A community for students.
Here's the question you clicked on:
 0 viewing
Kainui
 one year ago
In a square with integer side lengths, is it possible to find a point that's a rational number length away from all 4 corners?
Kainui
 one year ago
In a square with integer side lengths, is it possible to find a point that's a rational number length away from all 4 corners?

This Question is Closed

Kainui
 one year ago
Best ResponseYou've already chosen the best response.1I believe that it's not possible.

Kainui
 one year ago
Best ResponseYou've already chosen the best response.1Give an example, because this is an unproven problem in number theory. :)

freckles
 one year ago
Best ResponseYou've already chosen the best response.2dw:1438879607827:dw \[h=\frac{1}{\sqrt{2}}a \cancel{ \in} \mathbb{Q}\]

Kainui
 one year ago
Best ResponseYou've already chosen the best response.1dw:1438879604541:dw I was kinda thinking about trying to show that we can't shift the origin to any point and end up with these 4 vectors equal to rational numbers. I don't really know how to approach this though, just my first guess.

freckles
 one year ago
Best ResponseYou've already chosen the best response.2for some reason i thought we had to choose a point where all corners had the same rational length from this point we are trying to find but i think i misunderstood it like I took "a rational number length" to mean each corner had length b from point P where b is a rational number. that is not right right? like say the square is called ABCD where A,B,C,D are the vertices. We are just wanting to choose point P such that the following are all rational lengths not necessarily the same: AP BP CP DP

Kainui
 one year ago
Best ResponseYou've already chosen the best response.1Yeah, so like specifically for anyone else that might be confused, dw:1438880210962:dw So in my picture, a,b,c,d are all rational lengths. Is this possible?

freckles
 one year ago
Best ResponseYou've already chosen the best response.2dw:1438880394272:dw I found a point which made two of those rational but the other two aren't :p

freckles
 one year ago
Best ResponseYou've already chosen the best response.2i found one where 3 are rational one isn't though choosing one of the vertices to be our P

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1since scaling the square by the LCM of denominators gives a square with all integer lengths, I think the problem is equivalent to looking for a point inside a square which has integer distances from all the four corners

Kainui
 one year ago
Best ResponseYou've already chosen the best response.1Ohhh that's a cool observation I was thinking the other way around and looking for a point inside a unit square.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1its a trade off, we can choose either an unit square with rational distances from corners or some random square with integer distances..

Kainui
 one year ago
Best ResponseYou've already chosen the best response.1When you say it that way, that means something about stacking pythagorean triples together doesn't it?

Kainui
 one year ago
Best ResponseYou've already chosen the best response.1dw:1438880869802:dw All pythagorean triples. I remember a while back we proved something like all pythagorean triples have like they have to have at least one even numbered leg?

Kainui
 one year ago
Best ResponseYou've already chosen the best response.1Nevermind, that was a specific case I think you asked (not sure but I think): "If the length of one leg of a right triangle is 16 and the length of the other leg is odd, what are the lengths of all the sides of the triangle?"

Kainui
 one year ago
Best ResponseYou've already chosen the best response.1dw:1438881465435:dw Equivalently (?) can we stick 4 pythagorean triples together like this to form a quadrilateral?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1\(a^2+b^2=c^2\) if \((a,b,c)\) is primitive pythagorean triple, then \(a\not \equiv b \pmod{2}\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1Let \(s\) be the length of side of square. dw:1438881952775:dw I think we want to show that there exist \((x,y)\) such that all below expressions are perfect squares : \(x^2+y^2\) \((xs)^2+y^2\) \(x^2+(ys)^2\) \((xs)^2+(ys)^2\) and \(0\le x,y\le s\) all are integers

Kainui
 one year ago
Best ResponseYou've already chosen the best response.1Hmmmm I don't know anything about quadratic residues, do you know if they might be possibly useful here?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1yeah that seem to work, im able to find few squares with 3 integer distances https://i.gyazo.com/a2ddaf3248da9fb3781cdc5aa56d75fd.png

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1I'll need to review, but im pretty sure @oldrin.bataku has these on his finger tips!

Kainui
 one year ago
Best ResponseYou've already chosen the best response.1Hahaha true, I think the problem is getting that fourth one. That's the toughie to make them all line up hmmm.

freckles
 one year ago
Best ResponseYou've already chosen the best response.2http://mathworld.wolfram.com/RationalDistanceProblem.html yeah this conway and guy dude were only able to find an infinite number of solutions such that three of the distances were rational

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1Here is a fiddle (dumb, fast script) if you want to play with http://jsfiddle.net/ganeshie8/pdcgprc9/

Kainui
 one year ago
Best ResponseYou've already chosen the best response.1Ok thanks I'll definitely play with this and see what I can do.

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.0Is this equivalent to asking is it possible to draw four circles of integer radius with the center located at the four corners such that they intersect at one point?

Kainui
 one year ago
Best ResponseYou've already chosen the best response.1Possibly, could you maybe draw a rough picture to give me a better idea, I'm having some trouble visualizing it.
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.