Finding Middle Ground

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And here is another solution:

First, as above, we recognize the similar triangles. The top side of the rectangle is divided as $2a + b$ ; the bottom side as $a + 3b$ . Equating these, $2a + b = a + 3b\ \ \therefore\ \ a = 2b,$ and we see that the divisions are really 4:1 at the top and 2:3 at the bottom.

Now consider the truncated triangle formed by the green, blue, and pink sections. Its height is the height of the rectangle; its average width is the mean of $4/5$ and $3/5$ of the width of the rectangle; its area is therefore $7/10$ th of the area $U$ of the rectangle. We write $59 + A = \frac 7{10}U.$ Likewise, consider the combination of yellow, pink, and brown. Its average width is the mean of $1/5$ and $2/5$ of the width of the rectangle. Thus we have $11 + A = \frac 3{10}U.$

Compare the two areas as a ratio: $\frac{59+A}{11+A} = \frac 7 3 \ \ \ \ \therefore \ \ \ \ 177 + 3A = 77 + 7A\ \ \ \ \therefore\ \ \ \ 100 = 4A\ \ \ \ \therefore\ \ \ \ A = \boxed{25}.$

$\triangle AEG$ and $\triangle FDG$ are similar with areas in ratio of $32:8=4:1$ . Their respective lengths are therefore in the ratio $\sqrt{4}:1=2:1$ . So if the height of the $\triangle FDG$ perpendicular to $FD$ is $h$ , then the height of $\triangle AEG$ perpendicular to $AE$ is $2h$ and the height of $\triangle AEF$ perpendicular to $AE$ is $3h$ .

Since $\triangle AEF$ and $\triangle AEG$ have the same base $AE$ and height of $AEF$ is $\frac{3}{2}$ of the height of $\triangle AEG$ the area $A_{AEF}=\frac{3}{2}\times A_{AEG}=\frac{3}{2}\times32=48$ .

$A_{EGF}=A_{AEF}-A_{AEG}=48-32=16$

The ratio of $A_{FHC}:A_{BHE}=27:3=9:1$ so in the similar triangles $FHC$ and $BHE$ the ratio of sizes is $3:1$ .

$h_{FHC}=3\times h_{BHE}$ , $h_{FEC}=4\times h_{BHE}$

Therefore $A_{FEC}=\frac{4}{3}\times A_{FHC}=\frac{4}{3}\times 27=36$

$A_{FHE}=A_{FEC}-A_{FHC}=36-27=9$

$A_{GEHF}=A_{EGF}+A_{FHE}=16+9=25$

The Crossed Ladders Theorem tells us that $\frac{1}{h} \; = \; \frac{1}{h_1} + \frac{1}{h_2} \; = \; \frac{h_1 + h_2}{h_1h_2}$ and hence we have the following identities concerning the areas: $\frac{1}{D} \; = \; \frac{1}{A+D} + \frac{1}{C+D} \; = \; \frac{A+B+C+D}{(A+D)(C+D)}$ These equations can be rearranged to read $B \; = \; D \; = \; \sqrt{AC}$ Thus the area of the shaded region is $\sqrt{8\times32} + \sqrt{3\times27} = 16 + 9 = \boxed{25}$ .

Let the area of the required region be $X$

From the above figure we can see that ,

$\Delta AEH$ and $\Delta GDH$ are similar

$\begin{aligned}\dfrac{DG}{AE}&=\sqrt{\dfrac{\text{Area of }\Delta GDH}{\text{Area of }\Delta AEH}}\\ &=\sqrt{\dfrac{8}{32}}=\dfrac{1}{2}\\ \text{Let, DG}&=a\\ \implies AE&=2a\end{aligned}$

Similarly, $\Delta EBF$ and $\Delta CGF$ are similar

$\begin{aligned}\dfrac{EB}{GC}&=\sqrt{\dfrac{\text{Area of }\Delta EBF}{\text{Area of }\Delta CGF}}\\ &=\sqrt{\dfrac{3}{27}}=\dfrac{1}{3}\\ \text{Let, EB}&=b\\ \implies GC&=3b\end{aligned}$

$ABCD$ is a rectangle

$\begin{aligned}\implies AB&=DC\\ \implies 2a+b&=a+3b\hspace{7mm}\color{#3D99F6}\small AB=AE+EB,DC=DG+GC\\ \implies a&=2b\\ \implies AB&=2(2b)+b=5b\hspace{5mm}\color{#3D99F6}\small(1)\end{aligned}$

$\text{Let }BC=h$

from the figure it is easy to see that $h=h_{1}+h_{2}$

$\begin{aligned}\text{Area of }\Delta EBF&=\dfrac{1}{2}\cdot b \cdot h_{1}\\ h_{1}&=\dfrac{2\cdot \left(\text{Area of }\Delta EBF\right)}{b}\\ &=\dfrac{6}{b}\\ \text{Similarly, }\\ h_{2}&=\dfrac{2\cdot \left(\text{Area of }\Delta CGF\right)}{3b}\\ &=\dfrac{54}{3b}=\dfrac{18}{b}\\ h&=h_{1}+h_{2}=\dfrac{24}{b}\hspace{5mm}\color{#3D99F6}\small(2)\end{aligned}$

We have ,

$\begin{aligned}\text{Area of rectangle }ABCD&=AB\cdot BC\\ &=5b\cdot \dfrac{24}{b}\hspace{5mm}\color{#3D99F6}\small\text{From }(1) \text{ and }(2)\\ &=120\end{aligned}$

Now, we have ,

$\begin{aligned}\text{Area of }\Delta DEC=\text{Area of }\Delta AGB&=\dfrac{1}{2}\cdot(\text{Area of rectangle }ABCD)\\ \implies \text{Area of }\Delta DEC+\text{Area of }\Delta AGB&=\text{Area of rectangle }ABCD\\ (8+27+X)+(32+3+X)&=120\\ 70+2X&=120\\ X&=\color{#EC7300}\boxed{\color{#333333}25}\end{aligned}$

We have 2 pairs of similar triangles - $\triangle AGF$ and $\triangle BEG$ with area ratio $1:4$ , and $\triangle ECH$ and $\triangle FHD$ with area ratio $1:9$ . The corresponding length ratios are thus $1:2$ and $1:3$ .

$|BE| = 2|AF|, |FD| = 3|EC|$

$|BE|+|EC|=|AF|+|FD| \implies 2|AF|+|EC|=|AF|+3|EC| \implies |AF|=2|EC|$

$|BC|=|AD|=5|EC|$

$|HK|=3|JH| \implies |EI|=|JK|=|JH|+3|JH|=4|JH|$

$\frac {A_{AED}}{A_{ECH}} = \frac{\frac {1}{2} |AD||EI|}{\frac {1}{2} |EC||JH| }=\frac{|AD|}{|EC|} \times \frac{|EI|}{|JH|}=5 \times 4 = 20 \implies A_{AED}=A_{ECH} \times 20 = 3 \times 20 = 60$

$A_{GEHF}=60-8-27=\boxed {25}$