# Proof regarding distances from a point in a grid

1. Feb 6, 2013

### physicsernaw

1. The problem statement, all variables and given/known data

Given a grid, x by y, where x and y are integers both greater than or equal to one, prove that it is impossible for there to be a point (a,b) where the distance d from (a,b) to three distinct corners of the grid are equal only for these three corners (i.e. the point has to be equidistant to ONLY three corners; the point cannot also be equidistant to 2, or 4 corners).

Assume the grid is arranged in this manner:

top left corner: (0,0)
bottom left corner: (0, y)
top right corner: (x, 0)
bottom right corner: (x, y)

Remember:
x = horizontal distance of grid
y = vertical distance of grid
Both are integers that must be greater than or equal to one.

2. Relevant equations
None directly given, but use of the distance formula seems obvious.

3. The attempt at a solution

**This "proof" is extremely informal. I have very little experience in proof based maths. I am looking for help to make this "proof" more rigorous / formal, or perhaps even suggestions at a simpler way of attacking the proof**

- The center of a square grid is always (x/2, y/2). From the center point (a,b), all corners are equidistant away.
- In order for a point (a,b) to be equidistant from only two corners in the grid, one of the two must be true:
a = x/2 OR b = y/2 (if both were true, then the point would be at the center, and thus equidistant to all four corners)

- We will prove the above condition for both cases:
bottomRight: (x,y)
bottomLeft: (0,y)

distance to bottom right = sqrt[(x-a)^2 (y-b)^2]
distance to bottom left = sqrt[a^2 + (y-b)^2]

Make the two equations equal and solve for a:
(x-a)^2 + (y-b)^2 = a^2 + (y-b)^2
x^2 -2ax +a^2 = a^2
x^2 -2ax = 0
x(x - 2a) = 0; solu(x = 0)
x -2a = 0
x = 2a
a = x/2; solu(a = 0)

x cannot equal zero because we assumed from the beginning that the grid had integer dimensions where (x and y) > 0.
Thus, the only solution for the equation is: a = x/2, as we were trying to show.

Now, test any of the other two remaining corners to see if a third corner can be equidistant from the two prior corners. We will arbitrarily choose the distance of the top left corner from (a,b) and set it equal to the distance of (a,b) to the bottom left; the former distance is on the LHS, the latter on the RHS of the equation below.

a^2 + b^2 = a^2 + (y-b)^2
b^2 = (y-b)^2
b^2 = y^2 -2by + b^2
2by = y^2
2b = y
b = y/2

y/2 is the only solution for the bottom left corner and the top left corner. The three corners we were looking at were the bottom left, top left, and bottom right. We found that for a point (a,b) to be equidistant from the bottom left and bottom right, a=x/2. We now see that for the third corner, top left, to be equal to the bottom left, b=y/2. But taking both these solutions, we get a center point, i.e. (x/2, y/2), and as before stated, the center point is equidistant from all four corners.

Thus, there is not point (a,b) such that only 3 corners are equidistant from it.
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Last edited: Feb 6, 2013
2. Feb 6, 2013

3. Feb 6, 2013

### haruspex

Looks fine to me. Yes, a geometric version is probably a little simpler to lay out, but it's the same argument.

4. Feb 7, 2013

### Ray Vickson

You can simplify this a lot. I am going to change your x and y to a and b, because I need x and y for something else.

Say the side AB is from (0,0) to (a,0) and the side AC is from (0,0) to (0,b) (so AB is along the x axis and AC along the y axis). The fourth point, D, is at (a,b). If a point is equidistant from 3 of the 4 points, we can always find an equivalent setup in which a point is equidistant from A, B and C. Any point equidistant from A and B must lie on the line L1: x = a/2, and any point equidistant from A and C must lie on the line L2: y = b/2, so a point equidistant from A,B and C must lie on both lines, hence must be at (x,y) = (a/2,b/2). That point is also the same distance from D.