It's all about Areas!

Without the loss of generality, let the dimensions of this rectangle be $h\times 6$ . Then plot this figure onto the Cartesian plane. The equation of the straight line $BD$ is $x/6 + y/h = 1$ . $$ The equation of the straight line $EI$ is $\frac{y-0}{x-3} = \frac{h-0}{2-3} \Leftrightarrow y = -h(x-3)$ . $$ Likewise, the equation of the straight line $EF$ is $\frac{y-0}{x-3} = \frac{h-0}{4-3} \Leftrightarrow y = h(x-3)$ . $$ The coordinate of $G$ is simply the intersection point of the straight lines $BD$ and $EI$ . In other words, we need to solve a system of equations $\begin{cases} x/6 + y/h= 1 \\ y = -h(x-3) \end{cases} \implies G = (12/5, 3h/5)$ Analogously, we get $H = (24/7, 3h/7) .$ $$ Using the shoelace formula, the area of the triangle $GHI$ is: $$ $\frac12 \begin{vmatrix}{3} && {24/7} && {12/5} && {3} \\ {0} && {3h/5} && {3h/7} && {0}\end{vmatrix} = \frac12 \left [\left(\frac{9h}5 + \frac{36h}{35} \right) - \left( \frac{72h}{35} + \frac{9h}7 \right) \right ] = \frac{9h}{35}$ Since the area of the rectangle is $6h$ , the ratio in question is $\frac{9h}{35} \div 6h = \frac3{70}$ . The answer is $3 + 70 = \boxed{73} .$

Let the width and height of rectangle $ABCD$ be $w$ and $h$ respectively, and $GJ=h_1$ and $HK=h_2$ are perpendicular to $AD$ . We note that $\angle GIA = \angle FID = \tan^{-1} \dfrac {h}{\frac w2 - \frac w3} = \tan^{-1} \dfrac {6h}w$ and $\angle BDA = \tan^{-1} \dfrac hw$ . Then we have:

$\begin{aligned} JD & = JI + ID \\ GJ \cdot \cot \angle BDA & = GJ \cdot \cot \angle GIA + \frac w2 \\ h_1 \cdot \frac wh & = h_1 \cdot \frac w{6h} + \frac w2 \\ \frac {5h_1}{6h} & = \frac 12 \\ \implies h_1 & = \frac 35 h \end{aligned}$

Similarly,

$\begin{aligned} ID & = IK + KD \\ \frac w2 & = HK \cdot \cot \angle FID + HK \cdot \cot \angle BDA \\ \frac w2 & = h_2 \cdot \frac w{6h} + h_2 \cdot \frac wh \\ \frac 12 & = \frac {7h_2}{6h} \\ \implies h_2 & = \frac 37 h \end{aligned}$

Let the area of the rectangle $[ABCD]=wh = 1$ . Then $[HID] = \dfrac 12 \cdot \dfrac w2 \cdot \dfrac {3h}7 = \dfrac 3{28}$ , $[BEG] = \dfrac 12 \cdot \dfrac w3 \left(h-\dfrac {3h}5 \right) = \dfrac 1{15}$ , and

$\begin{aligned} [GHI] & = [ABD] - [ABGI] - [HID] \\ & = \frac 12wh - ([ABEI] - [BEG]) - \frac 3{28} \\ & = \frac 12 - \left(\frac {\frac w3+\frac w2}2 \cdot h - \frac 1{15} \right) - \frac 3{28} \\ & = \frac 12 - \frac 5{12} + \frac 1{15} - \frac 3{28} = \frac 3{70} \\ \implies \frac {[GHI]}{[ABCD]} & = \frac 3{70} \end{aligned}$

Therefore $\alpha + \beta = 3 + 70 = \boxed{73}$ .

I wanted to see if sympy could do the algebra for the general case, and almost got there, but ran into a python error I could not resolve, so I settled for a specific 3x2 rectangle. Here is an interactive demo of how the area ratio remains constant for the general rectangle. Just move the sliders.

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from sympy.geometry import *

# pick arbitrary rectangle
width, height = 3, 2

# define rectangle ABCD
A = Point(0,0)
B = Point(A.x + width, A.y)
C = Point(A.x + width, A.y + width)
D = Point(A.x, A.y + width)
rectangle = Polygon(A, B, C, D)

# compute boundry points, EFG
dc = D.distance(C)
E = Point(D.x + dc/3, D.y)
F = Point(D.x + 2*dc/3, D.y)
G = Point(A.x + dc/2, A.y)

# compute line segments
DB = Segment(D, B)
EG = Segment(E, G)
FG = Segment(F, G)

# define triangle HGI
H = list(DB.intersect(EG))[0]
I = list(DB.intersect(FG))[0]
triangle = Triangle(H, G, I)

ratio = triangle.area/rectangle.area
solution = ratio.numerator() + ratio.denominator()
print("area triangle / area rectangle = ", ratio)
print("solution =", solution)

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area triangle / area rectangle =  3/70
solution = 73

why is this ignored?

Let $A_{ABCD} = a$ with $\overline{BC} = 1$ and $\overline{AB} = a$

First we will obtain $h^{*}$ and $h$ using $\triangle{BEG} \sim \triangle{IGD}$ and $\triangle{BFH} \sim \triangle{HID}$ , then using the diagram above with $\triangle{GPD} \sim \triangle{ABD}$ and $\triangle{HQD} \sim\triangle{ABD}$ we obtain $x^{*}$ and $x$ .

$\dfrac{h^{*}}{a - h^{*}} = \dfrac{\dfrac{1}{2}}{\dfrac{1}{3}} = \dfrac{3}{2} \implies 2h^{*} = 3a - 3h^{*} \implies h^{*} = \dfrac{3a}{5}$

$\dfrac{h}{a - h} = \dfrac{\dfrac{1}{2}}{\dfrac{2}{3}} = \dfrac{3}{4} \implies 4h = 3a - 3h \implies h = \dfrac{3a}{7}$

Using $\triangle{GPD} \sim \triangle{ABD} \implies \dfrac{a}{h^{*}} = \dfrac{1}{x^{*}} \implies$ $ax^{*} = h^{*} = \dfrac{3a}{5} \implies x^{*} = \dfrac{3}{5}$ .

Using $\triangle{HQD} \sim \triangle{ABD} \implies \dfrac{a}{h} = \dfrac{1}{x} \implies$ $ax = h = \dfrac{3a}{7} \implies x = \dfrac{3}{7}$

$\overline{AP} = 1 - \dfrac{3}{5} = \dfrac{2}{5} \implies \overline{PI} = \dfrac{1}{2} - \dfrac{2}{5} = \dfrac{1}{10} \implies$

$A_{\triangle{GPI}} = \dfrac{1}{2}\overline{PI}h^{*} = \dfrac{1}{2}(\dfrac{1}{10})(\dfrac{3a}{5}) = \dfrac{3a}{100}$

and

$A_{\triangle{HID}} = \dfrac{1}{2}(\dfrac{1}{2})h = \dfrac{1}{4}(\dfrac{3a}{7}) = \dfrac{3a}{28}$

and

$A_{\triangle{GPD}} = \dfrac{1}{2}x^{*}h^{*} = \dfrac{1}{2}(\dfrac{3}{5})(\dfrac{3a}{5}) = \dfrac{9a}{50}$

$\implies$

$A_{R} = A_{\triangle{GPD}} - (A_{\triangle{HID}} + A_{\triangle{GPI}}) =$ $\dfrac{9a}{50} - (\dfrac{3a}{28} + \dfrac{3a}{100}) = (\dfrac{9}{50} - \dfrac{24}{175})a =$

$\dfrac{3}{70}a \implies \dfrac{A_{R}}{A_{ABCD}} = \dfrac{3}{70} = \dfrac{\alpha}{\beta} \implies$ $\alpha + \beta = \boxed{73}$ .

$\textbf{Below is an alternative approach using coordinate geometry:}$

Let $A_{ABCD} = a$ with $\overline{BC} = 1$ and $\overline{CD} = a$

$m_{FI} = 6a \implies y = 3a(2x - 1)$

$m_{BD} = - a \implies y = -a(x - 1)$

$\implies 3(2x - 1) = -(x - 1) \implies 6x - 3 = -x + 1 \implies x = \dfrac{4}{7} \implies$

$y = \dfrac{3a}{7} = h \implies \boxed{H(\dfrac{4}{7}, \dfrac{3a}{7})}$

$m_{EI} = -6a \implies y = -3a(2x - 1)$

$\overleftrightarrow{\rm BD} y = -a(x - 1)$

$\implies 3a(2x - 1) = a(x - 1) \implies 6x - 3 = x - 1 \implies x = \dfrac{2}{5} \implies y = \dfrac{3a}{5} = h^{*}$

$\implies \boxed{G(\dfrac{2}{5}, \dfrac{3a}{5})}$

$\overline{PI} = \dfrac{1}{2} - \dfrac{2}{5} = \dfrac{1}{10}$ and $\overline{IQ} = \dfrac{4}{7} - \dfrac{1}{2} = \dfrac{1}{14}$

$\implies A_{\triangle{HIQ}} = \dfrac{1}{2}(\dfrac{1}{14})(\dfrac{3a}{7}) = \dfrac{3a}{196}$

and

$A_{\triangle{GPI}} = \dfrac{1}{2}(\dfrac{1}{10})(\dfrac{3a}{5}) = \dfrac{3a}{100}$

and

$A_{PQHG} = \dfrac{1}{2}(\dfrac{3a}{7} + \dfrac{3a}{5})(\dfrac{1}{10} + \dfrac{1}{14}) =$ $\dfrac{3a}{2}(\dfrac{12}{35})(\dfrac{12}{70}) = \dfrac{432}{4900}a = \dfrac{108}{1225}a$

$\implies A_{R} = A_{PQHG} - (A_{\triangle{HIQ}} + A_{\triangle{GPI}}) =$ $(\dfrac{108}{1225} - (\dfrac{3}{196} + \dfrac{3}{100}))a =$

$(\dfrac{108}{1225} - \dfrac{111}{2450})a = \dfrac{3}{70}a \implies \dfrac{A_{R}}{A_{ABCD}} = \dfrac{3}{70} = \dfrac{\alpha}{\beta} \implies \alpha + \beta = \boxed{73}$

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