Solution:  4, 3, 8/3.  

The first thing to note is that the diameter of the circle is 2 units. Since the height of the triangle must be greater than the diameter of the circle, then a > 2. (Because we cannot logically distinguish between a and b due to symmetry, then b > 2, as well.)

Draw three radii, each perpendicular to a side of the triangle. This divides the triangle into three regions: a blue kite in the upper left, a green kite on the bottom right, and a square of side length 1. The blue kite consists of two triangles, each with height a – 1 and base 1, so it has area a – 1. Similarly, the green kite consists of two triangles with height b – 1 and base 1, so it has area b – 1. And the square, obviously, has area 1. Adding these pieces gives the following expression for the area of the triangle: a + b – 1.

However, we also know that a triangle with height a and base b has an area of ½ab. These two different expressions for the area lead to the following equation:

We can now test this equation for integer values of b > 2, which lead to the following:

At this point, we can stop. Remember that a > 2, so although there are values of a < 2 that satisfy the equation, they are not reasonable in the context of the problem. Therefore, we needn’t test any values of b > 6, and the three possible values of a are 4, 3, and 8/3.