Subfield: Geometry


Question: Let $A B C$ be a triangle with circumcircle $\Omega$. Let $S_{b}$ and $S_{c}$ respectively denote the midpoints of the $\operatorname{arcs} A C$ and $A B$ that do not contain the third vertex. Let $N_{a}$ denote the midpoint of arc $B A C$ (the arc $B C$ containing $A$ ). Let $I$ be the incentre of $A B C$. Let $\omega_{b}$ be the circle that is tangent to $A B$ and internally tangent to $\Omega$ at $S_{b}$, and let $\omega_{c}$ be the circle that is tangent to $A C$ and internally tangent to $\Omega$ at $S_{c}$. Show that the line $I N_{a}$, and the line through the intersections of $\omega_{b}$ and $\omega_{c}$, meet on $\Omega$.

The incentre of a triangle is the centre of its incircle, the circle inside the triangle that is tangent to all three sides.

Solution:
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Part I: First we show that A lies on the radical axis of $\omega_{b}$ and $\omega_{c}$.

We first note that the line through the intersections of two circles is the radical line of the two circles. Let the tangents to $\Omega$ at $S_{b}$ and $S_{c}$ intersect at $T$. Clearly $T$ is on the radical axis of $\omega_{b}$ and $\omega_{c}$ (and in fact is the radical centre of $\omega_{b}, \omega_{c}$ and $\Omega$ ).

We next show that $A$ lies on the radical axis of $\omega_{b}$ and $\omega_{c}$. Let $P_{b}$ denote the point of tangency of $\omega_{b}$ and $A B$, and let $P_{c}$ denote the point of tangency of $\omega_{c}$ and $A C$. Furthermore, let $U$ be the intersection of the tangent to $\Omega$ at $S_{b}$ with the line $A B$, and let $V$ be the intersection of the tangent to $\Omega$ at $S_{c}$ with the line $A C$. Then $T V A U$ is parallelogram. Morover, due to equality of tangent segments we have $\left|U S_{b}\right|=\left|U P_{b}\right|,\left|V P_{c}\right|=\left|V S_{c}\right|$ and $\left|T S_{b}\right|=\left|T S_{c}\right|$. It follows that

$$
\begin{aligned}
\left|A P_{b}\right| & =\left|U P_{b}\right|-|U A|=\left|U S_{b}\right|-|T V|=\left|T S_{b}\right|-|T U|-|T V| \\
& =\left|T S_{s}\right|-|T V|-|T U|=\left|V S_{c}\right|-|A V|=\left|V P_{c}\right|-|V A|=\left|A P_{c}\right| .
\end{aligned} \tag{11}
$$



But $\left|A P_{b}\right|,\left|A P_{c}\right|$ are exactly the square roots of powers of $A$ with respect to $\omega_{b}$ and $\omega_{c}$, hence $A$ is indeed on their radical axis.

Thus, the radical axis of $\omega_{b}, \omega_{c}$ is $A T$.

Part II: Consider the triangle $A S_{b} S_{c}$. Note that since $T$ is the intersection of the tangents at $S_{b}$ and $S_{c}$ to the circumcircle of $A S_{b} S_{c}$, it follows that $A T$ is the symmedian of $A$ in this triangle. Let $X$ denote the second intersection of the symmedian $A T$ with $\Omega$. We wish to show that $X$ is also on $I N_{a}$.

Note that $A N_{a}$ is the external angle bisector of angle $A$, and therefore it is parallel to $S_{b} S_{c}$. Let $M$ denote the midpoint of $S_{b} S_{c}$, and let $Y$ be the second intersection of $A M$ with $\Omega$. Since in $A S_{b} S_{c}, A X T$ is the symmedian and $A M Y$ is the median, it follows that $X Y$ is also parallel to $S_{b} S_{c}$. Thus, reflecting in the perpendicular bisector of $S_{b} S_{c}$ sends the line $A M Y$ to line $N_{a} M X$.

Next, consider the quadrilateral $A S_{b} I S_{c}$. From the trillium theorem we have $\left|S_{b} A\right|=\left|S_{b} I\right|$ and $\left|S_{c} A\right|=\left|S_{c} I\right|$, thus the quadrilateral is a kite, from which it follows that the reflection of the line $A M$ in $S_{b} S_{c}$ is the line $I M$. But previously we have seen that this is also the line $N_{a} M X$. Thus $M, I, N_{a}$ and $X$ are collinear, as we wanted to show.
<img_3378>

Part I: First we show that A lies on the radical axis of $\omega_{b}$ and $\omega_{c}$.

We first note that the line through the intersections of two circles is the radical line of the two circles. Let the tangents to $\Omega$ at $S_{b}$ and $S_{c}$ intersect at $T$. Clearly $T$ is on the radical axis of $\omega_{b}$ and $\omega_{c}$ (and in fact is the radical centre of $\omega_{b}, \omega_{c}$ and $\Omega$ ).

We next show that $A$ lies on the radical axis of $\omega_{b}$ and $\omega_{c}$. Let $P_{b}$ denote the point of tangency of $\omega_{b}$ and $A B$, and let $P_{c}$ denote the point of tangency of $\omega_{c}$ and $A C$. Furthermore, let $U$ be the intersection of the tangent to $\Omega$ at $S_{b}$ with the line $A B$, and let $V$ be the intersection of the tangent to $\Omega$ at $S_{c}$ with the line $A C$. Then $T V A U$ is parallelogram. Morover, due to equality of tangent segments we have $\left|U S_{b}\right|=\left|U P_{b}\right|,\left|V P_{c}\right|=\left|V S_{c}\right|$ and $\left|T S_{b}\right|=\left|T S_{c}\right|$. It follows that

$$
\begin{aligned}
\left|A P_{b}\right| & =\left|U P_{b}\right|-|U A|=\left|U S_{b}\right|-|T V|=\left|T S_{b}\right|-|T U|-|T V| \\
& =\left|T S_{s}\right|-|T V|-|T U|=\left|V S_{c}\right|-|A V|=\left|V P_{c}\right|-|V A|=\left|A P_{c}\right| .
\end{aligned} \tag{11}
$$



But $\left|A P_{b}\right|,\left|A P_{c}\right|$ are exactly the square roots of powers of $A$ with respect to $\omega_{b}$ and $\omega_{c}$, hence $A$ is indeed on their radical axis.

Thus, the radical axis of $\omega_{b}, \omega_{c}$ is $A T$.

We begin by showing that the radical axis of $\omega_{b}, \omega_{c}$ is $A T$ as above.

Part II: We introduce the point $S_{a}$ with the obvious meaning. Observe that the incentre $I$ of $A B C$ is the orthocentre of $S_{a} S_{b} S_{c}$ either because this is well-known, or because of an angle argument that $A$ reflects in $S_{b} S_{c}$ to $I$ (and similar results by cyclic change of letters). Therefore $A S_{a}$ is perpendicular to $S_{b} S_{c}$.

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Figure 7: A reflections argument for Solution

Let $M$ denote the midpoint of $S_{b} S_{c}$. Then $A$ is the reflection of $S_{a}$ in the diameter parallel to $S_{b} S_{c}$, so the reflection of $A$ in the diameter perpendicular to $S_{b} S_{c}$ is $N_{a}$, the antipode of $S_{a}$. Let the reflection of $X$ in $T M$ be $Y$, so $T Y$ passes through $N_{a}$ and is the reflection of $T X$ in $T M$.

Now $S_{b} S_{c}$ is the polar line of $T$ with respect to $\Omega$, so $A Y$ and $N_{a} X$ meet on this line, and by symmetry at its midpoint $M$. The line $N_{a} M X$ is therefore the reflection of the line $Y M A$ in $S_{b} S_{c}$, and so $N_{a} M X$ passes through $I$ (the reflection of $A$ in $S_{b} S_{c}$ ).

The triangle $A S_{c} S_{b}$ can be taken as generic, and from the argument above we can extract the fact that the symmedian point and the centroid are isogonal conjugates in that triangle.
Part I: First we show that A lies on the radical axis of $\omega_{b}$ and $\omega_{c}$.

We first note that the line through the intersections of two circles is the radical line of the two circles. Let the tangents to $\Omega$ at $S_{b}$ and $S_{c}$ intersect at $T$. Clearly $T$ is on the radical axis of $\omega_{b}$ and $\omega_{c}$ (and in fact is the radical centre of $\omega_{b}, \omega_{c}$ and $\Omega$ ).

We next show that $A$ lies on the radical axis of $\omega_{b}$ and $\omega_{c}$. Let $P_{b}$ denote the point of tangency of $\omega_{b}$ and $A B$, and let $P_{c}$ denote the point of tangency of $\omega_{c}$ and $A C$. Furthermore, let $U$ be the intersection of the tangent to $\Omega$ at $S_{b}$ with the line $A B$, and let $V$ be the intersection of the tangent to $\Omega$ at $S_{c}$ with the line $A C$. Then $T V A U$ is parallelogram. Morover, due to equality of tangent segments we have $\left|U S_{b}\right|=\left|U P_{b}\right|,\left|V P_{c}\right|=\left|V S_{c}\right|$ and $\left|T S_{b}\right|=\left|T S_{c}\right|$. It follows that

$$
\begin{aligned}
\left|A P_{b}\right| & =\left|U P_{b}\right|-|U A|=\left|U S_{b}\right|-|T V|=\left|T S_{b}\right|-|T U|-|T V| \\
& =\left|T S_{s}\right|-|T V|-|T U|=\left|V S_{c}\right|-|A V|=\left|V P_{c}\right|-|V A|=\left|A P_{c}\right| .
\end{aligned} \tag{11}
$$



But $\left|A P_{b}\right|,\left|A P_{c}\right|$ are exactly the square roots of powers of $A$ with respect to $\omega_{b}$ and $\omega_{c}$, hence $A$ is indeed on their radical axis.

Thus, the radical axis of $\omega_{b}, \omega_{c}$ is $A T$.

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Figure 8: Diagram to Solution

Part II: By the trillelium theorem, $S_{c} S_{b}$ bisects $A I$, and since $N_{a} A \| S_{b} S_{c}$, then $O T$ is a bisector of $A N_{a}$. This implies $\left|M N_{a}\right|=|M A|=|M I|$, since $M$ is the midpoint of $S_{c} S_{b}$ and lies also on $O T$. Hence, $M$ is the circumcentre of triangle $I A N_{a}$. But this triangle has a right angle at $A$ (since $A I$ and $A N_{a}$ are the inner and outer angle bisector at $A$ ), hence $M$ lies on $I N_{a}$.

Again, let $X$ be the second intersection of $T A$ and $\Omega$. By the above, it suffices to prove that $X$ lies on the line $N_{a} M$. From the power of point $T$ with respect to $\Omega$ we get $|T A| \cdot|T X|=\left|T S_{c}\right|^{2}$. Since $M$ is the foot of the altitude of right triangle $T S_{c} O$, we obtain $\left|T S_{c}\right|^{2}=|T M| \cdot|T O|$. Hence, $|T A| \cdot|T X|=|T M| \cdot|T O|$ so the points $O, M, A, X$ are concyclic. It follows that $\angle M X A=\angle M O A=\frac{1}{2} \angle N_{a} O A=\angle N_{a} X A$. Hence, $X$ lies on the line $N_{a} M$.

Remark. To show that $O M A X$ is cyclic, one can also invert the line $T A X$ in the circumcircle of the triangle $A B C$.

Solution 4. 
Part I: First we show that A lies on the radical axis of $\omega_{b}$ and $\omega_{c}$.

We first note that the line through the intersections of two circles is the radical line of the two circles. Let the tangents to $\Omega$ at $S_{b}$ and $S_{c}$ intersect at $T$. Clearly $T$ is on the radical axis of $\omega_{b}$ and $\omega_{c}$ (and in fact is the radical centre of $\omega_{b}, \omega_{c}$ and $\Omega$ ).

We next show that $A$ lies on the radical axis of $\omega_{b}$ and $\omega_{c}$. Let $P_{b}$ denote the point of tangency of $\omega_{b}$ and $A B$, and let $P_{c}$ denote the point of tangency of $\omega_{c}$ and $A C$. Furthermore, let $U$ be the intersection of the tangent to $\Omega$ at $S_{b}$ with the line $A B$, and let $V$ be the intersection of the tangent to $\Omega$ at $S_{c}$ with the line $A C$. Then $T V A U$ is parallelogram. Morover, due to equality of tangent segments we have $\left|U S_{b}\right|=\left|U P_{b}\right|,\left|V P_{c}\right|=\left|V S_{c}\right|$ and $\left|T S_{b}\right|=\left|T S_{c}\right|$. It follows that

$$
\begin{aligned}
\left|A P_{b}\right| & =\left|U P_{b}\right|-|U A|=\left|U S_{b}\right|-|T V|=\left|T S_{b}\right|-|T U|-|T V| \\
& =\left|T S_{s}\right|-|T V|-|T U|=\left|V S_{c}\right|-|A V|=\left|V P_{c}\right|-|V A|=\left|A P_{c}\right| .
\end{aligned} \tag{11}
$$



But $\left|A P_{b}\right|,\left|A P_{c}\right|$ are exactly the square roots of powers of $A$ with respect to $\omega_{b}$ and $\omega_{c}$, hence $A$ is indeed on their radical axis.

Thus, the radical axis of $\omega_{b}, \omega_{c}$ is $A T$.

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Figure 9: Diagram to Solution

Part II: We show that $A N_{a} \| S_{b} S_{c}$. In particular, $\angle N_{a} O T=\angle T O A$. The conclusion of the problem trivially holds if $|A B|=|A C|$, therefore we assume without loss of generality that $|A C|>|A B|$. Let $S_{a}$ be the midpoint of the $\operatorname{arc} B C$ which does not contain $A$. Then $N_{a} S_{a}$ is a diameter, so $\angle S_{a} B N_{a}=\frac{\pi}{2}=\angle O S_{c} T$. We also compute $\angle B N_{a} S_{a}=\angle B A S_{a}=\frac{1}{2} \angle B A C=\frac{1}{2} \angle S_{c} T S_{b}=\angle S_{c} T O$. It follows that the triangles $T S_{c} O$ and $N_{a} B S_{a}$ are similar. In particular,

$$
\frac{\left|N_{a} B\right|}{\left|T S_{c}\right|}=\frac{\left|S_{a} B\right|}{\left|O S_{c}\right|} \tag{12}
$$

Next we compute

$$
\angle I S_{a} B=\angle N_{a} S_{a} B-\angle N_{a} S_{a} I=\angle T O S_{c}-\frac{1}{2} \angle N_{a} O A=\angle T O S_{c}-\angle T O A=\angle A O S_{c} \tag{13}
$$



and

$$
\angle I B N_{a}=\angle C B N_{a}-\angle C B I=\frac{1}{2}\left(\pi-\angle B N_{a} C\right)-\frac{1}{2} \angle C B A=\frac{1}{2} \angle A C B=\angle A C S_{c}=\angle A S_{c} T \tag{14}
$$,

hence

$$
\angle S_{a} B I=\frac{\pi}{2}-\angle I B N_{a}=\frac{\pi}{2}-\angle A S_{c} T=\angle O S_{c} A .
$$

Together with (13) it follows that the triangles $I B S_{a}$ and $A S_{c} O$ are similar, so $\frac{\left|S_{a} B\right|}{\left|O S_{c}\right|}=$ $\frac{|I B|}{\left|A S_{c}\right|}$, and (12) implies $\frac{\left|N_{a} B\right|}{\left|T S_{c}\right|}=\frac{|I B|}{\left|A S_{c}\right|}$. Consequently, by (14) the triangles $T S_{c} A$ and $N_{a} B I$ are similar and therefore $\angle S_{c} T A=\angle B N_{a} I$. Now let $Q$ be the second intersection of $N_{a} I$ with $\Omega$. Then $\angle B N_{a} I=\angle B N_{a} Q=\angle B A Q$, so $\angle S_{c} T A=\angle B A Q$. Since $A B$ is parallel to $T S_{c}$, we get $A Q \| T A$, i.e. $A, T, Q$ are collinear.

Remark. After proving similarity of triangles $T S_{c} O$ and $N_{a} B S_{a}$ one can use spiral symmetry to show similarity of triangles $T S_{c} A$ and $N_{a} B I$.
Part I: First we show that $A$ lies on the radical axis between $\omega_{b}$ and $\omega_{c}$.

Let $T$ be the radical center of the circumcircle, $\omega_{b}$ and $\omega_{c}$; then $T S_{b}$ and $T S_{c}$ are common tangents of the circles, as shown in Figure 5a. Moreover, let $P_{b}=A B \cap S_{b} S_{c}$ and $P_{c}=$ $A C \cap S_{b} S_{c}$. The triangle $T S_{c} S_{b}$ is isosceles: $A B \| T S_{c}$ and $A C \| T S_{b}$ so

$$
\angle A P_{b} P_{c}=\angle T S_{c} S_{b}=\angle S_{c} S_{b} T=\angle P_{b} P_{c} A .
$$

From these angles we can see that $\omega_{b}$ passes through $P_{b}, \omega_{c}$ passes through $P_{c}$, and finally $A P_{b}$ and $A P_{c}$ are equal tangents to $\omega_{b}$ and $\omega_{c}$, so $A$ lies on the radical axis.

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Figure 5a

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Figure $5 b$



Part II. Let the radical axis $T A$ meet the circumcircle again at $X$, let $S_{a}$ be the midpoint of the $\operatorname{arc} B C$ opposite to $A$, and let $X I$ meet the circumcirlce again at $N$. (See Figure 2.) For solving the problem, we have prove that $N_{a}=N$.

The triples of points $A, I, S_{a} ; B, I, S_{b}$ and $C, I, S_{c}$ are collinear because they lie on the angle bisectors of the triangle $A B C$.

Notice that the quadrilateral $A S_{c} X S_{b}$ is harmonic, because the tangents at $S_{b}$ and $S_{c}$, and the line $A X$ are concurrent at $T$. This quadrilateral can be projected (or inverted) to the quadrilateral $S_{a} C N B$ through $I$. So, $S_{a} C N B$ also is a harmonic quadrilateral. Due to $S_{a} B=S_{a} C$, this implies $N B=N C$, so $N=N_{a}$. Done.
Part I: First let's show that this is equivalent to proving that $T A$ and $N_{a} I$ intersect in $\Omega$.

Lemma: Let's recall that if we have two circles $\omega_{1}$ and $\omega_{2}$ which are internally tangent at point $X$ and if we have a line $A B$ tangent to $\omega_{2}$ at $Y$. Let $M$ be the midpoint of the arc $A B$ not containing $Z$. We have that $Z, Y, M$ are collinear.

<img_3609>



Let $P_{b}=A B \cap \omega_{b}$ and $P_{c}=A C \cap \omega_{c}$. We can notice by the lemma that $S_{b}, P_{b}$ and $S_{c}$ are collinear, and similarly $S_{c}, P_{c}$ and $S_{b}$ are also collinear. Therefore $S_{c}, P_{b}, P_{c}$, and $S_{b}$ are collinear, and since $\angle A P_{b} P_{c}=\frac{\angle A B C}{2}+\frac{\angle A C B}{2}=\angle A P_{c} P_{b}$ then $A P_{b}=A P_{c}$ so $A$ is on the radical axis of $\omega_{b}$ and $\omega_{c}$. Let $T$ be the intersection of the tangent lines of $\Omega$ through $S_{c}$ and $S_{b}$. Since $T S_{c}=T S_{b}$ then $A T$ is the radical axis between $\omega_{b}$ and $\omega_{c}$.

Part II: $T A$ and $N_{a} I$ intersect in $\Omega$.



Let $\omega_{a}$ the $A$-mixtilinear incircle (that is, the circle internally tangent to $\Omega$, and tangent to both $A B$ and $A C$ ), and let $X=\Omega \cap \omega_{a}$. It is known that $N_{a}, I, X$ are collinear.

Let $M_{c}$ and $M_{b}$ be the tangent points of $\omega_{A}$ to $A B$ and $A C$ respectively, then by the lemma $X, M_{c}, S_{c}$ are collinear and $X, M_{b}, S_{b}$ are collinear. We can see that $S_{c} T S_{b}$ and $M_{c} A M_{b}$ are homothetic with respect to $X$; therefore $T$ and $A$ are homothetic with respect to $X$, impying that $T, A, X$ are collinear.

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Figure 10: Diagram to Solution