Triangle

Well, it's an easy problem except that the figure is totally confusing(not at all to scale, lol, that was good).

Let $\displaystyle [\text{Any Figure}]$ represent the area of the figure and we know area of a triangle is just $\displaystyle \frac{1}{2}*\text{base}*\text{height}$ , so for triangles with same heights, ratio of areas is just ratio of their respective bases.

Now, first of all,

Using similarity in $\displaystyle {A}_{1}B{C}_{1}$ and $ABC$ , we get $\displaystyle [{A}_{1}B{C}_{1}] = \frac{1}{4}[ABC]$ [Using Mid-point theorem], and as a result $\displaystyle {A}_{1}E = E{C}_{1}$

Therefore, $\displaystyle [{A}_{1}BE] = \frac{1}{8}[ABC]$

In $\displaystyle C, B, {B}_{1}, A$ and the figure made by it, we can use Appolonius theorem, so,

$\displaystyle \frac{AC}{{B}_{1}A}.\frac{{B}_{1}D}{BD}.\frac{{A}_{2}B}{{A}_{2}C} = 1$ [We have removed the negative sign because the magnitude is this only, we have not considered the vector case as it is just trivial]

Hence,

$\displaystyle \frac{{B}_{1}D}{BD} = \frac{1}{6}$

And so, $\displaystyle \frac{{B}_{1}D}{{B}_{1}B} = \frac{1}{7}$

Now $\displaystyle A{B}_{1}B$ and $A{B}_{1}D$ have the same height and $\displaystyle [A{B}_{1}B] = \frac{1}{2}[ABC] \Rightarrow [A{B}_{1}D] = \frac{1}{14}[ABC]$

We can also see that $\displaystyle [{A}_{2}AC] = \frac{1}{4}[ABC]$

Hence, $\displaystyle [{A}_{2}D{B}_{1}C] = (\frac{1}{4} - \frac{1}{14})[ABC]$

$\displaystyle [{A}_{2}D{B}_{1}C] = \frac{5}{28}[ABC]$

We know 2 more things that - $\displaystyle [{B}_{1}BC] = \frac{1}{2}[ABC]$ and $\displaystyle [{A}_{1}{A}_{2}DE] = [{B}_{1}BC] - [{A}_{1}BE] - [{A}_{2}D{B}_{1}C]$

As an immediate result,

$\displaystyle [{A}_{1}{A}_{2}DE] = (\frac{1}{2} - \frac{1}{8} - \frac{5}{28})[ABC]$

$\displaystyle [{A}_{1}{A}_{2}DE] = (\frac{11}{56})[ABC]$

$\displaystyle \frac{{\Delta}_{1}}{\Delta} = \frac{11}{56}$

P(A1A2DE)=P(DBA2)-P(EBA1),so we need to calculate areas of those two triangles.P(EBA1)=P(B1BC)/4 because they are similiar and BA1=BC/2 ,P(B1BC)=B1Cxh/2=ACxh/4=P(ABC)/2, they have same height. so we know that P(EBA1)=P(ABC)/8 and now we need to calculate P(DBA2). In order to do that we need to calculate A2A/A2D and we are going to do that using Menelaus' theorem aplied to vertices B,D,B1 and triangle AA2C. That means( AB1/B1C)(CB/BA2)(A2D/DA)=1,that means 1x4/3x(A2D/DA)=1, so DA=4/3xA2D,so because DA+DA2=AA2 we get that A2Dx(4/3+1)=AA2,so A2D=3/7AA2 so P(DBA2)=A2Dxh2/2=3/7AA2xh2/2=3/7P(ABA2)=3/7xABxh3/2=3/7xABx(3/4)h4/2=9/28xP(ABC) so P(A1A2DE)/P(ABC)=(9/28)-(1/8)=(18-7)/56=11/56,a =11,b=56 and a+b=67 is our answer