Subfield: Geometry


Question: Let the excircle of the triangle $A B C$ lying opposite to $A$ touch its side $B C$ at the point $A_{1}$. Define the points $B_{1}$ and $C_{1}$ analogously. Suppose that the circumcentre of the triangle $A_{1} B_{1} C_{1}$ lies on the circumcircle of the triangle $A B C$. Prove that the triangle $A B C$ is right-angled.

Solution:
Denote the circumcircles of the triangles $A B C$ and $A_{1} B_{1} C_{1}$ by $\Omega$ and $\Gamma$, respectively. Denote the midpoint of the $\operatorname{arc} C B$ of $\Omega$ containing $A$ by $A_{0}$, and define $B_{0}$ as well as $C_{0}$ analogously. By our hypothesis the centre $Q$ of $\Gamma$ lies on $\Omega$.

Lemma. One has $A_{0} B_{1}=A_{0} C_{1}$. Moreover, the points $A, A_{0}, B_{1}$, and $C_{1}$ are concyclic. Finally, the points $A$ and $A_{0}$ lie on the same side of $B_{1} C_{1}$. Similar statements hold for $B$ and $C$.

Proof. Let us consider the case $A=A_{0}$ first. Then the triangle $A B C$ is isosceles at $A$, which implies $A B_{1}=A C_{1}$ while the remaining assertions of the Lemma are obvious. So let us suppose $A \neq A_{0}$ from now on.

By the definition of $A_{0}$, we have $A_{0} B=A_{0} C$. It is also well known and easy to show that $B C_{1}=$ $C B_{1}$. Next, we have $\angle C_{1} B A_{0}=\angle A B A_{0}=\angle A C A_{0}=\angle B_{1} C A_{0}$. Hence the triangles $A_{0} B C_{1}$ and $A_{0} C B_{1}$ are congruent. This implies $A_{0} C_{1}=A_{0} B_{1}$, establishing the first part of the Lemma. It also follows that $\angle A_{0} C_{1} A=\angle A_{0} B_{1} A$, as these are exterior angles at the corresponding vertices $C_{1}$ and $B_{1}$ of the congruent triangles $A_{0} B C_{1}$ and $A_{0} C B_{1}$. For that reason the points $A, A_{0}, B_{1}$, and $C_{1}$ are indeed the vertices of some cyclic quadrilateral two opposite sides of which are $A A_{0}$ and $B_{1} C_{1}$.

Now we turn to the solution. Evidently the points $A_{1}, B_{1}$, and $C_{1}$ lie interior to some semicircle arc of $\Gamma$, so the triangle $A_{1} B_{1} C_{1}$ is obtuse-angled. Without loss of generality, we will assume that its angle at $B_{1}$ is obtuse. Thus $Q$ and $B_{1}$ lie on different sides of $A_{1} C_{1}$; obviously, the same holds for the points $B$ and $B_{1}$. So, the points $Q$ and $B$ are on the same side of $A_{1} C_{1}$.

Notice that the perpendicular bisector of $A_{1} C_{1}$ intersects $\Omega$ at two points lying on different sides of $A_{1} C_{1}$. By the first statement from the Lemma, both points $B_{0}$ and $Q$ are among these points of intersection; since they share the same side of $A_{1} C_{1}$, they coincide (see Figure 1).

<img_3398>

Figure 1



Now, by the first part of the Lemma again, the lines $Q A_{0}$ and $Q C_{0}$ are the perpendicular bisectors of $B_{1} C_{1}$ and $A_{1} B_{1}$, respectively. Thus

$$
\angle C_{1} B_{0} A_{1}=\angle C_{1} B_{0} B_{1}+\angle B_{1} B_{0} A_{1}=2 \angle A_{0} B_{0} B_{1}+2 \angle B_{1} B_{0} C_{0}=2 \angle A_{0} B_{0} C_{0}=180^{\circ}-\angle A B C,
$$

recalling that $A_{0}$ and $C_{0}$ are the midpoints of the $\operatorname{arcs} C B$ and $B A$, respectively.

On the other hand, by the second part of the Lemma we have

$$
\angle C_{1} B_{0} A_{1}=\angle C_{1} B A_{1}=\angle A B C .
$$

From the last two equalities, we get $\angle A B C=90^{\circ}$, whereby the problem is solved.
Let $Q$ again denote the centre of the circumcircle of the triangle $A_{1} B_{1} C_{1}$, that lies on the circumcircle $\Omega$ of the triangle $A B C$. We first consider the case where $Q$ coincides with one of the vertices of $A B C$, say $Q=B$. Then $B C_{1}=B A_{1}$ and consequently the triangle $A B C$ is isosceles at $B$. Moreover we have $B C_{1}=B_{1} C$ in any triangle, and hence $B B_{1}=B C_{1}=B_{1} C$; similarly, $B B_{1}=B_{1} A$. It follows that $B_{1}$ is the centre of $\Omega$ and that the triangle $A B C$ has a right angle at $B$.

So from now on we may suppose $Q \notin\{A, B, C\}$. We start with the following well known fact. Lemma. Let $X Y Z$ and $X^{\prime} Y^{\prime} Z^{\prime}$ be two triangles with $X Y=X^{\prime} Y^{\prime}$ and $Y Z=Y^{\prime} Z^{\prime}$.

(i) If $X Z \neq X^{\prime} Z^{\prime}$ and $\angle Y Z X=\angle Y^{\prime} Z^{\prime} X^{\prime}$, then $\angle Z X Y+\angle Z^{\prime} X^{\prime} Y^{\prime}=180^{\circ}$.

(ii) If $\angle Y Z X+\angle X^{\prime} Z^{\prime} Y^{\prime}=180^{\circ}$, then $\angle Z X Y=\angle Y^{\prime} X^{\prime} Z^{\prime}$.

Proof. For both parts, we may move the triangle $X Y Z$ through the plane until $Y=Y^{\prime}$ and $Z=Z^{\prime}$. Possibly after reflecting one of the two triangles about $Y Z$, we may also suppose that $X$ and $X^{\prime}$ lie on the same side of $Y Z$ if we are in case $(i)$ and on different sides if we are in case (ii). In both cases, the points $X, Z$, and $X^{\prime}$ are collinear due to the angle condition (see Fig. 2). Moreover we have $X \neq X^{\prime}$, because in case $(i)$ we assumed $X Z \neq X^{\prime} Z^{\prime}$ and in case $(i i)$ these points even lie on different sides of $Y Z$. Thus the triangle $X X^{\prime} Y$ is isosceles at $Y$. The claim now follows by considering the equal angles at its base.

<img_3275>

Figure $2(i)$

<img_3630>

Figure $2($ ii $)$

Relabeling the vertices of the triangle $A B C$ if necessary we may suppose that $Q$ lies in the interior of the arc $A B$ of $\Omega$ not containing $C$. We will sometimes use tacitly that the six triangles $Q B A_{1}, Q A_{1} C, Q C B_{1}, Q B_{1} A, Q C_{1} A$, and $Q B C_{1}$ have the same orientation.

As $Q$ cannot be the circumcentre of the triangle $A B C$, it is impossible that $Q A=Q B=Q C$ and thus we may also suppose that $Q C \neq Q B$. Now the above Lemma $(i)$ is applicable to the triangles $Q B_{1} C$ and $Q C_{1} B$, since $Q B_{1}=Q C_{1}$ and $B_{1} C=C_{1} B$, while $\angle B_{1} C Q=\angle C_{1} B Q$ holds as both angles appear over the same side of the chord $Q A$ in $\Omega$ (see Fig. 3). So we get

$$
\angle C Q B_{1}+\angle B Q C_{1}=180^{\circ}
\tag{1}
$$



We claim that $Q C=Q A$. To see this, let us assume for the sake of a contradiction that $Q C \neq Q A$. Then arguing similarly as before but now with the triangles $Q A_{1} C$ and $Q C_{1} A$ we get

$$
\angle A_{1} Q C+\angle C_{1} Q A=180^{\circ} .
$$

Adding this equation to (1), we get $\angle A_{1} Q B_{1}+\angle B Q A=360^{\circ}$, which is absurd as both summands lie in the interval $\left(0^{\circ}, 180^{\circ}\right)$.

This proves $Q C=Q A$; so the triangles $Q A_{1} C$ and $Q C_{1} A$ are congruent their sides being equal, which in turn yields

$$
\angle A_{1} Q C=\angle C_{1} Q A .
\tag{2}
$$

Finally our Lemma ( $i i$ i) is applicable to the triangles $Q A_{1} B$ and $Q B_{1} A$. Indeed we have $Q A_{1}=Q B_{1}$ and $A_{1} B=B_{1} A$ as usual, and the angle condition $\angle A_{1} B Q+\angle Q A B_{1}=180^{\circ}$ holds as $A$ and $B$ lie on different sides of the chord $Q C$ in $\Omega$. Consequently we have

$$
\angle B Q A_{1}=\angle B_{1} Q A .
\tag{3}
$$

From (1) and (3) we get

$$
\left(\angle B_{1} Q C+\angle B_{1} Q A\right)+\left(\angle C_{1} Q B-\angle B Q A_{1}\right)=180^{\circ},
$$

i.e. $\angle C Q A+\angle A_{1} Q C_{1}=180^{\circ}$. In light of (2) this may be rewritten as $2 \angle C Q A=180^{\circ}$ and as $Q$ lies on $\Omega$ this implies that the triangle $A B C$ has a right angle at $B$.

<img_3157>

Figure 3