Henry Ernest Dudeney/Puzzles and Curious Problems/244 - Sharing a Grindstone/Solution
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
- 1932: Henry Ernest Dudeney: Puzzles and Curious Problems ... (previous) ... (next): Solutions: $244$. -- Sharing a Grindstone
- 1968: Henry Ernest Dudeney: 536 Puzzles & Curious Problems ... (previous) ... (next): Answers: $291$. Sharing a Grindstone