Triangle-Ception IV

Using basic facts we can obtain the following:

$\angle ZOD= \angle WOZ=30^\circ ,\angle ADO=45^\circ, \overline{ED}=\sqrt{5}, \sin \widehat{EDO}=\frac{1}{\sqrt{5}}$ and $\overline{OW}=\overline{OZ}$ .

let $\angle EDO=\alpha \Rightarrow$

$\sin\widehat{OYD}=\sin(150^\circ-\alpha)=\frac{2+\sqrt{3}}{2\sqrt{5}}$

$\sin\widehat{OXD}=\sin(120^\circ-\alpha)=\frac{1+2\sqrt{3}}{2\sqrt{5}}.$

Applying the law of sines on the triangles OWD, OXD and OYD we get $\overline{OW}=\frac{4}{1+\sqrt{3}}$ ,

$\overline{OX}=\frac{4}{1+2\sqrt{3}}$ and $\overline{OY}=\frac{4}{2+\sqrt{3}}$ respectively.

The area of WXYZ=area of OWZ-area of XOZ = $\frac{1}{2} \times \sin 30^\circ \times \overline{OW} \times \overline{OZ}-\frac{1}{2} \times \sin 30^\circ \times \overline{OX} \times \overline{OY}$ $=\frac{2}{11} (38-21\sqrt{3})$

Conclusion 1. The answer is $\boxed{75}$ .

Conclusion 2. Writing a solution using my phone..never again.

I used coordinate geometry to solve the question, however my method is way too long and not appreciable enough to be shared. :|

First, lets break down the diagram into sections in order to get the area of $WXYZ$ . Let $[a]$ denote the area of $a$ .

We can tell by our diagram that

$[\triangle AOD]=2=[\triangle AWO]+[\triangle OZD]+[\triangle OXY]+[WXYZ]$

We can also place $O$ at the origin and let $OA$ lie on the $y-axis$ and $OD$ lie on the $x-axis$ .

We can represent every line that makes up $WXYZ$ on the diagram as:

$OB=y_{1}=\sqrt{3}x \\ OC=y_{2}=\frac{1}{\sqrt{3}}x \\ AD=y_{3}=2-x \\ DE=y_{4}=1-\frac{1}{2}x$

We can find the areas of these triangles by dropping down perpendiculars and using $Area=Base \times Height$

To find $[\triangle AWO]$ we need the x-value of $y_{1}=y_{3}$ , so we get

$\sqrt{3}x=2-x \\ x=\sqrt{3}-1$

And by $Area = Base \times Height$ , we get

$[\triangle AWO]=\frac{(\sqrt{3}-1)(2)}{2}=\sqrt{3}-1$

To Find $[\triangle OZD]$ we need the y-value of $y_{2}=y_{3}$ , so we get

$\frac{1}{\sqrt{3}}x=2-x=y_{2}=y_{3} \\ y_{2}=y_{3}=\sqrt{3}-1$

And so we can conclude that

$[\triangle AWO]=[\triangle OZD]=\sqrt{3}-1$

In order to find $[\triangle OXY]$ , we can see that

$[\triangle EOD]=1=[\triangle OXY]+[\triangle EXO]+[\triangle OYD]$

To find $\triangle EXO$ , we need the x-value of $y_{1}=y_{4}$ , so we get

$\sqrt{3}x=1-\frac{1}{2}x \\ x=\frac{2}{11}(2\sqrt{3}-1)$

So we can conclude that

$[\triangle EXO]=\frac{(\frac{2}{11}(2\sqrt{3}-1))(1)}{2}=\frac{1}{11}(2\sqrt{3}-1)$

Also, to find $[\triangle ODY]$ , we need the y-value of $y_{2}=y_{4}$ , so we get

$\frac{1}{\sqrt{3}}x=1-\frac{1}{2}x=y_{2}=y_{4} \\ y_{2}=y_{4}=4-2\sqrt{3}$

So we can conclude that

$[\triangle EXO]=4-2\sqrt{3}$

Now we have everything we need to finish the problem. We get that

$[WXYZ]=2-[\triangle AWO]-[\triangle OZD]-[\triangle OXY] \\ [WXYZ]=2-[\triangle AWO]-[\triangle OZD]-(1-[\triangle EXO]-[\triangle OYD]) \\ [WXYZ]=2-(\sqrt{3}-1)-(\sqrt{3}-1)-(1-(4-2\sqrt{3})-(\frac{1}{11}(2\sqrt{3}-1)))$

After some manipulation we get that

$[WXYZ]=\frac{2}{11}(38-21\sqrt{3})$

This satisfies all of our constraints for $m,n,p,q,r$ , so

$m+n+p+q+r=2+11+38+21+3=\boxed{75}$

Used the same method as Mahdi Al-kawaz . I was to give a little more detail. But I think his is sufficient.

However on second thought, coordinate geometry is also a nice method.
$\text{OD and OA, X and Y rectangular axis. Then,}\\ Y_{OB}=\sqrt 3 X,~~~~Y_{OC}=\dfrac 1 {\sqrt 3} X, ~~~~Y_{AD}= - X +2, ~~~ \dfrac{Y_{ED}} 1 + \dfrac{X_{ED}} 2 =1.\\ \text{Points of intersections give coordinates of W, X, Y, Z. So the area.}$