Calculate Triangular Region Area $ MNE $

Question

From the figure shown: $ M $ and $ N $ are tangency points, $ O $ and $ E $ are centers, and $ r = 2 $. Calculate Triangular Region Area $ MNE $

Attemp: The area is $2$. First and foremost I shall use the double angle formula for $\tan x$ which says $\tan{\frac{x}{2}}=\frac{\sin x}{1+\cos x}$. Applying it to $x=45^{\circ}$ we have $\tan 22.5^{\circ}=\sqrt{2}-1.$ Since $\text{NOV}\Delta \sim \text{MAO}\Delta$ thus $\frac{\text{NV}}{2}=\tan{22.5^{\circ}}=\sqrt{2}-1$ consequently $\text{NV}=2(\sqrt{2}-1).$ $\text{UV}=2+\text{NV}=2\sqrt{2}.$ In the next step we work out $R$. Since $\text{AWE}\Delta \sim \text{AUV}\Delta$ hence $\frac{\text{WE}}{\text{UV}}=\frac{\text{AW}}{AU}$ and it implies $\frac{R}{2\sqrt{2}}=\frac{2\sqrt{2}+4+R}{2\sqrt{2}+4}$ where from we have $R=2(\sqrt{2}+1).$ $PQ=2+\sqrt{2}$ and it gives $\text{EP}=\text{EQ}-\text{PQ}=\sqrt{2}$ furthermore $\text{MN}=2\sqrt{2}.$ From these we have$$\text{Area=}\frac{1}{2}\cdot\text{MN}\cdot\text{EP}=\frac{1}{2}\cdot2\sqrt{2}\cdot\sqrt{2}=2. $$

I think it has to show that $ MN $ is parallel to $ AQ $, right?

If you can also submit a smarter solution than this, thank you

Answer 1

The key step is to calculate the radius $R$, which may be done without 22.5$^\circ$. Note the circles of radii $r$ and $R$ are respectively the incircle and the excircle, which is related via,

$$Area_{ABC} = \frac12 r (AB+CB+AC) = \frac12 R (AB + AC - CB)\tag 1$$

For the isosceles right $\triangle ABC$, we have $AC = CB = \frac1{\sqrt2}AB$. Then, (1) reduces to

$$r(2+\sqrt2) = \sqrt2 R$$

Answer 2

Let $|AC|=|BC|=a$, then

\begin{align} |AB|&=\sqrt2\,a ,\\ S_{ABC}&=\tfrac12\,a^2 \tag{1}\label{1} ,\\ S_{ABC}&=\tfrac12\,(a+a+c)\,r =a\,r\,(1+\tfrac{\sqrt2}2) \tag{2}\label{2} , \end{align}

\begin{align} a&=r\,(2+\sqrt2) ,\\ S_{ABC} &= (3+2\sqrt2)\,r^2 ,\\ r_a&= \frac{2S}{-a+a+c} =r\,(1+\sqrt2) ,\\ |MN|&=r\sqrt2 , \end{align}

\begin{align} S_{EMN} &=\tfrac12\,|MN|\cdot h_{EMN} = \tfrac12\,r\sqrt2\cdot(r_a-r-r\,\tfrac{\sqrt2}2) \\ &=\tfrac12\,r^2=2 . \end{align}