Henry Ernest Dudeney/Puzzles and Curious Problems/188 - Squaring the Circle/Solution 2
Puzzles and Curious Problems by Henry Ernest Dudeney: $188$
- Squaring the Circle
- The problem of squaring the circle depends on finding the ratio of the diameter to the circumference.
- This cannot be found in numbers with exactitude,
- but we can get near enough for all practical purposes.
- But it is equally impossible, by Euclidean geometry, to draw a straight line equal to the circumference of a given circle.
- You can roll a penny carefully on its edge along a straight line on a sheet of paper and get a pretty exact result,
- but such a thing as a circular garden-bed cannot be so rolled.
- Now, the line below, when straightened out
- (it is bent for convenience in presentation),
- is very nearly the exact length of the circumference of the accompanying circle.
- The horizontal part of the line is half the circumference.
- Could you have found it by a simple method, using only pencil, compasses and ruler?
Solution
$AB$ is the diameter of the circle whose center is at $C$.
Bisect the semicircle $AB$ at $D$.
Construct $AE = AC$ and $BF = BC$.
Construct $DE$ and $DF$.
Let $DE$ and $DF$ intersect $AB$ and $G$ and $H$.
Then $DG + GH$ is one quarter the length of the circumference of the circle within a $\dfrac 1 {4000}$ part.
Proof
Notice that the above diagram is symmetrical.
Therefore $DC \perp AB$.
Let the radius of the circle be $r$.
Denote the foot of the perpendicular from $E$ to $AB$ be $P$.
Since $CA = CE = AC$, $\triangle ACE$ is equilateral.
Hence $AP = PC = \dfrac r 2$ and $EP = \dfrac {\sqrt 3} 2 r$.
$\triangle PEH \sim \triangle CDH$ as Equiangular Triangles are Similar.
Therefore $\dfrac {DC} {EP} = \dfrac {CH} {PH} = \dfrac 2 {\sqrt 3}$.
We can now obtain:
- $CH = \dfrac r 2 \times \dfrac 2 {2 + \sqrt 3} = \dfrac r {2 + \sqrt 3}$
- $DG = DH = \sqrt {\paren {\dfrac r {2 + \sqrt 3} }^2 + r^2} = \dfrac r {2 + \sqrt 3} \sqrt {8 + 4 \sqrt 3}$
Now we can calculate $DG + GH$:
\(\ds DG + GH\) | \(=\) | \(\ds DG + 2 CH\) | by symmetry | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac r {2 + \sqrt 3} \sqrt {8 + 4 \sqrt 3} + \dfrac {2 r} {2 + \sqrt 3}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {2 \paren {1 + \sqrt {2 + \sqrt 3} } } {2 + \sqrt 3} r\) |
The fraction:
- $\dfrac {2 \paren {1 + \sqrt {2 + \sqrt 3} } } {2 + \sqrt 3} \approx 1.571174 \ldots$
is a close approximation to:
- $\dfrac \pi 2 \approx 1.570796 \ldots$
$\blacksquare$
Sources
- 1932: Henry Ernest Dudeney: Puzzles and Curious Problems ... (previous) ... (next): Solutions: $188$. -- Squaring the Circle
- 1968: Henry Ernest Dudeney: 536 Puzzles & Curious Problems ... (previous) ... (next): Answers: $289$. Squaring the Circle