An Area Problem: Find $x$

Let $AC$ intersects $BD$ and $BE$ at $F$ and $G$ respectively. Let $GH=h$ be the altitude of $\triangle BCG$ ; then $HC=GH=h$ . We note that $\triangle BGH$ is similar to $\triangle BEC$ . Therefore we have:

$\begin{aligned} \frac {GH}{EC} & = \frac {BH}{BC} \\ \frac h{x-1} & = \frac {x-h}x \\ \implies x^2 - (2h+1)x + h & = 0 & ...(1) \end{aligned}$

Given that area of the shaded region is $1$ :

$\begin{aligned} [BFG] & = 1 \\ [BFC]-[BGC] & = 1 \\ \frac {x^2}4 - \frac {hx}2 & = 1 \\ \implies x^2 - 2hx & = 4 & ...(2) \end{aligned}$

Then we have $(2) - (1): \ x - h = 4 \implies h = x - 4$ and:

$\begin{aligned} (2): \quad x^2 - 2(x-4)x - 4 & = 0 \\ -x^2 + 8x - 4 & = 0 \\ x^2 - 8x + 4 & = 0 \end{aligned}$

$\begin{aligned} \implies x & = \frac {8+\sqrt{64-16}}2 = 4 + 2\sqrt 3 \end{aligned}$

Therefore $a+b+c = 4+2+3 = \boxed 9$ .

Taking the point $D$ as the origin, lines $\overline {DC}$ and $\overline {DA}$ along the $x$ - and the $y$ -axes respectively, we can write the vertices of the shaded triangle as $(\dfrac{x}{2},\dfrac{x}{2}), (\dfrac{x^2}{2x-1},\dfrac{x^2-x}{2x-1}), (x, x)$ . So the area of this triangle is $\dfrac{x^2}{4(2x-1)}$ . This is given to be $1$ . So $x=4+2\sqrt 3$ (since $x$ is greater than $|\overline {DE}|=1$ ), so that $a=4, b=2, c=3$ and $a+b+c=\boxed 9$