Wikipedia:Reference desk/Archives/Mathematics/2011 June 16

Welcome to the Wikipedia Mathematics Reference Desk Archives
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.

June 16

Geometry problem

Hello all, it's me from yesterday. This is the other question that I did not answer (pretty good, considering this was a 25 question test). Unlike the 90! one, I tried on this one but did not get it and so refrained from answering (guessing is penalized). It goes, Let R be a square region and n>3 an integer. A point P inside R is called n-ray partitional if there are n rays emanating from P divding R into n triangles of equal area (for example, there is one 4-ray partitional point at the center of the square). How many points are 100-ray partitional but not 60-ray partitional? I didn't even know where to begin on this one- can someone point me right? Thanks. 72.128.95.0 (talk) 00:48, 16 June 2011 (UTC)

Suppose wlog that R is the unit square. Orient R so that the sides are parallel to the coordinate axes and the lower left hand point is at the origin. Call the point (x,y). The rays from the point to the vertices of the square divide the square into four triangles of areas $x/2,y/2,(1-x)/2,(1-y)/2$. Now, (x,y) is a (2n)-ray point if and only if $x/2=a/2n$ and $y/2=b/2n$ for some integers $0\leq a,b\leq n$ (since these two triangles need to be divisible into equal pieces of area $1/2n$). Sławomir Biały (talk) 02:38, 16 June 2011 (UTC)

Angle bisectors of a triangle

prove that the angle bisectors of a triangle is concurrent. — Preceding unsigned comment added by Qwshubham (talk • contribs) 07:55, 16 June 2011 (UTC)

You need to put in a title otherwise it looks like it follows from the previous discussion. Have you had a look at triangle? Dmcq (talk) 08:42, 16 June 2011 (UTC)

Spot the error?

$0$
$=0+0+0+0+...$
$=(1-1)+(1-1)+(1-1)+(1-1)+...$
$=1-1+1-1+1-1+1-1+...$
$=1+(-1+1)+(-1+1)+(-1+1)+...$
$=1+0+0+0+...$
$=1$
Widener (talk) 13:29, 16 June 2011 (UTC)

The third and fourth equals signs are not justified. Indeed, the series $\sum (-1)^{n}$ is divergent. What this fallacious argument illustrates is that the summation of infinite series is not associative: one can't "insert and remove parentheses" into series in general. There is a discussion of this in Tom Apostol's textbook "Mathematical analysis", for instance. Sławomir Biały (talk) 13:47, 16 June 2011 (UTC)

See Grandi's series. Gandalf61 (talk) 13:52, 16 June 2011 (UTC)

…or the Eilenberg–Mazur swindle. — Fly by Night (talk) 14:58, 16 June 2011 (UTC)

There is an extra −1 at infinity plus one that you missed out. Sorry just joking ;-) Dmcq (talk) 17:28, 16 June 2011 (UTC)

See also section 3 of this article. Count Iblis (talk) 20:43, 16 June 2011 (UTC)

We have an article titled telescoping series that addresses this situation. Michael Hardy (talk) 22:43, 20 June 2011 (UTC)