Now let us fill a circle with different pieces: for example, with a half circle and two quarter circles. The result is 1 = 1/2 + 2/4. And in the inset itself it is shown that 1/2 = 2/4. If we should wish to fill the circle with the largest piece (1/2) combined with the fewest number of pieces possible, it would be necessary to withdraw the two quarter sectors and replace them by another half circle; result:
1 = 1/2 + 1/2 = 2/2 = 1.
Let us fill a circle with three 1/5 sectors and four 1/10 sectors:
1 = 3/5 + 4/10.
If the larger pieces are left in and the circle is then filled with the fewest number of pieces possible, it would necessitate replacing the four tenths by two fifths. Result:
1 = 3/5 + 2/5 = 5/5 = 1.
Let us fill the circle thus: 5/10 + 1/4 + 2/8 = 1.
Now try to put in the largest pieces possible by substituting for several small pieces a large piece which is equal to them. In the space occupied by the five tenths may be placed one half, and in that occupied by the two eighths, one fourth; then the circle is filled thus:
1 = 1/2 + 1/4 + 1/4 = 1/2 + 2/4.
We can continue to do the same thing, that is to replace the smaller pieces by as large a sector as possible, and the two fourths can be replaced by another half circle. Result: