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

From ProofWiki
Jump to navigation Jump to search

Puzzles and Curious Problems by Henry Ernest Dudeney: $244$

Sharing a Grindstone
Three men bought a grindstone $20$ inches in diameter.
How much must each grind off so as to share the stone equally,
making an allowance of $4$ inches off the diameter as waste for the aperture?
We are not concerned with the unequal value of the shares for practical use --
only with the actual equal quantity of stone each receives.


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:

$10^2 - {r_2}^2 = {r_2}^2 - {r_3}^2 = {r_3}^2 - 2^2$
\(\ds 10^2 - {r_2}^2\) \(=\) \(\ds {r_2}^2 - {r_3}^2\)
\(\ds \) \(=\) \(\ds {r_3}^2 - 2^2\)
\(\ds \leadsto \ \ \) \(\ds 100 + {r_3}^2\) \(=\) \(\ds 2 {r_2}^2\)
\(\ds {r_2}^2 + 4\) \(=\) \(\ds 2 {r_3}^2\)
\(\ds \leadsto \ \ \) \(\ds 100 + \dfrac { {r_2}^2 + 4} 2\) \(=\) \(\ds 2 {r_2}^2\)
\(\ds \leadsto \ \ \) \(\ds 100 + 2\) \(=\) \(\ds \dfrac {3 {r_2}^2} 2\)
\(\ds \leadsto \ \ \) \(\ds {r_2}^2\) \(=\) \(\ds 68\)
\(\ds \leadsto \ \ \) \(\ds {r_3}^2\) \(=\) \(\ds \dfrac {68 + 4} 2 = 36\)

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.


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Henry_Ernest_Dudeney/Puzzles_and_Curious_Problems/244_-_Sharing_a_Grindstone/Solution&oldid=562007"