Calculating Areas

Since $\triangle EIG$ , $\triangle BIG$ , $\triangle BID$ , and $\triangle DIF$ have the same base $t$ and the same height $h$ , they have the same area. Let the area be $A_1$ , then $3A_1 = t^2$ , $\implies A_1 = \frac 13 t^2$ . Note that $\triangle AEH$ and $\triangle CFH$ are congruent and each has an area of $A_2 = \frac 34 t^2$ .

Then the area of the black region is:

$\begin{aligned} [EHFI] & = [ABC] - [EIFB] - [AEH] - [CFH] \\ & = \frac 92 t^2 - 4A_1 - 2A_2 \\ & = \frac 92 t^2 - 4 \times \frac 13 t^2 - 2 \times \frac 34 t^2 \\ & = \boxed {\frac 53 t^2} \end{aligned}$

(Note: P is the intersection point between EF and IH.)

From isosceles right triangles $△AHB, △EBF, △EPB \\$

$BH = \frac{3t}{\sqrt{2}} , EF = 2t \sqrt{2} , BP = \frac{2t}{\sqrt{2}}$

From triangle \( △BIF \\

\frac{BI}{sin(arctan(\frac{1}{2}))} = \frac{2t}{sin(180º - 45º - arctan(\frac{1}{2}))}\)

$BI = \frac{2sin(arctan(\frac{1}{2}))t}{\sqrt{5}}$

Then,

$BH - BI = IH = \frac{3t}{\sqrt{2}} - \frac{2sin(arctan(\frac{1}{2}))t}{\sqrt{5}} \approx 1.179t \\$

Consider the heights of triangles $△EIF, △EHF \\$

By relation of triangles:

$\\PH = BH - BP = \frac{1}{3} BH = \frac{t}{\sqrt{2}}, \\PI = IH - PH = \frac{2t}{\sqrt{2}} - \frac{2sin(arctan(\frac{1}{2}))t}{\sqrt{5}} \approx 0.4714t$

The shaded area is

\(A = \frac{1}{2} ( EF * PH + EF * PI) \approx \frac{1}{2} ( 2t \sqrt{2}) ( \frac{t}{\sqrt{2}} + 0.4714t)\\

A \approx (1 + 0.4714 \sqrt{2} ) t^{2} \approx 1.66666 t^{2} \\

A = \boxed{\frac{5t^{2}}{3} } \)

The coordinates of all of the labeled points are trivial to find, except for point $I$ . Assuming we initially have all coordinates except for point $I$ , my process is:

1) Determine the coordinates of point $I$ by finding the intersection of segments $ED$ and $GF$
2) Form vectors from $I$ to $E$ and from $I$ to $F$
3) Take one half the magnitude of the cross product of the vectors from step 2 to get the area of the green triangle
4) Form vectors from $H$ to $E$ and from $H$ to $F$
5) Take one half the magnitude of the cross product of the vectors from step 4 to get the area of the red triangle
6) Add the areas of the two triangles together

I have 2 ways to do this problem: The second method is with integrals:

Note: for the second method, the vertex B of the triangle is located at the point (0,0)

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