Octagram and octagon

$It ~is~ clear~that~Squares~ EQTH~and~CLVM~both~have~sides~(\sqrt2-1)~ and~make~45^o ~with~ one~another.\\ So~diagonal ~of~one ~will~be~at~90^o~to~one~pair~of~sides~of~the~other.\\ In ~particular~diagonal~ET \bot MV ~and~=\sqrt2*(\sqrt2-1). \\ Let~MV~cut~ET~at~Y.~It ~is ~easy~ to~ see~that~Y~is~the~midpoint~of~MV. \implies~MY=\frac 1 2 *(\sqrt2-1).\\ Area~ EMT=\frac 1 2* (\sqrt2*(\sqrt2-1))*\frac 1 2 *(\sqrt2-1)\\$ $=\dfrac{3*\sqrt2 - 4} 4=\dfrac{\sqrt{34-24\sqrt2} }4= \dfrac{\sqrt{a-b\sqrt c}} 4\\ a+b+c=\large \color{#D61F06}{60}.$

In The Cartesian Plane, let $E=(0,0)$

Note that $AE=\dfrac{2-\sqrt{2}}{2}$ and $\angle WAU=45^\circ$

$\therefore E=(0,0),M=\left(0.5,\dfrac{1-\sqrt{2}}{2}\right),T=(\sqrt{2}-1,1-\sqrt{2})$

The area of $\triangle EMT=\dfrac{1}{2}\left|\dfrac{1-\sqrt{2}}{2}+\dfrac{3-2\sqrt{2}}{2}\right|=\dfrac{\sqrt{34-24\sqrt{2}}}{4}$

Because of symmetry, there is a circumcircle centered at $Z$ inscribing the octagram $CHMTVQLE$ (two squares). Therefore, $EZ = MZ = TZ = r$ , the radius of the circumcircle. By Pythagorean theorem , we have $MT^2 = EH^2+HT^2$ . Since $EH=HT=AB = \sqrt 2-1$ . $\implies MT = \sqrt 2 \left(\sqrt 2-1\right) = 2 - \sqrt 2 = 2r$ , $\implies r = \dfrac {2-\sqrt 2}2$ .

Let $MY$ be the perpendicular from $M$ to $ET$ . Then, we note that the area of $\triangle EMT$ ,

$\begin{aligned} [EMT] & = \frac 12 \times ET \times MY = \frac 12 (2r) \left(\frac r{\sqrt 2}\right) = \frac {r^2}{\sqrt 2} \\ & = \frac {\left(2-\sqrt 2\right)^2}{2^2\sqrt 2} = \frac {3\sqrt 2-4}4 = \frac {\sqrt{(3\sqrt 2-4)^2}}4 \\ & = \frac {\sqrt{34-24\sqrt 2}}4\end{aligned}$

$\implies a+b+c = 34+24+2 = \boxed{60}$

Observe that $ET = AB * \sqrt{2}$ and altitude of $\triangle EMT$ dropped from $M$ is $\dfrac{AB}{2}$ . This gives $\triangle EMT = AB^2 \dfrac{\sqrt{2} }{4}=\dfrac {3 \sqrt{2}-4}{4}=\dfrac{\sqrt{34-24\sqrt{2}}}{4}$ .

$EHTQ \equiv CMVL$ and they are square , $\Delta EFJ \equiv \Delta GHK \equiv \Delta OTS \equiv \Delta RQN$ and they are isosceles right-angled triangles

Let $EF =FC=CG=GH=HK=KM=MO=OT=TS=SV=VR=RQ=QN=NL=LJ=JE = x$

Let $FG = KO = RS = JN = y$

$\begin{cases} 2x+y = \sqrt{2}-1 \\ 2x^2 = y^2 \end{cases} \Rightarrow \begin{cases} x = \dfrac{3\sqrt{2}-4}{2} \\ y = 3-2\sqrt{2} \end{cases}$

$\angle TEM =45-\frac{180-(180-45)}{2} = 22.5^o$ , $\angle MTE = 45+\frac{180-(180-45)}{2} = 67.5^o$ , $\angle EMT = 180-22.5-67.5 =90^o$

$EM^2 = EG^2+GM^2 -2\cos(180^o-45^o) = 2(x+y)^2(1-\cos135^o)$

$MT^2 = MO^2+OT^2 -2\cos(180^o-45^o) = 2x^2(1-\cos135^o)$

$[EMT] = \dfrac{EM\times MT}{2} = \dfrac{\sqrt{4x^2(x+y)^2(1-\cos135^o)^2}}{2} = \dfrac{\sqrt{34-24\sqrt{2}}}{4}$

$\Rightarrow a+b+c = 34+24+2 = 60$