and these increments will be in proportion to the angles of arc, viz. 55ô¯ã₤22ÿ£¢ãýã₤:ã₤34ô¯ã₤38ÿ£¢ãý, or as ôñ96ã₤:ã₤ôñ60, i.e. as 8ã₤:ã₤5. And accordingly, if we may assume (and the assumption is a very plausible one), that, just as the quadrant itself divided into two halves after it got to a certain size, so each of its two halves will reach the same size before again dividing, it is obvious that the triangular portion will be doubled in size, and therefore ready to divide, a considerable time before the quadrilateral part. To work out the problem in detail would lead us into troublesome mathematics; but if we simply assume that the increments are proportional to the increasing radii of the circle, we have the following equations:ã
Let us call the triangular cell T, and the quadrilateral, Q (Fig. 151); let the radius, OA, of the original quadrantal cell =ã₤a =ã₤1; and let the increment which is required to add on a portion equal to T (such as PPÿ£¢ãýAÿ£¢ãýA) be called x, and let that required, similarly, for the doubling of Q be called xÿ£¢ãý.
Then we see that the area of the original quadrant
while the area of T
The area of the enlarged sector, pÿ£¢ãýOAÿ£¢ãý,
and the area OPA
Therefore the area of the added portion, Tÿ£¢ãý,
And this, by hypothesis,
We get, accordingly, since a =ã₤1,
and, solving,
Working out xÿ£¢ãý in the same way, we arrive at the apôÙproxôÙiôÙmate value, xÿ£¢ãýã₤+ã₤1 =ã₤1ôñ517. {369}
This is as much as to say that, supposing each cell tends to divide into two halves when (and not before) its original size is doubled, then, in our flattened disc, the triangular cell T will tend to divide when the radius of the disc has increased by about a third (from 1 to 1ôñ345), but the quadrilateral cell, Q,