Subfield: Geometry


Question: Points $A_{1}, B_{1}, C_{1}$ are chosen on the sides $B C, C A, A B$ of a triangle $A B C$, respectively. The circumcircles of triangles $A B_{1} C_{1}, B C_{1} A_{1}, C A_{1} B_{1}$ intersect the circumcircle of triangle $A B C$ again at points $A_{2}, B_{2}, C_{2}$, respectively $\left(A_{2} \neq A, B_{2} \neq B, C_{2} \neq C\right)$. Points $A_{3}, B_{3}, C_{3}$ are symmetric to $A_{1}, B_{1}, C_{1}$ with respect to the midpoints of the sides $B C, C A, A B$ respectively. Prove that the triangles $A_{2} B_{2} C_{2}$ and $A_{3} B_{3} C_{3}$ are similar.

Solution:
We will work with oriented angles between lines. For two straight lines $\ell, m$ in the plane, $\angle(\ell, m)$ denotes the angle of counterclockwise rotation which transforms line $\ell$ into a line parallel to $m$ (the choice of the rotation centre is irrelevant). This is a signed quantity; values differing by a multiple of $\pi$ are identified, so that

$$
\angle(\ell, m)=-\angle(m, \ell), \quad \angle(\ell, m)+\angle(m, n)=\angle(\ell, n)
$$

If $\ell$ is the line through points $K, L$ and $m$ is the line through $M, N$, one writes $\angle(K L, M N)$ for $\angle(\ell, m)$; the characters $K, L$ are freely interchangeable; and so are $M, N$.

The counterpart of the classical theorem about cyclic quadrilaterals is the following: If $K, L, M, N$ are four noncollinear points in the plane then

$$
K, L, M, N \text { are concyclic if and only if } \angle(K M, L M)=\angle(K N, L N) \text {. }\tag{1}
$$

Passing to the solution proper, we first show that the three circles $\left(A B_{1} C_{1}\right),\left(B C_{1} A_{1}\right)$, $\left(C A_{1} B_{1}\right)$ have a common point. So, let $\left(A B_{1} C_{1}\right)$ and $\left(B C_{1} A_{1}\right)$ intersect at the points $C_{1}$ and $P$. Then by $(1)$

$$
\begin{gathered}
\angle\left(P A_{1}, C A_{1}\right)=\angle\left(P A_{1}, B A_{1}\right)=\angle\left(P C_{1}, B C_{1}\right) \\
=\angle\left(P C_{1}, A C_{1}\right)=\angle\left(P B_{1}, A B_{1}\right)=\angle\left(P B_{1}, C B_{1}\right) .
\end{gathered}
$$

Denote this angle by $\varphi$.

The equality between the outer terms shows, again by (1), that the points $A_{1}, B_{1}, P, C$ are concyclic. Thus $P$ is the common point of the three mentioned circles.

From now on the basic property (1) will be used without explicit reference. We have

$$
\varphi=\angle\left(P A_{1}, B C\right)=\angle\left(P B_{1}, C A\right)=\angle\left(P C_{1}, A B\right)\tag{2}
$$

<img_3832>



Let lines $A_{2} P, B_{2} P, C_{2} P$ meet the circle $(A B C)$ again at $A_{4}, B_{4}, C_{4}$, respectively. As

$$
\angle\left(A_{4} A_{2}, A A_{2}\right)=\angle\left(P A_{2}, A A_{2}\right)=\angle\left(P C_{1}, A C_{1}\right)=\angle\left(P C_{1}, A B\right)=\varphi
$$

we see that line $A_{2} A$ is the image of line $A_{2} A_{4}$ under rotation about $A_{2}$ by the angle $\varphi$. Hence the point $A$ is the image of $A_{4}$ under rotation by $2 \varphi$ about $O$, the centre of $(A B C)$. The same rotation sends $B_{4}$ to $B$ and $C_{4}$ to $C$. Triangle $A B C$ is the image of $A_{4} B_{4} C_{4}$ in this map. Thus

$$
\angle\left(A_{4} B_{4}, A B\right)=\angle\left(B_{4} C_{4}, B C\right)=\angle\left(C_{4} A_{4}, C A\right)=2 \varphi
$$

Since the rotation by $2 \varphi$ about $O$ takes $B_{4}$ to $B$, we have $\angle\left(A B_{4}, A B\right)=\varphi$. Hence by $(2)$

$$
\angle\left(A B_{4}, P C_{1}\right)=\angle\left(A B_{4}, A B\right)+\angle\left(A B, P C_{1}\right)=\varphi+(-\varphi)=0
$$

which means that $A B_{4} \| P C_{1}$.
<img_3550>

Let $C_{5}$ be the intersection of lines $P C_{1}$ and $A_{4} B_{4}$; define $A_{5}, B_{5}$ analogously. So $A B_{4} \| C_{1} C_{5}$ and, by (3) and (2),

$$
\angle\left(A_{4} B_{4}, P C_{1}\right)=\angle\left(A_{4} B_{4}, A B\right)+\angle\left(A B, P C_{1}\right)=2 \varphi+(-\varphi)=\varphi ;
$$

i.e., $\angle\left(B_{4} C_{5}, C_{5} C_{1}\right)=\varphi$. This combined with $\angle\left(C_{5} C_{1}, C_{1} A\right)=\angle\left(P C_{1}, A B\right)=\varphi$ (see (2)) proves that the quadrilateral $A B_{4} C_{5} C_{1}$ is an isosceles trapezoid with $A C_{1}=B_{4} C_{5}$.

Interchanging the roles of $A$ and $B$ we infer that also $B C_{1}=A_{4} C_{5}$. And since $A C_{1}+B C_{1}=$ $A B=A_{4} B_{4}$, it follows that the point $C_{5}$ lies on the line segment $A_{4} B_{4}$ and partitions it into segments $A_{4} C_{5}, B_{4} C_{5}$ of lengths $B C_{1}\left(=A C_{3}\right)$ and $A C_{1}$ (=BC $C_{3}$ ). In other words, the rotation which maps triangle $A_{4} B_{4} C_{4}$ onto $A B C$ carries $C_{5}$ onto $C_{3}$. Likewise, it sends $A_{5}$ to $A_{3}$ and $B_{5}$ to $B_{3}$. So the triangles $A_{3} B_{3} C_{3}$ and $A_{5} B_{5} C_{5}$ are congruent. It now suffices to show that the latter is similar to $A_{2} B_{2} C_{2}$.

Lines $B_{4} C_{5}$ and $P C_{5}$ coincide respectively with $A_{4} B_{4}$ and $P C_{1}$. Thus by (4)

$$
\angle\left(B_{4} C_{5}, P C_{5}\right)=\varphi
$$

Analogously (by cyclic shift) $\varphi=\angle\left(C_{4} A_{5}, P A_{5}\right)$, which rewrites as

$$
\varphi=\angle\left(B_{4} A_{5}, P A_{5}\right)
$$



These relations imply that the points $P, B_{4}, C_{5}, A_{5}$ are concyclic. Analogously, $P, C_{4}, A_{5}, B_{5}$ and $P, A_{4}, B_{5}, C_{5}$ are concyclic quadruples. Therefore

$$
\angle\left(A_{5} B_{5}, C_{5} B_{5}\right)=\angle\left(A_{5} B_{5}, P B_{5}\right)+\angle\left(P B_{5}, C_{5} B_{5}\right)=\angle\left(A_{5} C_{4}, P C_{4}\right)+\angle\left(P A_{4}, C_{5} A_{4}\right) .
$$

On the other hand, since the points $A_{2}, B_{2}, C_{2}, A_{4}, B_{4}, C_{4}$ all lie on the circle $(A B C)$, we have

$$
\angle\left(A_{2} B_{2}, C_{2} B_{2}\right)=\angle\left(A_{2} B_{2}, B_{4} B_{2}\right)+\angle\left(B_{4} B_{2}, C_{2} B_{2}\right)=\angle\left(A_{2} A_{4}, B_{4} A_{4}\right)+\angle\left(B_{4} C_{4}, C_{2} C_{4}\right)
$$

But the lines $A_{2} A_{4}, B_{4} A_{4}, B_{4} C_{4}, C_{2} C_{4}$ coincide respectively with $P A_{4}, C_{5} A_{4}, A_{5} C_{4}, P C_{4}$. So the sums on the right-hand sides of (5) and (6) are equal, leading to equality between their left-hand sides: $\angle\left(A_{5} B_{5}, C_{5} B_{5}\right)=\angle\left(A_{2} B_{2}, C_{2} B_{2}\right)$. Hence (by cyclic shift, once more) also $\angle\left(B_{5} C_{5}, A_{5} C_{5}\right)=\angle\left(B_{2} C_{2}, A_{2} C_{2}\right)$ and $\angle\left(C_{5} A_{5}, B_{5} A_{5}\right)=\angle\left(C_{2} A_{2}, B_{2} A_{2}\right)$. This means that the triangles $A_{5} B_{5} C_{5}$ and $A_{2} B_{2} C_{2}$ have their corresponding angles equal, and consequently they are similar.