Local Mathematical Competitions

We start with a reformulation of a well-known statement on harmonic cyclic quadruples $(K_1,K_2,K_3,K_4)$, also provable by polar transformation (projective methods). LEMMA. Being given four pairwise non-parallel lines ${\ell }_i$, $i=1,2,3,4$, tangent to a circle $\omega$ at points $K_i$, and such that lines ${\ell }_1,{\ell }_3$ and $K_2{K}_4$ are concurrent, then lines ${\ell }_2,{\ell }_4$ and $K_1{K}_3$ are also concurrent.

Proof. Let $O$ be the center of $\omega$, $X={\ell }_1\cap {\ell }_3\cap K_2{K}_4$, $Y={\ell }_2\cap {\ell }_4$. We have $OX\perp K_1{K}_3$ and $OY\perp K_2{K}_4$. Let $Z=OX\cap K_1{K}_3$, $T=OY\cap K_2{K}_4$. Notice that triangles $△O{K}_{3}X$ and $△OZ{K}_3$ are similar, and also similar are triangles $△O{K}_{4}Y$ and $△OT{K}_4$, hence $OY\cdot OT=O{K}_4^2=O{K}_3^2=OX\cdot OZ$. This means that triangles $△OXT$ and $△OYZ$ are similar, hence $YZ\perp OX$, and so $Y\in K_1{K}_3$. $\square$

Suppose now lines $A_2{A}_3$, $A_4{A}_1$ and $T_1{T}_3$ are concurrent at a point $P$. Let $T_4^{\prime },T_2^{\prime }$ be the tangency points of lines $A_4{A}_1$, respectively $A_2{A}_3$, to circle ${\omega }_1$, and let $T_3^{\prime }$ be the second meeting point of line $T_1{T}_3$ and circle ${\omega }_1$. Let the tangent to ${\omega }_1$ at $T_3^{\prime }$ meet the lines $A_4{A}_1$, $A_2{A}_3$ at points $A_4^{\prime }$, respectively $A_3^{\prime }$. The (direct) homothety of center $P$ that takes ${\omega }_1$ to ${\omega }_3$ maps $T_3^{\prime }$ to $T_3$, hence $A_3{A}_4\parallel A_3^{\prime }{A}_4^{\prime }$. Let $Q=A_1{A}_2\cap A_3{A}_4$, $Q^{\prime }=A_1{A}_2\cap A_3^{\prime }{A}_4^{\prime }$. Applying the LEMMA to circle ${\omega }_1$ and lines $A_2{A}_3$, $A_3^{\prime }{A}_4^{\prime }$, $A_4{A}_1$, $A_1{A}_2$, yields that points $Q^{\prime }$, $T_2^{\prime }$, $T_4^{\prime }$ are collinear. The (inverse) homothety of center $A_1$ that takes $△Q{A}_1{A}_4$ to $△Q^{\prime }{A}_1{A}_4^{\prime }$ maps ${\omega }_4$ to ${\omega }_1$, so maps $T_4$ to $T_4^{\prime }$, hence $Q^{\prime }{T}_4^{\prime }\parallel Q{T}_4$. Similarly, the (inverse) homothety of center $A_2$ that takes $△Q{A}_2{A}_3$ to $△Q^{\prime }{A}_2{A}_3^{\prime }$ maps ${\omega }_2$ to ${\omega }_1$, so maps $T_2$ to $T_2^{\prime }$, hence also $Q^{\prime }{T}_2^{\prime }\parallel Q{T}_2$. Since points $Q^{\prime }$, $T_2^{\prime }$, $T_4^{\prime }$ are collinear, it follows points $Q,T_2,T_4$ are also collinear.

The converse implication is done in a similar way, due to the cyclic nature of the notations used (just increase each index by 1).

---

Alternative solution.

Alternative Solution. (D. Şerbănescu) Suppose $Q,T_2,T_4$ are collinear. We will show $P,T_1,T_3$ are collinear. We will use the notations of the solution above, but also let $S_1^{\prime },S_1^{\prime \prime }$ be the tangency points of line $A_1{A}_2$ to circle ${\omega }_2$, respectively ${\omega }_4$, and let $S_3^{\prime },S_3^{\prime \prime }$ be the tangency points of line $A_3{A}_4$ to circle ${\omega }_2$, respectively ${\omega }_4$. Let $T_4^{\prime \prime }$ be the (other than $T_2$) meeting point of line $Q{T}_2{T}_4$ and circle ${\omega }_2$, and let the tangent line to ${\omega }_2$ at $T_4^{\prime \prime }$ (parallel to $A_1{A}_4$) meet $A_2{A}_3$ at $P^{\prime }$ (via the (direct) homothety of center $Q$ that takes ${\omega }_4$ to ${\omega }_2$).

Clearly $△P{T}_2{T}_4\sim △P^{\prime }{T}_2{T}_4^{\prime \prime }$ and $△P^{\prime }{T}_2{T}_4^{\prime \prime }$ is isosceles, so $P{T}_2=P{T}_4$ (in other words, if $Q,T_2,T_4$ are collinear then $P{T}_2=P{T}_4$; the other implication trivially also holds, but is irrelevant here). From $P{T}_2=P{T}_4$ and $P{T}_2^{\prime }=P{T}_4^{\prime }$ follows $T_2^{\prime }{T}_2=T_4^{\prime }{T}_4$. As external tangents, $T_2^{\prime }{T}_2=T_1{S}_1^{\prime }$ and $T_4^{\prime }{T}_4=T_1{S}_1^{\prime \prime }$, hence $T_1$ is the midpoint of $S_1^{\prime }{S}_1^{\prime \prime }$. Similarly, $T_3$ is the midpoint of $S_3^{\prime }{S}_3^{\prime \prime }$. It follows that $P,T_1,T_3$ lie on the radical axis of the circles ${\omega }_2$ and ${\omega }_4$, hence are collinear.