pentagon inscribed in triangle

Let the radius of the regular pentagon be $1$ . This means that we have fixed the area of the pentagon to $A_p = \dfrac 52 \sin 72^\circ = 5 \sin 36^\circ \cos 36^\circ$ ; and we are looking for the smallest inscribing isosceles triangle.

The area of the triangle is $A_\triangle = AC\cdot BC$ . Let the apex angle of the isosceles triangle be $\theta$ . Then

$\begin{aligned} AC & = AD + DC \\ & = DE \cdot \cot \frac \theta 2 + EF \cdot \cos 18^\circ \\ & = \cos 18^\circ \cot \frac \theta 2 + 2 \sin 36^\circ \cos 18^\circ \\ & = \cos 18^\circ \left(\cot \frac \theta 2 + 2 \sin 36^\circ \right) \end{aligned}$

And $BC = AC \cdot \tan \frac \theta 2$ . Therefore,

$\begin{aligned} A_\triangle & = \cos^2 18^\circ \left(\cot \frac \theta 2 + 2 \sin 36^\circ \right)^2 \tan \frac \theta 2 \\ & = \cos^2 18^\circ \left(1 + 2 \sin 36^\circ \tan \frac \theta 2 \right)\left(\cot \frac \theta 2 + 2 \sin 36^\circ \right) \\ & = \cos^2 18^\circ \left(\blue{\cot \frac \theta 2} + 4 \sin 36^\circ + \blue{4 \sin^2 36^\circ \tan \frac \theta 2} \right) & \small \blue{\text{By AM-GM inequality}} \\ & \blue \ge \cos^2 18^\circ \left(\blue{4 \sin 36^\circ} + 4 \sin 36^\circ \right) & \small \blue{\text{Equality occurs when }\cot \frac \theta 2 = 2 \sin 36^\circ} \end{aligned}$

$\implies \min(A_\triangle) = 8\cos^2 18^\circ \sin 36^\circ$

Then we have:

$\begin{aligned} \frac {A_p}{\min(A_\triangle)} & = \frac {5\sin 36^\circ \cos 36^\circ}{8\cos^2 18^\circ \sin 36^\circ} = \frac {5 \cos 36^\circ}{4\cos 36^\circ + 4} & \small \blue{\text{Note that }\cos 36^\circ = \frac {\sqrt 5 + 1}4} \\ & = \frac {5(\sqrt 5+1)}{4(\sqrt 5 + 1 +4)} = \frac {5(\sqrt 5+1)}{4(\sqrt 5 + 5)} \\ & = \frac {5(\sqrt 5+1)}{4\sqrt 5(\sqrt 5 + 1)} = \frac {\sqrt 5}4 \end{aligned}$

Therefore $a+b = 5+4 = \boxed 9$ .

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Reference: AM-GM inequality