Henry Ernest Dudeney/Puzzles and Curious Problems/244 - Sharing a Grindstone/Solution

by : $244$

 * Sharing a Grindstone

Solution
The first man can use approximately $1.754$ inches of the grindstone.

The second man can use approximately $2.246$ inches of what remains.

The third man is left with $4$ usable inches plus the aperture.

Proof
To simplify our thinking, let us discuss the radius rather than the diameter of the grindstone.

Thus we have a grindstone $10$ inches in radius with an aperture of $2$ inches.

Thus the usable radius is $8$ inches

Let $r_1$, $r_2$ and $r_3$ be the radii in inches of the grindstone as it is received by the three men in turn.

We are of course given that $r_1 = 10$.

From Area of Circle, the total quantity $q$ of grindstone of a given inches $r$ is proportional to $r^2$.

That is:
 * $q = k r^2$

where $k = \pi$, but this is irrelevant.

We have that:
 * $10^2 - {r_2}^2 = {r_2}^2 - {r_3}^2 = {r_3}^2 - 2^2$

Thus:
 * $r_2 = \sqrt {68} \approx 8.246$
 * $r_3 = \sqrt {36} = 6$

Hence:
 * the first man can use $10 - \sqrt {68} \approx 1.754$ inches of the grindstone
 * the second man can use $\sqrt {68} - 6 \approx 2.246$ inches of what remains
 * the third man has $4$ usable inches of what remains.