Trisectors!

Let the intersection of $AG$ and $BH$ be $I$ and the intersection of $EC$ and $BH$ be $J$ . Now, we shall set up a coordinate system where $A(0,a),B(0,0),C(a,0),D(a,a)$ . The side length of the square is thus $a$ . From the question, we have $E\left(0,\frac{2}{3}a\right),H\left(\frac{2}{3}a,a\right),G\left(a,\frac{1}{3}a\right)$ . From this, we obtain:

$\mbox{Equation of BH: }y=\frac{3}{2}x\\\mbox{Equation of AG: }y=-\frac{2}{3}x+a\\\mbox{Equation of AG: }y=-\frac{2}{3}x+\frac{2}{3}a$

Solving for the coordinates of $I,J$ , we get $I\left(\frac{6}{13}a,\frac{9}{13}a\right),J\left(\frac{4}{13}a,\frac{6}{13}a\right)$

This implies that $|IJ|=\sqrt{1}{13}a=1$ Solving, $a=\sqrt{13}$ So the answer is $1+13=14$ .

Let the name of the small square formed be MNOP. Clearly, MP || NO .'. AM || EO

In triangle ABM , AM || EO ,

.'. $\frac{BN}{MN}$ = $\frac{EB}{AE}$ {By the Thales' theorem}

=> $\frac{BN}{1}$ = $\frac{2}{1}$ { .'. MN = 1 and EB= 2AE}

=> BN = 2 Similarly , from △ ADP we get : AM = 2

Now , In △ ABM , AB^2 = BM^2 + AM ^2 {By Pythagoras' theorem}

=> AB^2 = (BN + MN)^2 + AM^2

=> AB^2 = (2+1)^2 + 2^2 {Putting the values which have been found above}

=> AB^2 = 3^2 + 2^2 => AB = √(13)
------------------------------an easy and complete solution by Vishwash Kumar

Square within Trisected square

$\tan\alpha = \frac{2}{3}$ giving $\cos\alpha = \frac{1}{\sqrt{1+{(\frac{2}{3}})^2}}= \frac{3}{\sqrt{13}}$

Let AB = 3a giving AE = a and $AE \cos\alpha = AE \frac{3}{\sqrt{13}} = \frac{AB}{\sqrt{13}} = 1$

$\textbf AB = \sqrt{13}$