Subfield: Geometry


Question: Let the sides $A D$ and $B C$ of the quadrilateral $A B C D$ (such that $A B$ is not parallel to $C D$ ) intersect at point $P$. Points $O_{1}$ and $O_{2}$ are the circumcenters and points $H_{1}$ and $H_{2}$ are the orthocenters of triangles $A B P$ and $D C P$, respectively. Denote the midpoints of segments $O_{1} H_{1}$ and $O_{2} H_{2}$ by $E_{1}$ and $E_{2}$, respectively. Prove that the perpendicular from $E_{1}$ on $C D$, the perpendicular from $E_{2}$ on $A B$ and the line $H_{1} H_{2}$ are concurrent.

Solution:
We keep triangle $A B P$ fixed and move the line $C D$ parallel to itself uniformly, i.e. linearly dependent on a single parameter $\lambda$ (see Figure 1). Then the points $C$ and $D$ also move uniformly. Hence, the points $\mathrm{O}_{2}, \mathrm{H}_{2}$ and $E_{2}$ move uniformly, too. Therefore also the perpendicular from $E_{2}$ on $A B$ moves uniformly. Obviously, the points $O_{1}, H_{1}, E_{1}$ and the perpendicular from $E_{1}$ on $C D$ do not move at all. Hence, the intersection point $S$ of these two perpendiculars moves uniformly. Since $H_{1}$ does not move, while $H_{2}$ and $S$ move uniformly along parallel lines (both are perpendicular to $C D$ ), it is sufficient to prove their collinearity for two different positions of $C D$.

<img_3726>

Figure 1

Let $C D$ pass through either point $A$ or point $B$. Note that by hypothesis these two cases are different. We will consider the case $A \in C D$, i.e. $A=D$. So we have to show that the perpendiculars from $E_{1}$ on $A C$ and from $E_{2}$ on $A B$ intersect on the altitude $A H$ of triangle $A B C$ (see Figure 2).



<img_3804>

Figure 2

To this end, we consider the midpoints $A_{1}, B_{1}, C_{1}$ of $B C, C A, A B$, respectively. As $E_{1}$ is the center of FEUERBACH's circle (nine-point circle) of $\triangle A B P$, we have $E_{1} C_{1}=E_{1} H$. Similarly, $E_{2} B_{1}=E_{2} H$. Note further that a point $X$ lies on the perpendicular from $E_{1}$ on $A_{1} C_{1}$ if and only if

$$
X C_{1}^{2}-X A_{1}^{2}=E_{1} C_{1}^{2}-E_{1} A_{1}^{2}
$$

Similarly, the perpendicular from $E_{2}$ on $A_{1} B_{1}$ is characterized by

$$
X A_{1}^{2}-X B_{1}^{2}=E_{2} A_{1}^{2}-E_{2} B_{1}^{2}
$$

The line $H_{1} H_{2}$, which is perpendicular to $B_{1} C_{1}$ and contains $A$, is given by

$$
X B_{1}^{2}-X C_{1}^{2}=A B_{1}^{2}-A C_{1}^{2}
$$

The three lines are concurrent if and only if

$$
\begin{aligned}
0 & =X C_{1}^{2}-X A_{1}^{2}+X A_{1}^{2}-X B_{1}^{2}+X B_{1}^{2}-X C_{1}^{2} \\
& =E_{1} C_{1}^{2}-E_{1} A_{1}^{2}+E_{2} A_{1}^{2}-E_{2} B_{1}^{2}+A B_{1}^{2}-A C_{1}^{2} \\
& =-E_{1} A_{1}^{2}+E_{2} A_{1}^{2}+E_{1} H^{2}-E_{2} H^{2}+A B_{1}^{2}-A C_{1}^{2}
\end{aligned}
$$

i.e. it suffices to show that

$$
E_{1} A_{1}^{2}-E_{2} A_{1}^{2}-E_{1} H^{2}+E_{2} H^{2}=\frac{A C^{2}-A B^{2}}{4}
$$

We have

$$
\frac{A C^{2}-A B^{2}}{4}=\frac{H C^{2}-H B^{2}}{4}=\frac{(H C+H B)(H C-H B)}{4}=\frac{H A_{1} \cdot B C}{2}
$$

Let $F_{1}, F_{2}$ be the projections of $E_{1}, E_{2}$ on $B C$. Obviously, these are the midpoints of $H P_{1}$,



$H P_{2}$, where $P_{1}, P_{2}$ are the midpoints of $P B$ and $P C$ respectively. Then

$$
\begin{aligned}
& E_{1} A_{1}^{2}-E_{2} A_{1}^{2}-E_{1} H^{2}+E_{2} H^{2} \\
& =F_{1} A_{1}^{2}-F_{1} H^{2}-F_{2} A_{1}^{2}+F_{2} H^{2} \\
& =\left(F_{1} A_{1}-F_{1} H\right)\left(F_{1} A_{1}+F_{1} H\right)-\left(F_{2} A_{1}-F_{2} H\right)\left(F_{2} A_{1}+F_{2} H\right) \\
& =A_{1} H \cdot\left(A_{1} P_{1}-A_{1} P_{2}\right) \\
& =\frac{A_{1} H \cdot B C}{2} \\
& =\frac{A C^{2}-A B^{2}}{4}
\end{aligned}
$$

which proves the claim.
Let the perpendicular from $E_{1}$ on $C D$ meet $P H_{1}$ at $X$, and the perpendicular from $E_{2}$ on $A B$ meet $P H_{2}$ at $Y$ (see Figure 3). Let $\varphi$ be the intersection angle of $A B$ and $C D$. Denote by $M, N$ the midpoints of $P H_{1}, P H_{2}$ respectively.

<img_3910>

Figure 3

We will prove now that triangles $E_{1} X M$ and $E_{2} Y N$ have equal angles at $E_{1}, E_{2}$, and supplementary angles at $X, Y$.

In the following, angles are understood as oriented, and equalities of angles modulo $180^{\circ}$.

Let $\alpha=\angle H_{2} P D, \psi=\angle D P C, \beta=\angle C P H_{1}$. Then $\alpha+\psi+\beta=\varphi, \angle E_{1} X H_{1}=\angle H_{2} Y E_{2}=\varphi$, thus $\angle M X E_{1}+\angle N Y E_{2}=180^{\circ}$.

By considering the FEUERBACH circle of $\triangle A B P$ whose center is $E_{1}$ and which goes through $M$, we have $\angle E_{1} M H_{1}=\psi+2 \beta$. Analogous considerations with the FEUERBACH circle of $\triangle D C P$ yield $\angle H_{2} N E_{2}=\psi+2 \alpha$. Hence indeed $\angle X E_{1} M=\varphi-(\psi+2 \beta)=(\psi+2 \alpha)-\varphi=\angle Y E_{2} N$.

It follows now that

$$
\frac{X M}{M E_{1}}=\frac{Y N}{N E_{2}}
$$

Furthermore, $M E_{1}$ is half the circumradius of $\triangle A B P$, while $P H_{1}$ is the distance of $P$ to the orthocenter of that triangle, which is twice the circumradius times the cosine of $\psi$. Together



with analogous reasoning for $\triangle D C P$ we have

$$
\frac{M E_{1}}{P H_{1}}=\frac{1}{4 \cos \psi}=\frac{N E_{2}}{P H_{2}}
$$

By multiplication,

$$
\frac{X M}{P H_{1}}=\frac{Y N}{P H_{2}}
$$

and therefore

$$
\frac{P X}{X H_{1}}=\frac{H_{2} Y}{Y P}
$$

Let $E_{1} X, E_{2} Y$ meet $H_{1} H_{2}$ in $R, S$ respectively.

Applying the intercept theorem to the parallels $E_{1} X, P H_{2}$ and center $H_{1}$ gives

$$
\frac{H_{2} R}{R H_{1}}=\frac{P X}{X H_{1}}
$$

while with parallels $E_{2} Y, P H_{1}$ and center $H_{2}$ we obtain

$$
\frac{H_{2} S}{S H_{1}}=\frac{H_{2} Y}{Y P}
$$

Combination of the last three equalities yields that $R$ and $S$ coincide.