\(\mathbb{R}^3\) is coloured so that each point is either red, blue or green.
Prove that for the set of points of at least one of the colours it is the case that for all \(r \in \mathbb{R}^3\) there are points in the set separated by distance \(r\).
============================================================
Suppose otherwise. Then for each colour set \(R, B, G\) there is an \(r_r, r_b, r_g\ge0\) such that there are no points of that colour separated by that distance. Also suppose that without loss of generality that \(r_r\ge r_b \ge r_g\) and so \(r_r>0\).
Choose any point \(x_r\) in \(R\) and consider the sphere centred on \(x_r\) of radius \(r_r\). This sphere (which to be clear is the 2D surface of the ball of radius \(r_r\) centred at \(x_r\) ) has no points coloured red, in fact it must have points coloured blue and others coloured green. Choose a point \(x_b\) on the sphere coloured blue, and consider the intersection of the first sphere with a sphere of radius \(r_b\) centred at \(x_b\). This circle cannot be coloured red or blue so must be green.
Now choose any point on this green circle \(x_g\) and consider the sphere of radius \(r_g\) centred there, this meets the green circle at two points, which therefore cannot be green - a contradiction.
Tuesday, 27 March 2012
Draft:CaptainBlacks PoW #2 for MathHelpBoards
Posted on MathHelpBoards
This problem I think is a bit tedious and can be hard work. It comes from the Purdue Maths Dept PoW, only slightly modified.
Given that \(\cos(36^\circ)=\frac{1}{4}(1+\sqrt{5}) \) find \( \tan^2(18^\circ)\, \tan^2(54^\circ) \)
=============================================================
Solution 1 (this is mine and I must admit I used Maxima to handle the algebra):
First observe that from the quadrant corresponding to te angles here:
\[\tan(A/2)=\sqrt{\frac{1-\cos(A)}{1+\cos(A)}}\]
and:
\[ \tan(3A/2)=\frac{3\tan(A/2)-\tan^3(A/2)}{1-3\tan^2(A/2)} \]
So if we put \( A=36^\circ \) and allowing Maxima to do the algebra we get:
\[ \tan^2(18^\circ)\, \tan^2(54^\circ)=\frac{1}{5} \]
--------------------------------------------------------------------
Solution 2 (this is the solution give on the originating site):
\[ \cos(72^\circ)=2\cos^2(36^\circ)-1=\frac{1}{4}(\sqrt{5}-1) \]
and:
\(\displaystyle \phantom{xxxx} \tan^2(18^\circ)\, \tan^2(54^\circ) =\frac{\sin^2(54^\circ)\sin^2(18^\circ)}{\cos^2(54^\circ)\cos^2(18^\circ)} =\left[ \frac{(1/2)(\cos(36^\circ)-\cos(72^\circ))}{(1/2)(\cos(36^\circ)+\cos(72^\circ))} \right]^2\)
\(\displaystyle \phantom{xxxx} \phantom{\tan^2(18^\circ)\, \tan^2(54^\circ)}=\left[ \frac{\frac{\sqrt{5}+1}{4} -\frac{\sqrt{5}-1}{4}}{\frac{\sqrt{5}+1}{4}+\frac{\sqrt{5}-1}{4}} \right] =\frac{1}{5}\)
This problem I think is a bit tedious and can be hard work. It comes from the Purdue Maths Dept PoW, only slightly modified.
Given that \(\cos(36^\circ)=\frac{1}{4}(1+\sqrt{5}) \) find \( \tan^2(18^\circ)\, \tan^2(54^\circ) \)
=============================================================
Solution 1 (this is mine and I must admit I used Maxima to handle the algebra):
First observe that from the quadrant corresponding to te angles here:
\[\tan(A/2)=\sqrt{\frac{1-\cos(A)}{1+\cos(A)}}\]
and:
\[ \tan(3A/2)=\frac{3\tan(A/2)-\tan^3(A/2)}{1-3\tan^2(A/2)} \]
So if we put \( A=36^\circ \) and allowing Maxima to do the algebra we get:
\[ \tan^2(18^\circ)\, \tan^2(54^\circ)=\frac{1}{5} \]
--------------------------------------------------------------------
Solution 2 (this is the solution give on the originating site):
\[ \cos(72^\circ)=2\cos^2(36^\circ)-1=\frac{1}{4}(\sqrt{5}-1) \]
and:
\(\displaystyle \phantom{xxxx} \tan^2(18^\circ)\, \tan^2(54^\circ) =\frac{\sin^2(54^\circ)\sin^2(18^\circ)}{\cos^2(54^\circ)\cos^2(18^\circ)} =\left[ \frac{(1/2)(\cos(36^\circ)-\cos(72^\circ))}{(1/2)(\cos(36^\circ)+\cos(72^\circ))} \right]^2\)
\(\displaystyle \phantom{xxxx} \phantom{\tan^2(18^\circ)\, \tan^2(54^\circ)}=\left[ \frac{\frac{\sqrt{5}+1}{4} -\frac{\sqrt{5}-1}{4}}{\frac{\sqrt{5}+1}{4}+\frac{\sqrt{5}-1}{4}} \right] =\frac{1}{5}\)
Monday, 19 March 2012
Draft: CaptainBlack's Problem of the Week #1 for MHB
This is the draft for the problem posted on MathHelpBoards
===========================================================
Let \(x_1, ..., x_n\) be real numbers such that:
\(\sum_{i=1}^n x_i=0\) and \(\sum_{i=1}^n x_i^2=1\)
Prove that for some \( k,l \) both in \( \{1, .. , n\} \) that \(x_k x_l\le -1/n\)
CB
===========================================================
Solution
Let \(a=\max(x_1, ... , x_n) \) and \( b=\min(x_1, ... , x_n) \), and put:
\[ f(x)=(x-a)(x-b)=x^2-(a+b)\; x+ab \]
Then by construction \( f(x_i)\le 0,\;i=1, ... , n \) . Now consider:
\[ \sum_{i=1}^n f(x_i)=\sum_{i=1}^n x^2 -(a+b) \sum_{i=1}^n x_i +n\;ab = 1+n\;ab \le 0\]
hence:
\[ab \le -\frac{1}{n} \]
QED
===========================================================
Let \(x_1, ..., x_n\) be real numbers such that:
\(\sum_{i=1}^n x_i=0\) and \(\sum_{i=1}^n x_i^2=1\)
Prove that for some \( k,l \) both in \( \{1, .. , n\} \) that \(x_k x_l\le -1/n\)
CB
===========================================================
Solution
Let \(a=\max(x_1, ... , x_n) \) and \( b=\min(x_1, ... , x_n) \), and put:
\[ f(x)=(x-a)(x-b)=x^2-(a+b)\; x+ab \]
Then by construction \( f(x_i)\le 0,\;i=1, ... , n \) . Now consider:
\[ \sum_{i=1}^n f(x_i)=\sum_{i=1}^n x^2 -(a+b) \sum_{i=1}^n x_i +n\;ab = 1+n\;ab \le 0\]
hence:
\[ab \le -\frac{1}{n} \]
QED
Sunday, 18 March 2012
Math Help Boards Progress
After MathHelpBoards (www.mathhelpboards.com ) was established the former site came back online under new ownership. The new owner moved it to a new server and reorganised it to his liking. Discussions over a potential merger between the new owner and leading members MHB did not lead anywhere so both sites continue in existence. The new owner has in the past made zero contribution to the work of the site, and has little idea how it used to be run.
The former site has the advantage of a higher position in any Google search for maths help, but retains fewer of the former regulars. MHBs has the advantage of a large number of experienced helpers. As a result the former site has a larger number of new posts per day, but also a large number of unanswered or slowly answered posts. MHB has a smaller number of new posts per day but few go unanswered for very long and is better moderated.
If I were in the business of making predictions I would expect both sites to remain in business for some time with the same relative traffic levels, but the former site will probably fail in the longer term when the new owner loses interest.
The former site has the advantage of a higher position in any Google search for maths help, but retains fewer of the former regulars. MHBs has the advantage of a large number of experienced helpers. As a result the former site has a larger number of new posts per day, but also a large number of unanswered or slowly answered posts. MHB has a smaller number of new posts per day but few go unanswered for very long and is better moderated.
If I were in the business of making predictions I would expect both sites to remain in business for some time with the same relative traffic levels, but the former site will probably fail in the longer term when the new owner loses interest.