# Proof of the existence of Schiffler Point using complex numbers

Let $$ABC$$ be a triangle with incenter $$I$$. Prove that the Euler lines of triangles $$AIB, BIC, CIA,$$ and $$ABC$$ are concurrent (called the Schiffler point of $$ABC$$).

Here is my approach :

Set the circumcircle of $$ABC$$ as the unit circle, hence we have $$\overrightarrow{OH}=a+b+c$$, where $$O$$ is the circumcenter and $$H$$ is the orthocenter of triangle $$ABC$$. Next, I attempted to find the Euler line of triangle $$AIB$$, but I will find the line connecting the circumcenter and the centroid of triangle $$AIB$$ instead (since the centroid also lies on the Euler line). Now let $$z_{1}$$ and $$g_{1}$$ denote the circumcenter and the centroid of triangle $$AIB$$ respectively. We then have $$g_{1}=\frac{a+b+i}{3}$$ and $$z_{1}=\frac{\begin{vmatrix} a&|a|^2&1\\ i&|i|^2&1\\ b&|b|^2&1 \end{vmatrix}}{\begin{vmatrix} a&\bar{a}&1 \\ i&\bar{i}&1 \\ b&\bar{b}&1 \end{vmatrix}}=\frac{|i|^2 - 1}{\bar{i}-\frac{a+b}{ab}+\frac{i}{ab}}$$ after some manipulations. But I need to find $$\overrightarrow{Z_{1}G_{1}}$$, which will be too ugly if I simply do $$g_{1}-z_{1}$$. The key point is that, I need to find some $$k \in \mathbb{R}$$ such that $$k(a+b+c)$$ lies on line $$Z_{1}G_{1}$$, and the same for the other two triangles. Can anyone find a better proof of this, or perhaps make the calculations simpler? Thanks for any help!

• This sort of question is best done in the real or complex projective plane as in E.A. Maxwell's elegeant book The methods of plane projective geometry based on the use of general hoogeneous coordinates. Find the 3 lines. Show that the determinant of the coefficients is 0. Jan 21 '20 at 19:42

Let's set $$(ABC)$$ as the unit circle. Then, let's set $$a=u^2,b=v^2,c=w^2$$ for complex numbers $$u,v,w$$. We characterise the intersection of the Euler Lines of triangles $$AIB, BIC, CIA$$ as the intersection of the lines joining the orthocentre and centroid of each triangle.
Now by the Incentre-Excentre Lemma it follows that the circumcentre of $$AIB$$ say $$O_1$$ is the midpoint of $$\overset{ \frown}{AB}$$ not containig C.
Thus we've $$o_1=-uv$$. Then the centroid (say $$G_1$$), whose existence is an affine property is just the arithmetic mean of the vertices i.e. $$g_1=(a+b+i)/3$$. Now by the theorem on complex incentre we've $$i=-(uv+vw+wu)$$. Therefore we've $$g_1=u^2+v^2-uv-vw-wu$$.

Now we note that the as the orthocentre of $$ABC$$ is $$u^2+v^2+w^2$$ then the it's Euler line can be defined as all points $$k(u^2+v^2+w^2)$$ for all real numbers $$k$$

Now we find $$k$$ such that $$k(u^2+v^2+w^2),O_1,G_1$$ are collinear. Thus

\begin{align*} \frac{-uv-k(u^2+v^2+w^2)}{-uv-\frac13(u^2+v^2-uv-vw-wu)}&=\overline{\left(\frac{-uv-k(u^2+v^2+w^2)}{-uv-\frac13(u^2+v^2-uv-vw-wu)}\right)}\\ \frac{uv+k(u^2+v^2+w^2)}{u^2+v^2+2uv-vw-wu}&=\frac{\frac1{uv}+k\left(\frac1{u^2}+\frac1{v^2}+\frac1{w^2}\right)}{\frac1{u^2}+\frac1{v^2}+\frac2{uv}-\frac1{vw}-\frac1{wu}}\\ \frac uv+\frac vu+2-\frac uw-\frac vw+k(u^2+v^2+w^2)\left(\frac1{u^2}+\frac1{v^2}+\frac2{uv}-\frac1{vw}-\frac1{wu}\right)&=\frac uv+\frac vu+2-\frac wu-\frac wv+k\left(\frac1{u^2}+\frac1{v^2}+\frac1{w^2}\right)\left(u^2+v^2+2uv-vw-wu\right)\\ \end{align*}
Further computation gives
\begin{align*} k&=\frac{\frac uw+\frac vw-\frac wu-\frac wv}{(u^2+v^2+w^2)\left(\frac1{u^2}+\frac1{v^2}+\frac2{uv}-\frac1{vw}-\frac1{wu}\right)-\left(\frac1{u^2}+\frac1{v^2}+\frac1{w^2}\right)\left(u^2+v^2+2uv-vw-wu\right)}\\ &=\frac{\frac uw+\frac vw-\frac wu-\frac wv}{\frac{2w^2}{vu}-\frac{u^2}{vw}-\frac{v^2}{wu}+\frac{wv}{u^2}+\frac{w^2}{u^2}+\frac{wu}{v^2}+\frac{w^2}{v^2}-\frac{u^2}{w^2}-\frac{2uv}{w^2}-\frac{v^2}{w^2}}\\ &=\frac{uvw(u+v)(uv-w^2)}{(u+v)(uv-w^2)(uvw-u^2v-uv^2-v^2w-vw^2-w^2u-wu^2)}\\ &=\frac{uvw}{uvw-u^2v-uv^2-v^2w-vw^2-w^2u-wu^2},\\ \end{align*}
As it's symmetric in $$u,v,w$$ it follows that
$$\frac{uvw(u^2+v^2+w^2)}{uvw-u^2v-uv^2-v^2w-vw^2-w^2u-wu^2}$$
lies on all four Euler Lines i.e. of the triangles $$ABC,AIB,BIC,CIA$$ and we're done.