Area Problem I

Let $a = BC$ , $b = AC$ , and $c = AB$ . Since $\triangle ABC$ is a right triangle, $a^2 + b^2 = c^2$ , and since $I$ is at the intersection of the angle bisectors of $\triangle ABC$ , it is the incenter of $\triangle ABC$ with an incircle radius of $r = \frac{ab}{a + b + c}$ . The area of $\triangle IAB$ is then $A_{\triangle IAB} = \frac{1}{2}cr = \frac{abc}{2(a + b + c)}$ .

Now place and rotate the diagram on a coordinate graph so that $C$ is on the origin, $A$ is on the $x$ -axis, and $B$ is on the $y$ -axis.

Then the line through $B(0, a)$ and $I(\frac{ab}{a + b + c}, \frac{ab}{a + b + c})$ has an equation of $y = -\frac{a + c}{ab}x + a$ , so that $E$ is the $x$ -intercept at $E(\frac{ab}{a + c}, 0)$ , which means $CE = \frac{ab}{a + c}$ . By a similar argument, $CD = \frac{ab}{b + c}$ .

The area of quadrilateral $ABDE$ is $A_{ABDE} = A_{\triangle ABC} - A_{CDE} = \frac{1}{2}ab - \frac{1}{2}\frac{ab}{a + c}\frac{ab}{b + c}$ , which simplifies to $A_{ABDE} = \frac{abc}{a + b + c} = 2A_{\triangle IAB}$ .

Since $A_{\triangle IAB} = 6$ , $A_{ABDE} = 2 \cdot 6 = \boxed{12}$ .

Let $\angle A = \theta$ ; then $\angle B = 90^\circ - \theta$ , $\angle IAB = \dfrac \theta 2$ and $\angle IBA = 45^\circ - \theta$ . Let $AB=c$ and $IF$ be perpendicular to $AB$ with length $r$ . Then

$\begin{aligned} c & = r \cot \frac \theta 2 + r \cot \left(45^\circ - \frac \theta 2 \right) & \small \color{#3D99F6} \text{Let }t = \tan \frac \theta 2 \\ & = r\left( \frac 1t + \frac {1+t}{1-t} \right) \\ \implies r & = \frac {ct(1-t)}{1+t^2} & \small \color{#3D99F6} \text{Since }[IAB] = \frac {rc}2 = 6 \\ rc & = \color{#3D99F6} \frac {c^2t(1-t)}{1+t^2} = 12 \end{aligned}$

Now the area of $\triangle ABC$ ,

$\begin{aligned} [ABC] & = \frac {\overline{BC} \times \overline{AC}}2 \\ & = \frac {c^2}2 \sin \theta \cos \theta \\ & = \frac {c^2}2 \times \frac {2t}{1+t^2} \times \frac {1-t^2}{1+t^2} \\ & = \frac {{\color{#3D99F6}c^2t(1-t)}(1+t)}{{\color{#3D99F6}(1+t^2)} (1+t^2)} & \small \color{#3D99F6} \text{Recall } \frac {c^2t(1-t)}{1+t^2} = 12 \\ & = \frac {{\color{#3D99F6}12}(1+t)}{1+t^2} \end{aligned}$

And the area of $\triangle CDE$ .

$\begin{aligned} [CDE] & = \frac {\overline{CD} \times \overline{CE}}2 \\ & = \frac {\overline{AC} \tan \frac \theta 2 \times \overline{BC} \tan \left(45^\circ - \frac \theta 2\right)} 2 \\ & = {\color{#3D99F6}\frac {c^2}2 \sin \theta \cos \theta} \tan \frac \theta 2 \tan \left(45^\circ - \frac \theta 2\right) & \small \color{#3D99F6} \text{Recall } \frac {c^2}2 \sin \theta \cos \theta = \frac {12(1+t)}{1+t^2} \\ & = {\color{#3D99F6} \frac {12(1+t)}{1+t^2}} \times \frac {t(1-t)}{1+t} \end{aligned}$

Therefore, the area of quadrilateral $ABDE$ ,

$\begin{aligned} [ABDE] & = [ABC] - [CDE] \\ & = \frac {12(1+t)}{1+t^2} \left(1- \frac {t(1-t)}{1+t}\right) \\ & = \frac {12(1+t)}{1+t^2} \left(\frac {1+t^2}{1+t}\right) \\ & = \boxed{12} \end{aligned}$