Subfield: Geometry


Question: Let $A B C$ be a triangle, and let $\ell_{1}$ and $\ell_{2}$ be two parallel lines. For $i=1,2$, let $\ell_{i}$ meet the lines $B C, C A$, and $A B$ at $X_{i}, Y_{i}$, and $Z_{i}$, respectively. Suppose that the line through $X_{i}$ perpendicular to $B C$, the line through $Y_{i}$ perpendicular to $C A$, and finally the line through $Z_{i}$ perpendicular to $A B$, determine a non-degenerate triangle $\Delta_{i}$.

Show that the circumcircles of $\Delta_{1}$ and $\Delta_{2}$ are tangent to each other.

Solution:
Throughout the solutions, $\sphericalangle(p, q)$ will denote the directed angle between lines $p$ and $q$, taken modulo $180^{\circ}$.

Let the vertices of $\Delta_{i}$ be $D_{i}, E_{i}, F_{i}$, such that lines $E_{i} F_{i}, F_{i} D_{i}$ and $D_{i} E_{i}$ are the perpendiculars through $X, Y$ and $Z$, respectively, and denote the circumcircle of $\Delta_{i}$ by $\omega_{i}$.

In triangles $D_{1} Y_{1} Z_{1}$ and $D_{2} Y_{2} Z_{2}$ we have $Y_{1} Z_{1} \| Y_{2} Z_{2}$ because they are parts of $\ell_{1}$ and $\ell_{2}$. Moreover, $D_{1} Y_{1} \| D_{2} Y_{2}$ are perpendicular to $A C$ and $D_{1} Z_{1} \| D_{2} Z_{2}$ are perpendicular to $A B$, so the two triangles are homothetic and their homothetic centre is $Y_{1} Y_{2} \cap Z_{1} Z_{2}=A$. Hence, line $D_{1} D_{2}$ passes through $A$. Analogously, line $E_{1} E_{2}$ passes through $B$ and $F_{1} F_{2}$ passes through $C$.

<img_3731>

The corresponding sides of $\Delta_{1}$ and $\Delta_{2}$ are parallel, because they are perpendicular to the respective sides of triangle $A B C$. Hence, $\Delta_{1}$ and $\Delta_{2}$ are either homothetic, or they can be translated to each other. Using that $B, X_{2}, Z_{2}$ and $E_{2}$ are concyclic, $C, X_{2}, Y_{2}$ and $F_{2}$ are concyclic, $Z_{2} E_{2} \perp A B$ and $Y_{2}, F_{2} \perp A C$ we can calculate

$$
\begin{aligned}
\sphericalangle\left(E_{1} E_{2}, F_{1} F_{2}\right) & =\sphericalangle\left(E_{1} E_{2}, X_{1} X_{2}\right)+\sphericalangle\left(X_{1} X_{2}, F_{1} F_{2}\right)=\sphericalangle\left(B E_{2}, B X_{2}\right)+\sphericalangle\left(C X_{2}, C F_{2}\right) \\
& =\sphericalangle\left(Z_{2} E_{2}, Z_{2} X_{2}\right)+\sphericalangle\left(Y_{2} X_{2}, Y_{2} F_{2}\right)=\sphericalangle\left(Z_{2} E_{2}, \ell_{2}\right)+\sphericalangle\left(\ell_{2}, Y_{2} F_{2}\right) \\
& =\sphericalangle\left(Z_{2} E_{2}, Y_{2} F_{2}\right)=\sphericalangle(A B, A C) \not \equiv 0,
\end{aligned}
\tag{1}
$$

and conclude that lines $E_{1} E_{2}$ and $F_{1} F_{2}$ are not parallel. Hence, $\Delta_{1}$ and $\Delta_{2}$ are homothetic; the lines $D_{1} D_{2}, E_{1} E_{2}$, and $F_{1} F_{2}$ are concurrent at the homothetic centre of the two triangles. Denote this homothetic centre by $H$.

For $i=1,2$, using (1), and that $A, Y_{i}, Z_{i}$ and $D_{i}$ are concyclic,

$$
\begin{aligned}
\sphericalangle\left(H E_{i}, H F_{i}\right) & =\sphericalangle\left(E_{1} E_{2}, F_{1} F_{2}\right)=\sphericalangle(A B, A C) \\
& =\sphericalangle\left(A Z_{i}, A Y_{i}\right)=\sphericalangle\left(D_{i} Z_{i}, D_{i} Y_{i}\right)=\sphericalangle\left(D_{i} E_{i}, D_{i} F_{i}\right),
\end{aligned}
$$



so $H$ lies on circle $\omega_{i}$.

The same homothety that maps $\Delta_{1}$ to $\Delta_{2}$, sends $\omega_{1}$ to $\omega_{2}$ as well. Point $H$, that is the centre of the homothety, is a common point of the two circles, That finishes proving that $\omega_{1}$ and $\omega_{2}$ are tangent to each other.
As in the first solution, let the vertices of $\Delta_{i}$ be $D_{i}, E_{i}, F_{i}$, such that $E_{i} F_{i}, F_{i} D_{i}$ and $D_{i} E_{i}$ are the perpendiculars through $X_{i}, Y_{i}$ and $Z_{i}$, respectively. In the same way we conclude that $\left(A, D_{1}, D_{2}\right),\left(B, E_{1}, E_{2}\right)$ and $\left(C, F_{1}, F_{2}\right)$ are collinear.

The corresponding sides of triangles $A B C$ and $D_{i} E_{i} F_{i}$ are perpendicular to each other. Hence, there is a spiral similarity with rotation $\pm 90^{\circ}$ that maps $A B C$ to $D_{i} E_{i} F_{i}$; let $M_{i}$ be the centre of that similarity. Hence, $\sphericalangle\left(M_{i} A, M_{i} D_{i}\right)=\sphericalangle\left(M_{i} B, M_{i} E_{i}\right)=\sphericalangle\left(M_{i} C, M_{i} F_{i}\right)=90^{\circ}$. The circle with diameter $A D_{i}$ passes through $M_{i}, Y_{i}, Z_{i}$, so $M_{i}, A, Y_{i}, Z_{i}, D_{i}$ are concyclic; analogously $\left(M_{i}, B, X_{i}, Z_{i}, E_{i}\right)$ and $\left(M_{i}, C, X_{i}, Y_{i}, F_{i}\right)$ are concyclic.

By applying Desargues' theorem to triangles $A B C$ and $D_{i} E_{i} F_{i}$ we conclude that the lines $A D_{i}, B E_{i}$ and $B F_{i}$ are concurrent; let their intersection be $H$. Since $\left(A, D_{1} . D_{2}\right),\left(B, E_{1} \cdot E_{2}\right)$ and $\left(C, F_{1} . F_{2}\right)$ are collinear, we obtain the same point $H$ for $i=1$ and $i=2$.

<img_3371>

By $\sphericalangle(C B, C H)=\sphericalangle\left(C X_{i}, C F_{i}\right)=\sphericalangle\left(Y_{i} X_{i}, Y_{i} F_{i}\right)=\sphericalangle\left(Y_{i} Z_{i}, Y_{i} D_{i}\right)=\sphericalangle\left(A Z_{i}, A D_{i}\right)=$ $\sphericalangle(A B, A H)$, point $H$ lies on circle $A B C$.

Analogously, from $\sphericalangle\left(F_{i} D_{i}, F_{i} H\right)=\sphericalangle\left(F_{i} Y_{i}, F_{i} C\right)=\sphericalangle\left(X_{i} Y_{i}, X_{i} C\right)=\sphericalangle\left(X_{i} Z_{i}, X_{i} B\right)=$ $\sphericalangle\left(E_{i} Z_{i}, E_{i} B\right)=\sphericalangle\left(E_{i} D_{i}, E_{i} H\right)$, we can see that point $H$ lies on circle $D_{i} E_{i} F_{i}$ as well. Therefore, circles $A B C$ and $D_{i} E_{i} F_{i}$ intersect at point $H$.

The spiral similarity moves the circle $A B C$ to circle $D_{i} E_{i} F_{i}$, so the two circles are perpendicular. Hence, both circles $D_{1} E_{1} F_{1}$ and $D_{2} E_{2} F_{2}$ are tangent to the radius of circle $A B C$ at $H$.