction. The vertices are colored with $r$ colors, so we can find an arithmetic progression. The vertices corresponding to the numbers in arithmetic progression form an isosceles triangle.

P.S.

I think this has been discussed before. I seem to remember Pierre giving the exact same solution a while back.

**Posted:** Thu Jan 06, 2005 9:24 am

Myth

Ok! Let's color circle in countable many colors! Mr. Green

**Posted:** Thu Jan 06, 2005 9:26 am

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pbornsztein

Are you asking, or do you know about it?

Pierre.

**Posted:** Thu Jan 06, 2005 1:00 pm

grobber

I think I can find a coloring of $\mathbb R$ with countably many colors s.t. no three numbers with the same color form an arithmetic progression, but I'm not sure about it. Maybe someone comes up with something a little bit easier to explain.

Is the problem we have here significantly tougher than the one with $\mathbb R$ ? It doesn't look like it is..

**Posted:** Thu Jan 06, 2005 1:11 pm

Myth

Show us your example. I am too lazy today to think about it.

**Posted:** Thu Jan 06, 2005 1:53 pm

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Myth is out of here

grobber

I'll write it, but it's too complicated to be Ok. I must have gone astray somewhere Smile

. Just be kind and gentle when you discover my mistakes Mr. Green

.

Let $(a_i)_{i\in I}$ be a basis of $\mathbb R$ over $\mathbb Q$ . We may assume all the elements of the basis are positive (don't know if we need it Smile

). Each real can be written as $\sum_{j=1}^k r_ja_{i_j}$ , where $a_{i_j}$ are elements of the basis listed in increasing order, and $r_j\ne 0$ are rationals (assume $k=0$ in the decomposition of $0$ ).

Now just color each element of the basis according to its corresponding $k$ -tuple $(r_1,r_2,\ldots,r_k)$ (there are countably many, because they're finite sets of rationals). By this I mean that two reals have the same color iff their corresponding $k+1$ -tuples $(k,r_1,r_2,\ldots,r_k)$ are identical.

Does it work?

**Posted:** Thu Jan 06, 2005 2:16 pm

Myth

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Myth is out of here

grobber

Could you just tell me what's wrong? I'm sooo tired of trying to find counterexamples... Smile

Pleeease?

**Posted:** Thu Jan 06, 2005 2:26 pm

Myth

I think, I was wrong. Your construction contains advanced details, so it is correct. I will remove my remark.

**Posted:** Thu Jan 06, 2005 2:34 pm

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Myth is out of here

pbornsztein

I didn't know what you was refusing in Grobber's construction.

Pierre.

**Posted:** Thu Jan 06, 2005 2:41 pm

grobber

Myth asked me it this thread about the construction for the circle.

By the bijection $f:[0,1)\to [0,\pi)$ given by $f(x)=e^{\pi ix}$ color the points on the upper semi-circle according to the coloring of $[0,1)$ given in the message about the coloring of $\mathbb R$ . Then color the lower semi-circle $e^{\pi ix}$ corresponding to $x\in [1,2)$ according to the same scheme, only that each color on the lower semi-circle is different from each color on the upper semi-circle.

I think this takes care of the circle. First of all, since the colors on the lower semi-circle are different from those on the upper semi-circle, any monochromatic isosceles triangle must either be situated in the upper or lower semi-circle. However, on the upper semi-circle, for example, thare are no monochromatic triangles $e^{\pi ia},e^{\pi ib},e^{\pi ic}$ , because that would make $a,b,c\in [0,1)$ three numbers in arithmetic progression on the real line, which are colored the same, and our coloring of $\mathbb R$ does not allow that.

Is it Ok?

**Posted:** Mon Jan 24, 2005 1:22 am

pbornsztein

Looks cool Smile

Pierre.

**Posted:** Mon Jan 24, 2005 3:23 am