A dizzy spiral

Let the spiral to be on the complex plain with $A$ being the origin. Let complex number $z_1 = x_1 + y_1i = \vec {AB}$ , $z_2 = x_2 + y_2i = \vec {AC}$ , $z_3 = x_3 + y_3i = \vec {AD}$ and so on, such that $B(x_1, y_1)$ , $C(x_2, y_2)$ , $D(x_3, y_3)$ and so on.

We note that a turning $90^\circ$ clockwise is equivalent to multiplying by $-i$ , therefore, we have $z_n = 1 + 2(-i) +3(-i)^2 + 4(-i)^3 + \cdots + n(-i)^{n-1} = \sum_{k=1}^n k(-i)^{k-1}$ , which is the sum of an arithmetic-geometrical progression with initial term $a=1$ , common difference $d=1$ and common ratio $r=-i$ . Therefore,

$\begin{aligned} z_n & = \frac {1-n(-i)^n}{1+i} + \frac {(-i)\left(1-(-i)^{n-1}\right)}{(1+i)^2} \\ & = \frac {\left(1-n(-i)^n \right)(1-i)}{(1+i)(1-i)} + \frac {(-i)\left(1-(-i)^{n-1}\right)}{2i} \\ & = \frac {(-i)^{n-1}(n+1+ni)-i}2 \\ & = \frac {(-i)^n(-n +(n+1)i))-i}2 \\ \implies z_n & = \begin{cases} - \dfrac n2 + \dfrac n2 i & \text{for } n \bmod 4 = 0 \\ \dfrac {n+1}2 + \dfrac {n-1}2 i & \text{for } n \bmod 4 = 1 \\ \dfrac n2 - \dfrac {n+2}2 i & \text{for } n \bmod 4 = 2 \\ -\dfrac {n+1}2 - \dfrac {n+1}2 i & \text{for } n \bmod 4 = 3 \end{cases} \end{aligned}$

Now, we have

$\begin{cases} \vec {AE} = z_4 = -2 + 2i & \implies E(-2,2) \\ \vec {AM} = z_{12} = -6 + 6i & \implies M(-6,6) \\ \vec {AT} = z_{19} = -10 - 10i & \implies T(-10,-10) \end{cases}$

From the coordinates, we can find the area of $\triangle EMT$ as follows:

$\begin{aligned} [EMT] & = \left| \frac {6+2}2 (-6+2) + \frac {-10+6}2 (-10+6) + \frac {2-10}2 (-2+10) \right| \\ & = \left| 4 (-4) -2(-4) - 4(8) \right| \\ & = \left| -16 +8 -32 \right| \\ & = \boxed{40} \end{aligned}$