Pentagon Problem

Let $P$ be the intersection of $CE$ and $BM$ . Let $|AB|=x$ and denote $\varphi=\dfrac{1+\sqrt{5}}{2}$ .

Claim: $|CE|=\varphi \times x$ .
Proof: We will apply Ptolemy's Theorem:
Inscribe a regular pentagon $ABCDE$ inside a circle. WLOG, let its side length be $1$ . If we let the length of the diagonal be $x$ , then using Ptolemy's Theorem on ABCD tells us that $x^{2}=x+1$ . Using the quadratic formula and taking the positive root tells us that $x=\dfrac{1+\sqrt{5}}{2} = \phi$ . Therefore, the length of the diagonal is $\phi$ multiplied by the side length.

Note that $\triangle AMB\cong\triangle EMP$ by $ASA,$ hence $|EP|=|AB|=x$ .

Now note that $\triangle BZA\sim\triangle PZC$ by $AAA$ and $|AB|=x$ while $|CP|=|CE|+|EP|=(1+\varphi)(x)$ .

It follows that $|CZ|=(1+\varphi)|AZ|=3+3\varphi$ . Hence $|AC|=6+3\varphi$ and $|AB|=\dfrac{|AC|}{\varphi}=\dfrac{6+3\varphi}{\varphi}=\boxed{3\sqrt{5}}$ .

Let the side length of the regular pentagon $ABCDE$ be $a$ . By sine rule , we have:

$\begin{aligned} \frac {\sin \angle ABM}{AM} & = \frac {\sin \angle AMB}{AB} \\ \frac {\sin {\color{#3D99F6}\angle ABM}}{\frac a2} & = \frac {\sin (180^\circ - \angle EAB - {\color{#3D99F6}\angle ABM})}a & \small {\color{#3D99F6} \text{Let }\angle ABM = \theta} \\ 2 \sin {\color{#3D99F6}\theta} & = \sin (180^\circ - 108^\circ - {\color{#3D99F6}\theta}) \\ 2 \sin \theta & = \sin (72^\circ - \theta) \\ 2 \sin \theta & = \sin 72^\circ \cos \theta - \sin \theta \cos 72^\circ \\ \tan \theta & = \frac {\sin 72^\circ}{2+\cos 72^\circ} \end{aligned}$

By sine rule again,

$\begin{aligned} \frac {\sin \angle ABM}{AZ} & = \frac {\sin \angle AZB}{AB} \\ \frac {\sin \theta}3 & = \frac {\sin (180^\circ - 36^\circ - \theta)}a \\ a \sin \theta & = 3 \sin (36^\circ + \theta) \\ a \sin \theta & = 3 \sin 36^\circ \cos \theta +3 \sin \theta \cos 36^\circ \\ \tan \theta & = \frac {3 \sin 36^\circ}{a-3\cos 36^\circ} \end{aligned}$

Therefore,

$\begin{aligned} \implies \frac {3 \sin 36^\circ}{a-3\cos 36^\circ} & = \frac {\sin 72^\circ}{2+\cos 72^\circ} \\ \frac {a-3\cos 36^\circ}{3 \sin 36^\circ} & = \frac {2+\cos 72^\circ}{\sin 72^\circ} \\ \frac {a-3\cos 36^\circ}3 & = \frac {2+ 2\cos^2 36^\circ-1}{2\cos 36^\circ} \\ \implies a & = \frac {3+ 6\cos^2 36^\circ}{2\cos 36^\circ} + 3\cos 36^\circ \\ & = \frac 3{2\color{#3D99F6}\cos 36^\circ} + 6\color{#3D99F6}\cos 36^\circ \quad \quad \small \text{See Note.} \\ & = \frac 3{2\color{#3D99F6}\left(\frac {1+\sqrt 5}4\right)} + 6\color{#3D99F6}\left(\frac {1+\sqrt 5}4\right) \\ & = \frac 32 \left(\frac 4{1+\sqrt 5} + 1+\sqrt 5 \right) \\ & = \boxed{3\sqrt 5} \end{aligned}$

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Note: Using the identity below:

$\begin{aligned} \cos \frac \pi 5 + \cos \frac {3\pi}5 & = \frac 12 \\ 2\cos \frac \pi 5 + 2\cos \frac {3\pi}5 & = 1 \\ 2\cos \frac \pi 5 - 2\cos \frac {2\pi}5 & = 1 \\ 2\cos \frac \pi 5 - 4\cos^2 \frac \pi 5 + 2 & = 1 \\ 4\cos^2 \frac \pi 5 - 2\cos \frac \pi 5 - 1 & = 0 \\ \implies \cos \frac \pi 5 & = \frac {1+\sqrt 5}4 \end{aligned}$

Let $[XYZ]$ be the area of triangle $\triangle XYZ$ . We have $\angle ZAB=36^\circ$ , $\angle MAZ=72^\circ$ and $[ABM]=[ABZ]+[AZM]$ . Let $AB=x$ , then: $\dfrac{x \cdot \frac{x}{2}\sin 108^\circ}{2}=\dfrac{3x \sin 36^\circ}{2}+\dfrac{3\cdot \frac{x}{2} \sin 72^\circ}{2}$ Solve for $x$ , using the identity $\sin 108^\circ=\sin 72^\circ$ : $x=3\left(\dfrac{2\sin 36^\circ+\sin 72^\circ}{\sin 72^\circ}\right)\\ x=3\left(\dfrac{2\sin 36^\circ}{2\sin 36^\circ \cos 36^\circ}+1\right)\\ x=3\left(\dfrac{1}{\cos 36^\circ}+1\right)\\ x=3\left(\sqrt{5}-1+1\right)\\ x=\boxed{3\sqrt{5}}$