Pentagon construction

Vignesh and Amrit from Hymers College in the UK sent in proofs which use the same reasoning.

This is Vignesh's diagram, with the names of the points added.

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This is the beginning of Amrit's proof:

Let the radius of the circle $C_1$ (black in Vignesh's diagram) with centre at point $\text A$ passing through point $\text B$ be $1.$

Vignesh made a different choice:

$R_0 = \text{AB}$ (radius of black circle)

Let $R_0 = x$

Vignesh calls the distance $\text{AB}$ $x$, because it is unknown, but Amrit says it is $1.$ Changing the length $\text{AB}$ will rescale the diagram, but it won't affect whether the pentagon is regular, so it doesn't matter how long $\text{AB}$ is. This is why it is safe for Amrit to choose that $\text{AB}=1.$ Vignesh also chose this length at the beginning of the constuction, because it starts with two points $\text{A}$ and $\text{B}$ - and Vignesh had to choose how far apart to put them.

Using $1$ instead of $x$ makes the proof simpler, so here is the rest of Amrit's proof: 

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 The right angled triangle $\text{FAD}$ was apparent to me when I completed the construction. By Pythagoras' theorem,

Since point F is the midpoint of the radius $\text{AB},$ $\text{FA}=\frac{1}{2}.$ Since $\text{DA}$ is a radius, $\text{DA}=1.$ Substituting these values into the above equation gives



Then, I saw the right angled triangle $\text{ADG}$. By Pythagoras' theorem,

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But $\text{AG}=\text{GF}+\text{FA}$

And, since $\text{DF}$ and $\text{GF}$ are both radii of the circle with centre $\text F$ passing through the point $\text D$ (red), $\text{GF}=\text{DF}=\frac{\sqrt5}{2},$ so $\text{AG}=\frac{\sqrt5}{2}+\frac{1}{2}=\frac{\sqrt5+1}{2}.$ $\text{DA}$ is again equal to $1.$ Substituting these values into the above equation, we obtain

$\text{DI = DJ = DG},$ since these line segments are all radii of the circle with centre $\text D$ passing through $\text G$ (yellow).

The next right angled triangle I observed was triangle $\text{AHD.}$ By Pythagoras' theorem,

            $\text{DH}^2=\text{AH}^2+\text{DA}^2$

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But $\text{AH}=\text{HF}-\text{FA}$

And $\text{HF}=\text{DF}=\frac{\sqrt5}{2}$, since they are both radii of the circle with centre $\text F$ that passes through $\text D$ (red). So $\text{AH}=\frac{\sqrt5}{2}-\frac{1}{2}=\frac{\sqrt5-1\ }{2}.$ $\text{DA}$ is again equal to $1.$ Substituting these values into the above equation, we obtain

$\text{DL = DK = DH,}$ since these line segments are all radii of the circle with centre $\text D$ passing through $\text H$ (green).

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 We can now find the size of $\angle \text{ DAL}$ using the cosine rule:

$\text{LA = DA}=1,$ since these line segments are both radii of the circle $C_1$ (black), and it was deduced above that $\text{DL = DH}=\dfrac{\sqrt{10-2\sqrt5}}{2}.$ Substituting these values into the above equation, we obtain

Now, triangles $\text{LAD}$ and $\text{DAK}$ are congruent. This is because $\text{LA = KA}$ (radii of [black] circle), $\text{AD = DA}$ (common side of both triangles) and $\text{DL = DK}$ (deduced above), so the triangles are congruent by the SSS theorem. In particular, this means that

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 I then considered triangle $\text {ALK.}$

$\angle\text{ LAK}=\angle\text{ DAL}+\angle\text{ DAK}=144^\circ$

From the cosine rule, we can find the length of $\text{LK}:$



$\text{LA = KA}=1,$ since they are the radii of the circle $C_1$ (black), and we know that $\cos{\angle\text {LAK}}=\cos{144^\circ}=\frac{-1-\sqrt5}4.$ Substituting these values into the above equation, we obtain



So $\text{LK = DI = DJ.}$ We see then that triangle $\text{ALK}$ is congruent to triangles $\text{DAI}$ and $\text{DAJ}$ by the SSS theorem, since one side of one triangle is equal to another side of the other two triangles ($\text{LK = DI = DJ}$), and the other two sides of one triangle are equal to the other two sides of the other two triangles, because they are all radii of the circle $C_1$ (black).

In particular then, $\angle\text{ DAJ}=\angle\text{ DAI}=144^\circ.$ Also,



So $\angle\text{ JAI}=72^\circ.$

Now, $\angle\text{ LAI}=\angle\text{ DAI}-\angle\text{ DAL}=72^\circ.$ Therefore, $\angle \text { KAJ}=360^\circ-\left(\angle \text{ DAK}+\angle \text{ DAL}+\angle \text{ LAI}+\angle \text{ JAI}\right)=72^\circ.$

Therefore, by the SAS congruence axiom, the triangles $\text{AIJ, AJK, AKD, ADL, ALI}$ are all congruent to each other and, in particular, $\text{IJ = JK = KD = DL = LI}$ and all the angles in one of the above triangles are equal to the angles in the corresponding four triangles, therefore $\angle\text{ I}=\angle\text{ J}=\angle\text{ K}=\angle\text{ D}=\angle\text{ L},$ where these angles are the interior angles of the shape $\text{IJKDL,}$ making this shape a regular pentagon.

It is also possible to prove that the construction gives a pentagon using coordinate geometry:

Using Cartesian coordinates with $A$ as the origin and $B$ at $(1,0)$.

Then, $D$ is at $(0,1)$, since it is the same distance away from $A$ as $B$ is but on a perpendicular line. $F$ is the midpoint of $AB$, so is at $(\frac 12,0)$.

Pythagoras' Theorem tells us that the distance $DF = \sqrt{1^2 + (\frac 12)^2} = \frac{\sqrt{5}}{2}$. Therefore $G$ is at $(\frac{1+\sqrt 5}2,0)$ and $H$ is at $(\frac{1-\sqrt 5}2,0)$.

Then, $DG = \sqrt{1^2 + \left( \frac{1+\sqrt 5}2 \right)^2} = \sqrt{1+\frac{6+2\sqrt 5}{4}} = \frac{\sqrt{10+2\sqrt{5}}}{2}$.

Then, looking at the equations of the two circles (black and yellow), we have:

\begin{eqnarray}

x^2 + y^2 &=& 1\\

x^2 + (y-1)^2 &=& \frac{5+\sqrt 5}2 \\

\end{eqnarray}

Subtracting one from the other yields:$$\begin{eqnarray}

y^2-(y-1)^2&=&1-\frac{5+\sqrt 5}{2}\\

\Rightarrow y^2-\left(y^2-2y+1\right) &=&1-\frac{5+\sqrt 5}{2}\\

\Rightarrow 2y-1 &=& 1-\frac{5+\sqrt 5}{2}\end{eqnarray}$$This has solution:$$y = -\frac{1+\sqrt 5}{2} = \cos 144^\circ$$

So $x =\pm \sin 144^\circ$, since $\left( \pm\sin 144^\circ \right) ^2 + \cos ^2 144^\circ = 1$

To use right-angled triangles, use the sine and cosine graphs to see how $\sin 144^\circ$ and $\cos 144^\circ$ compare to angles less than $90^\circ$:

 

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$\sin 144^\circ = \sin (90^\circ-(144^\circ-90^\circ)) = \sin 36^\circ,\cos 144^\circ = -\cos(180^\circ-144^\circ)=-\cos36^\circ$

So the points $I$ and $J$ can be plotted as shown:



Meanwhile, $DH = \sqrt{1^2 + \left( \frac{1-\sqrt 5}2 \right)^2} = \sqrt{1+\frac{6-2\sqrt 5}{4}} = \frac{\sqrt{10-2\sqrt{5}}}{2}$.

Then, looking at the equations of the two circles, we have:

\begin{eqnarray}

x^2 + y^2 &=& 1\\

x^2 + (y-1)^2 &=& \frac{5-\sqrt 5}2 \\

\end{eqnarray}

Subtracting one from the other yields:$$2y-1 = 1-\frac{5-\sqrt 5}{2}$$This has solution:$$y = \frac{\sqrt 5-1}{2} = \cos 72^\circ$$

So $x =\pm \sin 72^\circ$, since $\left( \pm\sin 72^\circ \right) ^2 + \cos ^2 72^\circ = 1$

So the points $K$ and $L$ can be plotted as shown: