A Complex Number Aided Proof in Plane Geometry

The proof in my post “A Computer Algebra Aided Proof in Plane Geometry” can be significantly simplified if complex numbers are used.

Consider complex numbers v_1, v_2, w_1, w_2, p in FIg. 1.

Fig. 1


v_1=p-(-a)= p+a \overset{p=x+\bold{i}y}{=} x+\bold{i}y+a \implies v_1 = x+a+\bold{i}y,\quad\quad\quad(1)

we have,

\bold{i}v_1 = w_1-(-a) \implies w_1=-a+\bold{i} v_1 \overset{(1)}{=} -a +\bold{i}(x+a+\bold{i}y)=-a-y+\bold{i}(x+a).


\begin{cases} x_1 = -a-y \\ y_1=x+a\end{cases}\quad\quad\quad(*)


v_2 = p-a = (x+\bold{i}y)-a=x-a+\bold{i}y\quad\quad\quad(2)


-\bold{i}v_2=w_2-a\implies w_2 = a + (-\bold{i} v_2) \overset{(2)}{=} a + (-\bold{i}(x-a+\bold{i}y)) = a+y+\bold{i}(a-x)


\begin{cases} x_3 = a+y \\ y_3 = a-x \end{cases}\quad\quad\quad(**)

Let k_1, k_3 denote the slope of line connecting (0, a) and (x_1, y_1), (x_3, y_3) respectively.


k_1=\frac{y_1-a}{x_1-0} \overset{(*)}{=} \frac{x+a-a}{-a-y} = \frac{-x}{a+y},

k_3=\frac{y_3-a}{x_3-0} \overset{(**)}{=} \frac{a-x-a}{a+y-0} = \frac{-x}{a+y},

we see that

k_1 = k_3\implies (x_1, y_1), (0, a), (x_3, y_3) lie on the same line.

Notice that a+y \ne 0 since y>0, a>0.

See also “Treasure Hunt with Complex Numbers“.


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