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Hypotenuse Lattice Points

Age 14 to 16
Challenge Level

The triangle OMN has vertices M(0,m) and N(n,0) with whole number co-ordinates and we have to find how many lattice points there are on the hypotenuse MN.

The hint here was "First test what happens when m and n are co-prime (have no common factor)", so this tells you that somehow the problem involves common factors.

Absolutely the best way to learn is to make mistakes (generally if you make none then the maths has not got enough challenge in it for you).

A solution was sent in with a graph drawn for m=27 and n=20, and the line looked as if it went through the point (3, 23) but it actually goes through (3, 22.95). You have to check things like this. One way is to draw the line yourself, decide what you think about it, then check using a graphic calculator and using the trace function which will give you readings of the co-ordinates of the points on the line. ... When you go 20 squares in the x-direction you have to go DOWN 27 squares in the y direction. You can't divide this big step into smaller steps going across and down by whole number changes. [If you use other numbers which have a common factor, instead of 20 and 27, then there will be smaller steps between points on the line which have whole number co-ordinates].

Think about the line extended on and on in the example above. It has equation:
27 x + 20 y = 540
and we are looking for whole number solutions to this equation (a Diophantine equation).

We know that x = 0, y = 27 is a solution and if there is another solution with a bigger x then it will go with a smaller y .

Looking at this equation we see that 20 divides exactly into 540 - 20y so x has to be a multiple of 20. Similarly y has to be a multiple of 27. Solutions are given by points on the line with integer co-ordinates and they are:

. . , (-20, 54), (0,27), (20, 0), (40, -27), . . .

When you go 20 squares in the x -direction you have to go DOWN 27 squares in the y direction. There are no solutions in the first quadrant apart from the points on the axes.

The following is a short general proof of this result:

If m and n are co-prime then there are no lattice points on the line between M(0, m) and N(n, 0) in the first quadrant.

The line MN has equation mx + ny = mn
Suppose (p,q) is the lattice point on MN and we have
mp + nq = mn (1)
We see from (1) that n must divide nq so that, if m,n are co-prime, then m must divide q so q = 0 or q = m
Similarly n divides mp so p = 0 or p = n. The only two lattice points are the points M and N.
In the example given with m = 27 and n = 20. The only 2 lattice points are the points M and N.
In the example given with m = 27 and n = 20 the general points with whole number co-ordinates are given by:
x = 20t, y = 27 - 27t (where t is any integer).

Now you will need to do some work to check that when the greatest common divisor of m and n is the number 2, so that m = 2r and n = 2s where r and s are co-prime, this introduces one lattice point between M and N.

Next, if the greatest common divisor is 3, there will be two lattice points between M and N, and so on.

We use the notation gcd(m,n) for the greatest common divisor of m and n, so for example gcd(30, 70) = 10.

The general answer is that the number of lattice points (including M and N on the axes) is 1 + gcd(m,n).