Henry Ernest Dudeney/Puzzles and Curious Problems/188 - Squaring the Circle/Solution 2

by : $188$

 * Squaring the Circle

Solution

 * Dudeney-Puzzles-and-Curious-Problems-188-solution.png

$AB$ 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:
 * $CH = \dfrac r 2 \times \dfrac 2 {2 + \sqrt 3} = \dfrac r {2 + \sqrt 3}$

By Pythagoras' Theorem:
 * $DG = DH = \sqrt {\paren {\dfrac r {2 + \sqrt 3} }^2 + r^2} = \dfrac r {2 + \sqrt 3} \sqrt {8 + 4 \sqrt 3}$

Now we can calculate $DG + GH$:

The fraction:
 * $\dfrac {2 \paren {1 + \sqrt {2 + \sqrt 3} } } {2 + \sqrt 3} \approx 1.571174 \ldots$

is a close approximation to:
 * $\dfrac \pi 2 \approx 1.570796 \ldots$