Solution

We know that in a cyclic quadrilateral, the opposite angles sum to 180, so we need to consider the different ways that the angles of the cyclic quadrilateral could be paired up to achieve this.

Either x + kx = 180, or x + mx = 180 or kx + mx = 180.

In the first case, solving x + kx = 180, we get x = 180/(k+1), so the angles of the quadrilateral must be:

180/(k+1), 180k/(k+1), 180m/(k+1), and lastly 180 – 180m/(k+1), which simplifies to

180(k – m + 1)/(k+1).

Since m is at least 1, k/(k+1) must be greater or equal to (k-m + 1)/(k+1), so the largest angle is max (180k/(k+1), 180m/(k+1)).

If 180k/(k+1) = 160 then k/ (k + 1) = 8/9 so k = 8

Now checking what m would be, for 180(k – m + 1)/(k+1) to give a sensible answer, we need m to be less than or equal to k, so m could be any of the integers from 1 to 8 inclusive.

If 180m/(k+1) = 160 we know is that 9m = 8k+ 8, and for 180k/(k+1) not to exceed 160, we must have k less than or equal to 8.

However, not all of these values for k will give an integer value for m, so we need to check:

Using 9m = 8(k+1), we can see that k + 1 needs to be divisible by 9 so the only acceptable value for k is k = 8 (so m = 8) which is already counted in what we worked out before.

Therefore, for this way of pairing up the angles we get the following eight solutions

k	8	8	8	8	8	8	8	8
m	1	2	3	4	5	6	7	8

In the second case, if, x + mx = 180 by following a similar argument, the angles must be

180/(m+1), 180m/(m+1), 180k/(m+1), and 180(m – k + 1)/(m+1).

This time the largest angle will be max (180k/(m+1), 180m/(m+1)).

We can use the same arguments as in the first case, but with k and m swapped round.

Therefore, for this way of pairing up the angles we get the following solutions

k	1	2	3	4	5	6	7	8
m	8	8	8	8	8	8	8	8

We’ve already counted k = 8, m= 8 so there are seven new possibilities here.

In the third case, kx + mx = 180 gives us x = 180/(k + m), so the angles of the quadrilateral must be:

180/(k + m), 180k/(k + m), 180m/(k + m) and lastly 180 - 180/(k + m) which simplifies to

180(k + m – 1)/(k + m).

Since m is at least 1, (k + m - 1)/(k + m) must be greater or equal to each of k/(k + m), m/(k + m), and 1/(k + m) so the largest angle is 180(k + m - 1)/(k + m)

If 180(k + m – 1)/(k + m) = 160, then we can simplify the equation to get k + m = 9_.

Therefore, for this way of pairing up the angles we get the following integer solutions

k	1	2	3	4	5	6	7	8
m	8	7	6	5	4	3	2	1

We’ve already counted k = 8, m= 1 and k = 1, m = 8, so there are six new possibilities here.

This makes 21 possible pairs of values for k and m.

Challenge 239: Quadrilateral Quandary