Similar Rectangles

Since rectangle $GHIJ$ is similar to rectangle $ABCD$ , $\frac{GH}{HI}=\frac{AB}{BC}=\frac{3}{2}$ Simple angle chasing gives, $\angle DGH=\angle AHI=\angle FIJ=\angle EJG$ Combining this with $GH=JI\;$ and $\;GJ=HI$ , we get the following relations: $\triangle GDH\cong \triangle IFJ\;,\;\;\; \triangle HAI\cong \triangle JEG$ $\triangle GDH\sim \triangle IFJ\sim \triangle HAI\sim \triangle JEG$ Let $DH=x\implies AH=4-x \;$ and $\;AI=y\implies FI=GD=5-y$ .

Since $\triangle GDH\sim \triangle HAI$ , $\;\;\;\;\;\;\;\;\frac{x}{y}=\frac{5-y}{4-x}=\frac{3}{2}$ $\implies x=\frac{6}{5}\;,\;y=\frac{4}{5}$ Using the Pythagorean Theorem , $GH^2=GD^2+DH^2=x^2+(5-y)^2=\frac{477}{25}$ The ratio of the area of $GHIJ$ to the area of $ABCD$ is, $\frac{[GHIJ]}{[ABCD]}=\left (\frac{GH}{AB}\right )^2=\frac{\large \frac{477}{25}}{36}=\frac{53}{100}$

Therefore, $p=53, q=100\implies p+q=\boxed{153}.$

Let $HG = IJ = x$ and $\angle JIF = \angle GJE = \theta$ . Then $HI = GJ = \dfrac 23 x$ and

$\begin{cases} AI + IF = AF & \implies \dfrac 23 x \sin \theta + x \cos \theta = 5 & \implies 2 x \sin \theta + 3x \cos \theta = 15 & ...(1) \\ EJ + JF = EF & \implies \dfrac 23 x \cos \theta + x \sin \theta = 4 & \implies 2 x \cos \theta + 3x \sin \theta = 12 & ...(2) \end{cases}$

From $\dfrac {(1)}{(2)}$ :

$\begin{aligned} \frac {2 x \sin \theta + 3x \cos \theta}{2 x \cos \theta + 3x \sin \theta} & = \frac {15}{12} \\ \frac {2 \sin \theta + 3 \cos \theta}{2 \cos \theta + 3 \sin \theta} & = \frac 54 \\ 8 \sin \theta + 12 cos \theta & = 10 \cos \theta + 15 \sin \theta \\ 2 \cos \theta & = 7 \sin \theta \\ \implies \tan \theta & = \frac 27 & \small \blue{\implies \sin \theta = \frac 2{\sqrt{53}}, \cos \theta = \frac 7{\sqrt{53}}} \end{aligned}$

From $(1)$ : $\dfrac 4{\sqrt{53}}x + \dfrac {21}{\sqrt{53}}x = 15 \implies \dfrac {25}{\sqrt{53}}x = 15 \implies x = \dfrac {3 \sqrt{53}}5$ . Therefore $\dfrac {[GHIJ]}{[ABCD]} = \dfrac {x^2}{6^2} = \dfrac {53}{100}$ and $p+q = 53 + 100 = \boxed{153}$ .

Let $\angle AHI = \theta, HI = 2x$ , then:

$IJ = 2x \times 6 \div 4 = 3x \\ 2x \sin \theta + 3x \cos \theta = 5 \quad (1) \\ 2x \cos \theta + 3x \sin \theta = 4 \quad (2) \\ 3 \times (1) - 2 \times (2): \\ \implies 5x \cos \theta = 7 \quad (3) \\ 3 \times (2) - 2 \times (1): \\ \implies 5x \sin \theta = 2 \quad (4) \\ (3)^2 + (4)^2 : \\ \implies x^2 = \dfrac {53}{25} \\ \dfrac {S_{GHIJ}}{S_{ABCD}} = \dfrac {2x \times 3x}{4 \times6} = \dfrac {x^2}{4} = \dfrac {53}{100}$

Since $GHIJ$ is similar to $ABCD$ , let $IJ = HG = 3k$ and $HI = GJ = 2k$ . The ratio of the area of $GHIJ$ to the area of $ABCD$ is $r = \frac{2k \cdot 3k}{4 \cdot 6} = \frac{1}{4}k^2$ .

By angle chasing the four right triangles are similar by AA similarity. Let $\theta = \angle JIF = \angle GJE = \angle HGD = \angle AHI$ , then $IF = GD = 3k \cos \theta$ , $FJ = DH = 3k \sin \theta$ , $HA = EJ = 2k \cos \theta$ , and $AI = EG = 2k \sin \theta$ , so that from $AI + IF = AF$ :

$2k \sin \theta + 3k \cos \theta = 5 \text{ } ... \text{ } (1)$

and from $DH + HA = DA$ :

$3k \sin \theta + 2k \cos \theta = 4 \text{ } ... \text{ } (2)$

From $(1)^2 + (2)^2$ we obtain $13k^2 + 24 k^2 \sin \theta \cos \theta = 41$ , or

$\sin \theta \cos \theta = \frac{1}{24k^2}(41 - 13k^2)\text{ } ... \text{ } (3)$

The area of $GHIJ$ is the same as the difference between the area of $ADEF$ and the four right triangles, so $4 \cdot 5 - (\frac{9}{2}k^2\sin \theta \cos \theta + 2k^2 \sin \theta \cos \theta + \frac{9}{2}k^2\sin \theta \cos \theta + 2 k^2\sin \theta \cos \theta) = 2k \cdot 3k$ , or

$20 - 13k^2 \sin \theta \cos \theta = 6k^2\text{ } ... \text{ } (4)$

Substituting $(3)$ into $(4)$ and simplifying gives $k^2 = \frac{53}{25}$ .

Therefore, the ratio is $r = \frac{1}{4}k^2 = \frac{53}{100}$ , so $p = 53$ , $q = 100$ , and $p + q = \boxed{153}$ .