IMO Problem Shortlist

We keep triangle $ABP$ fixed and move the line $CD$ 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 $O_2,H_2$ and $E_2$ move uniformly, too. Therefore also the perpendicular from $E_2$ on $AB$ moves uniformly. Obviously, the points $O_1,H_1,E_1$ and the perpendicular from $E_1$ on $CD$ 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 $CD$), it is sufficient to prove their collinearity for two different positions of $CD$.

Figure 1

Let $CD$ pass through either point $A$ or point $B$. Note that by hypothesis these two cases are different. We will consider the case $A\in CD$, i.e. $A=D$. So we have to show that the perpendiculars from $E_1$ on $AC$ and from $E_2$ on $AB$ intersect on the altitude $AH$ of triangle $ABC$ (see Figure 2).

Figure 2

To this end, we consider the midpoints $A_1,B_1,C_1$ of $BC,CA,AB$, respectively. As $E_1$ is the center of FEUERBACH's circle (nine-point circle) of $△ABP$, 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 $$\begin{array}{rl}0& =X{C}_1^2-X{A}_1^2+X{A}_1^2-X{B}_1^2+X{B}_1^2-X{C}_1^2\\ & =E_1{C}_1^2-E_1{A}_1^2+E_2{A}_1^2-E_2{B}_1^2+A{B}_1^2-A{C}_1^2\\ & =-E_1{A}_1^2+E_2{A}_1^2+E_1{H}^2-E_2{H}^2+A{B}_1^2-A{C}_1^2\end{array}$$ 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 $$\frac{A{C}^2-A{B}^2}{4}=\frac{H{C}^2-H{B}^2}{4}=\frac{(HC+HB)(HC-HB)}{4}=\frac{H{A}_1\cdot BC}{2}.$$ Let $F_1,F_2$ be the projections of $E_1,E_2$ on $BC$. Obviously, these are the midpoints of $H{P}_1$, $H{P}_2$, where $P_1,P_2$ are the midpoints of $PB$ and $PC$ respectively. Then $$\begin{array}{rl}& E_1{A}_1^2-E_2{A}_1^2-E_1{H}^2+E_2{H}^2\\ & =F_1{A}_1^2-F_1{H}^2-F_2{A}_1^2+F_2{H}^2\\ & =\left(F_1{A}_1-F_{1}H\right)\left(F_1{A}_1+F_{1}H\right)-\left(F_2{A}_1-F_{2}H\right)\left(F_2{A}_1+F_{2}H\right)\\ & =A_{1}H\cdot \left(A_1{P}_1-A_1{P}_2\right)\\ & =\frac{A_{1}H\cdot BC}{2}\\ & =\frac{A{C}^2-A{B}^2}{4},\end{array}$$ which proves the claim.

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Alternative solution.

Let the perpendicular from $E_1$ on $CD$ meet $P{H}_1$ at $X$, and the perpendicular from $E_2$ on $AB$ meet $P{H}_2$ at $Y$ (see Figure 3). Let $\phi$ be the intersection angle of $AB$ and $CD$. Denote by $M,N$ the midpoints of $P{H}_1,P{H}_2$ respectively.

Figure 3

We will prove now that triangles $E_{1}XM$ and $E_{2}YN$ 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 $\alpha =\angle H_{2}PD$, $\psi =\angle DPC$, $\beta =\angle CP{H}_1$. Then $\alpha +\psi +\beta =\phi$, $\angle E_{1}X{H}_1=\angle H_{2}Y{E}_2=\phi$, thus $\angle MX{E}_1+\angle NY{E}_2=180^{\circ }$. By considering the Feuerbach circle of $△ABP$ 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 $△DCP$ yield $\angle H_{2}N{E}_2=\psi +2\alpha$. Hence indeed $\angle X{E}_{1}M=\phi -(\psi +2\beta )=(\psi +2\alpha )-\phi =\angle Y{E}_{2}N$. It follows now that $$\frac{XM}{M{E}_1}=\frac{YN}{N{E}_2}.$$ Furthermore, $M{E}_1$ is half the circumradius of $△ABP$, 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 $△DCP$ we have $$\frac{M{E}_1}{P{H}_1}=\frac{1}{4\cos \psi }=\frac{N{E}_2}{P{H}_2}.$$ By multiplication, $$\frac{XM}{P{H}_1}=\frac{YN}{P{H}_2},$$ and therefore $$\frac{PX}{X{H}_1}=\frac{H_{2}Y}{YP}.$$ 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 $$\frac{H_{2}R}{R{H}_1}=\frac{PX}{X{H}_1},$$ while with parallels $E_{2}Y,P{H}_1$ and center $H_2$ we obtain $$\frac{H_{2}S}{S{H}_1}=\frac{H_{2}Y}{YP}$$ Combination of the last three equalities yields that $R$ and $S$ coincide.