/sci/ - Science & Math

The first thing I can come up with is that to maximize the area given the other constraint, the diagonal would be 8 and the sides 4 each, with the sides perpendicular to the diagonal. That has an area of 32. Do any other such quadrilaterals even exist? It's easy enough to see that the other diagonal would be <span class="math">8\sqrt2[/spoiler]

>>4006215
>discouraging actual content
GTFO

>2011
>posting actual content on /sci/
ISHYGDDT

No, seriously, if the quadrilateral is split in half, for a given diagonal length the area is maximized by having the measured side be perpendicular to that diagonal and the area is (a+b)(16-a-b)/2. This is maximized by a + b = 8, and so the area of such a quadrilateral can be at most 32, right? All of these constraints - the two measured sides being perpendicular to the measured diagonal, the measured diagonal being length 8 - they apply, so there's only one other diagonal length possible.

Shit, 'm probably gonna have to find another problem.

>>4006219
nice contribution bro

>>4006232
by "split in half" I should say that the quadrilateral is divided by the diagonal we're constraining, and it makes two triangles, with the constrained sides being other sides of the triangles.

>>4006232
I'm not sure if we know the minimum length yet. That's to say, is it possible to build a triangle of area 32 with two sides 8 and 8? Yes indeed: that triangle's area is at most half the area of a square of side 8: 1/2 * 64 = 32. Thus it's possible to have a sides 0,8,8 and 8sqrt(2) so that the diagonal is 8, the sum of the diagonal, one of the "8" sides and the "0" side is 16, and the area is 32.

Fine, done.

rage, okay, here's another problem, and I don't know the answer. It's from two days ago, and looks like it's from Putnam 1975.
"Find 1975 points on the circumference of a unit circle such that the distance between each pair is rational, or prove it impossible."

>>4006240
Yeah, no matter what you do, the other diagonal is the same. I'm now imagining an animated version with the diagonal sliding up and down, but always staying the same length.

>>4006245
all points on same point.
distance 0.
check and mate. next.

>>4006245
Consider a regular polygon inscribed in the unit circle of 2^n > 1975 sides (n=11 should give 2048). Aren't its diagonals rational?

(total intuition, nothing to back this up)

>>4006262
Ok, doesn't work for n=2. Never mind that.

>>4006254
clever

>>4006262
Doesn't work for n=5, either.

You can do it for 3 - can we do it for 4?

>>4006280
by 5, I meant the pentagon. Wasn't reading closely.

In polar coords, the metric is:

<span class="math">a/b = \sqrt{r_{1}^{2} + r_{2}^{2} -2r_{1}r_{2}cos(\theta_{1} - \theta_{2} )}[/spoiler]

= <span class="math"> \sqrt{2} \sqrt{(1-cos(\theta_{1} - \theta_{2} ))}[/spoiler]

Okay, let's try to see if it's possible that 3 points A, B and C on the unit circle are such that:
AB=AC,
AB, AC and BD are rational.
If not, the distances must all be distinct.


Let's work with complex representations.
I take A=1 wlog.
Let us write <span class="math">x=|A-B|=|A-C|[/spoiler] and write <span class="math">B=1-xe^{ib}[/spoiler].
As <span class="math">B[/spoiler] is on the unit circle, <span class="math">|1-xe^{ib}|=1[/spoiler].
<span class="math">|1-xe^{ib}|=(1-xe^{ib})(1-xe^{-ib})=1+x^2-2x\cos(b)=1[/spoiler]

Thus,
<span class="math">x^2=2x\cos(b)[/spoiler]
<span class="math">\cos(b)=\frac{x}{2}[/spoiler]

<span class="math">|BC|=2\sin(b)=2\sin\left(\frac{x}{2}\right)[/spoiler]

So you can have 3 points with two pairs equally spaced if the distance <span class="math">x[/spoiler] is so that <span class="math">\sin(x/2)[/spoiler] is rational.
...well that unfortunately didn't help much.

>>4006341
Okay, now use any pythagorean triple to generate a rational sine.

>>4006345
Okay, so assume we take any of the <span class="math">x[/spoiler] that works. Can we just space the points around the circle evenly so that point number <span class="math">i[/spoiler] is distant from point number <span class="math">i+1[/spoiler] by <span class="math">x[/spoiler]?

Let's see with 4 points, to begin with.

I have a group of four points that satisfies the condition - (1,0),(-1,0),(7/25, 24/25),(7/25,-24/25)

>>4006352
If we take my above example, (-1,0),(7/25,24/25),(7/25,-24/25), we can space one out by the same distance.

>>4006345
rational sines is the key

you then use the trig identity formula to show other distances are rational

i can't be arsed with the detail

>>4006363
sigh, can you work out what this bloody thing is? I don't even want to, really. BRB, pen.

>>4006399
For equally spaced points, such that the distance between point <span class="math">i[/spoiler] and point <span class="math">i+k[/spoiler] is <span class="math">x_k[/spoiler], I have that there is an angle <span class="math">\alpha[/spoiler] (angle at the center between two successive points) so that
<div class="math">x_k=\frac{\sin(k\alpha)}{\cos\left(\frac{k\alpha}{2}\right)}</div>

I guess applying trig identities to that formula would work, indeed.

>>4006424
That simplifies to
<div class="math">x_k=2\sin\left(\frac{k\alpha}{2}\right)</div>

I'm not sure about how you're supposed to find <span class="math">\alpha[/spoiler] so that all of these are rational and distinct modulo <span class="math">2\pi[/spoiler], though. Some theory on rational sines answers this?

>>4006404
good lord, I hate arithmetic.
Okay, if you consider the 3,4,5 triangle, resized so the hypotenuse is on the x axis between (-1,0) and (1,0) the point opposite the hypotenuse is (7/25,24/25). If you follow that point down and around, you get to (7/25,-24/25). If you follow the point one more time, you have a tilted triangle, and you build the next point, and I got something like (352/625,392/625) as the next point around the circle. The distance, here, seems to be a little broken.

>>4006424
well, if 2 sin (alpha/2) is rational and its opposing cosine also is, then 2 sin(alpha/2) cos(alpha/2) also is, obviously, and so is cos^2 - sin^2. Then we have that sin(alpha)cos(alpha/2) and cos(alpha)sin(alpha/2) as well as cos(alpha)cos(alpha/2) - sin(alpha)sin(alpha/2) are all rational, and by induction all of the sin(alpha*k/2) are rational.

Since the angle is an irrational multiple of pi, we're fine with overlaps.

Sorry about being too lazy to LaTeX it.

>>4006435
I probably did that wrong.

>>4006450
Too lazy to read across the exact formulas used but as they are all closed under rationality, that must be true.

nice of you guys to do the legwork for me

not that i checked it

>>4006459
Yeah, since your thing is correct, and the sines and cosines are all correct, and since they are irrational angles, we've solved it! Yay!

Someone else find a problem?

>>4006459
to mathgimp and TN5, so it's easier to read:
if sin(a) and cos(a) are rational then that forms the base case of an induction, and
sin(ka) = sin(ka-a)cos(a) + cos(ka-a)sin(a)
is rational, and
cos(ka) = cos(ka-a)cos(a) - sin(ka-a)sin(a)
is rational.

MOAR
I SURVIVE ONLY ON FUN MATH PROBLEMS.

>>4007657
Where you here when I posted the one about aligned points?

>>4007690
No, but I was here for the thread where you were trying to cover the k-size subsets with disjoint covers made of k+1-size subsets.

>>4007713
That was actually related to the small memory problem I asked. I was trying to extend it to n cells, writing data in t steps. I investigated the restriction that at step k (1<=k<=t), the maximum number of cells at "1" is k. See what I mean? :-)

>>4007719
Sort of, but I don't have an intuition as to why that would be a good thing.

>>4007719
I mean, I do see now how it would allow you to come up with a value, but I don't see how that's beneficial in analysis of the system's properties.

>>4007724
Because it simplifies the problem a lot. It's actually quite hard to treat the general case, so figuring out a restriction that doesn't look too dumb and treating it might produce valuable results.

Also, about the aligned points:
Let <span class="math">E[/spoiler] be a finite set of points the Euclidean plane so that the following property holds:
For all distinct <span class="math">x[/spoiler] and <span class="math">y[/spoiler] in <span class="math">E[/spoiler], there exists <span class="math">z\in E[/spoiler] distinct from <span class="math">x[/spoiler] and <span class="math">y[/spoiler] so that <span class="math">x[/spoiler], <span class="math">y[/spoiler] and <span class="math">z[/spoiler] are on a straight line.
Prove that <span class="math">E[/spoiler] is a subset of a line.

>>4007741
Interesting!

>>4007767
Let me know if you want the solution, I have it in a pic. Actually, you might find a better one. Mine is cool but maybe it's possible to find a shorter one.

>>4007775
I should solve it.

>>4007795
I looked at it for the first time about 4 years ago, together with a friend, before we gave it as a small blackboard test to his students (guiding them through it, of course). It took us about 15 minutes, which puts it by far over the others that were in the same book, in terms of difficulty, as it was aimed at 1st year students.

>>4007811
I'm trying to come up with something that would show that such a set would be unbounded.

>>4007827
Yeah, that's more or less what my proof does. I use the fact that N is well-ordered.

We could rotate it so that no horizontal or vertical line contained more than one point, and then sort it by x-coordinate or y-coordinate.

>>4007888
But where would the contradiction come from, after that? I can see what you mean by the rotation but I can't see it helps (well it gives you a way to go through the points in a "simple" order indeed, but you have a finite set so you can already do that).

>>4007935
This could take me a long while more. Also, I keep going back to my homework, which is coding in Perl.

>>4007975
Just let me know if you want a hint. Or just let it haunt you forever ;)

Also, Perl... Efficient in terms of characters written, worst non-esoteric language ever in terms of readability. I wouldn't like to have homeworks in Perl.

>>4007998
Yeah, but it's for coding IBM Model 1 (a machine translation algorithm) and while it will take a while, at least I won't have to deal with strings. Perl's really good at that.

I do want a hint. My curiosity has gotten the better of me.

>>4008020
Wait, I posted a hint...
Well, posting it again since it didn't work last time. Crappy net here, I should give it time to actually post before refreshing pages.

Anyway:
Consider the smallest non-degenerate triangle whose vertices are points of <span class="math">E[/spoiler]. Prove that there's a smaller one. Contradiction, win.

"Smallest" is voluntary vague. Hint: there's a particular property that I am thinking of which is very small almost-degenerate triangles. Second hint: it's a distance property.

Isn't is this theorem?
http://en.wikipedia.org/wiki/Sylvester%E2%80%93Gallai_theorem

>>4008796
Cool! I didn't know it had a name.
And also, Cool! The proof is roughly the one my friend and I found when we solved this :D

>>4008804
I think I had heard of it as a hard (putnam style) problem a while ago, but that might have been something else. I mostly know it from this
http://en.wikipedia.org/wiki/Proofs_from_THE_BOOK
which gives two proofs in different places, I think the same two mentioned in the wiki article on the theorem.

>>4008814
I guess it could have been asked in putnam indeed. Anyway, here's the proof I wrote, which is basically the 1st proof in wikipedia (messier, but on the same basic idea).