Bilkent University problem of the month
 

http://www.fen.bilkent.edu.tr/~cvmath/Problem/0703q.pdf

Interesting problem . My guess is the answer is n= 3 but I have to prove it .

Re: Bilkent University problem of the month
 

The answer is going to be > 2.
For n = 1, let P = A_1. The distance is 0, which is rational. So the n cannot be 1.

For n = 2, let A_1 = (x_1, y_1) and A_2 = (x_2, y_2). I will show that there exist P such that both distances (from A_1 to P and from A_2 to P) are rational.
Case 1: dist(A_1, A_2) rational
Let M = midpoint(A_1, A_2). Take P so that the line segment P, M is perpendicular to the line segment A_1, A_2 such that P is 4 / 3 * dist(A_1,M) away from M. Then dist(P, A_1) = dist(P, A_2) = 5 / 3 * dist(A_1,M) by the Pythagorean Theorem. Since dist(A_1, A_2) rational, dist(A_1,M) rational, and therefore dist(P, A_1) = dist(P, A_2) rational.

Case 2: dist(A_1, A_2) irrational
Let P = some point distance x so that the line segment P, M is perpendicular to the segment A_1, M. Then dist(P, A_1) = (dist^2(M, A_1) + x^2) ^(1/2). We can find x (irrational I think) so that dist(P, A_1) = dist(P, A_2) is rational. So if we want dist(P, A_1) to be the smallest rational number greater than dist(M, A_1), call it Q_1. Then x = (Q_1^2 - dist^2(M, A_1)) ^ (1/2)
Hence, n cannot be 2.

Re: Bilkent University problem of the month
 

For n=2 , just pick any equilateral triangle with a1= (0,0)
a2= (2,0) p(1,sqrt3) . The distance (p,a1) and (p,a2) = 2

For n=1 is obvious . Just pick a1(0,0) and p(0,x/y)

I already figured this problem out but I won't post solutions just yet .

Re: Bilkent University problem of the month
 

I think you're proof of n=2 is not correct.
You have to start with two points and find a point that is a rational distance from both of them.
I might be reading incorrectly what you are doing, but it seems like you start with two particular points instead of grabbing two arbitrary ones.

The proof the other guy mentions is the same one I have (pick a point in the perpendicular to the midpoint. Along it there will be some point that will make the hypotenuse rational and since it is the same distance from both sides, it works).

Re: Bilkent University problem of the month
 

You're right Enrique . I only showed the case when the distance between a1 and a2 is rational . Then certainly we can find a P that is rational to both points and an equilateral triangle works .

** Spoiler **
I think I have the solution below so if you don't want to know , then don't read it .





Take three points equally spaced from each other and there distances are apart are rational . Label the three points A,B,C and we have AB =BC

I claim that P to at least one of the points will be an irrational distance . Suppose the opposite is true .
Lets use the cosine law .
(AP)^2 = (AB)^2 + (PB)^2 -2*(AB)(PB)cos<apb (1)
Also , (PC)^2 = (PB)^2 +(BC)^2 -2(PB)(BC)cps<pbc (2)

Lets simplify things using the fact that AB=BC and that cos<abp = - cos<pbc

Add 1 and 2 to get (AP)^2 + (PC)^2=2(AB)^2 + 2(PB)^2
We get an immediate contradiction since there is an odd number of 2's on the rhs and there must be an even number on the lhs . This means AP , PC or PB must be irrational .

Let me know if the solution is incomplete or if I missed something out .

Re: Bilkent University problem of the month
 

Here is my final solution with corrections .

Answer: n=3

Take A(0,0) B(sqrt2,0) c(2sqrt2,0)

P is any arbitrary point on the same plane and I claim that dist(p,(A,B,C)) will be irrational no matter what .

case1 (P is not on the x axis ).

Denote dist(PA)=x ; dist(PB)=y; dist(PC)=z

from cosine law on triangle PAB we get

x^2=2+y^2-2*ycos<PAB (1)
cosine on triangle PBC

z^2= 2 + y^2 -2*ycos<PBC (2)

Again we use the fact that cos<PAB=-cos<PBC
Add 1 and 2

x^2+z^2= 4+2y^2

If we use a parity argument , it's impossible for x,y,z to all be rational since there is an odd number of factors of 2 on the rhs and an even number on the lhs .

case 2 (P is on the x axis) .

This is easy to verify that x,y,z cannot all be rational and we may use a parity argument as before .

Re: Bilkent University problem of the month
 

Aren't A, B & C collinear here? If so, what's the definition dist(p, (A,B,C))? The length of the perpendicular? If so, your claim is certainly false. Take p(0,1).

Re: Bilkent University problem of the month
 

Hey djames , thx for the reply . Yes A,B,C are collinear and on the x-axis .

What I mean is that the distance PA , PB, or PC will contain at least one irrational length no matter where P is on the xy plane .

If we take P(0,1) like you said , then PA=1
PB= sqrt(2+1)=sqrt3 which is irrational which suffices since we only need to show that there exists at least one irrational distance which is PB .

We've already established that it's impossible for n=2 or n=1 but this shows that for n=3 we can find three points and a third point P which contains at least one irrational length to any of the three points .

Re: Bilkent University problem of the month
 

Actually we don't even need to consider two cases .

The cosine law handles the situation where P is on the x axis as well with <PAB = 0 and cos<PAB = 1

If either x,y or z =0 then certainly p is irrational to one of the distances .

ie x=0 , then PB= sqrt 2