Problem 1301

Let the center of the incircle and the circumcircle of the original triangle be denoted by $K^{*}$ and $L^{*}$, respectively, and the centroid of the triangle by $S$.

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1. Figure

We will prove that (1. Figure)
a) $L^{*}$ bisects the segment $K^{*} L$,
b) $S$ is the point of the segment $L^{*} M$ that is closer to $L^{*}$ and divides it in the ratio 1:2,
c) $S$ is the point of the segment $K K^{*}$ that is closer to $K$ and divides it in the ratio 1:2.

According to statements b) and c), $K L^{*} \| K^{*} M$ and $K L^{*} = K^{*} M / 2$, since the triangles $K L^{*} S$ and $K^{*} M S$ are similar, and the ratio of their corresponding sides is $K S : S K^{*} = 1 : 2$ (2. Figure).

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2. Figure

Accordingly, the lines $K^{*} L^{*}$ and $M K$ intersect, and for their intersection point $L^{\prime}$, it holds that $L^{*}$ bisects the segment $L^{\prime} K^{*}$ and $K$ bisects the segment $L^{\prime} M$. Since the first property can only hold for the reflection of $K^{*}$ over $L^{*}$, and according to a), $L$ has this property, $L$ is identical to $L^{\prime}$, so $K$ bisects the segment $L M$, which is what we wanted to prove.

Therefore, it is sufficient to prove our statements a), b), and c). For this, we will use the following two statements (3-4. Figures):

I. In a triangle $H_{0} = A_{0} B_{0} C_{0}$, the projections of the vertices on the opposite sides are $A_{1}, B_{1}, C_{1}$, and the midpoints of the segments $B_{0} C_{0}$, $C_{0} A_{0}$, $A_{0} B_{0}$ are $A_{2}, B_{2}, C_{2}$, respectively. These points lie on a circle $k_{0}$, whose center $F_{0}$ is the midpoint of the segment $L_{0} M_{0}$, where $M_{0}$ is the orthocenter of $H_{0}$, $L_{0}$ is the circumcenter of $H_{0}$, and $S_{0}$ is the centroid of $H_{0}$.

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II. The centroid $S_{0}$ of the triangle $H_{0}$ is the point of the segment $L_{0} M_{0}$ that is closer to $L_{0}$ and divides it in the ratio 1:2.

The circle $k_{0}$ mentioned in statement I is called the Feuerbach circle of the triangle $H_{0}$, and the line passing through the points $M_{0}, S_{0}, L_{0}$ is called the Euler line of the triangle. Although proofs of these statements can be found in many places, it is worth repeating the simple proof here.

I. Let the midpoints of the segments $A_{0} M_{0}$, $B_{0} M_{0}$, $C_{0} M_{0}$ be denoted by $A_{3}$, $B_{3}$, $C_{3}$, respectively. The segment $A_{3} C_{2}$ is the midline of the triangle $A_{0} B_{0} M_{0}$, so it is half the length of the segment $M_{0} B_{0}$ and parallel to it. Similarly, we find that $C_{3} A_{2}$ is also half the length of $M_{0} B_{0}$, and $A_{3} C_{3}$ and $C_{2} A_{2}$ are half the length of $A_{0} C_{0}$. Since $M_{0} B_{0}$ is perpendicular to $A_{0} C_{0}$, it follows that the quadrilateral $A_{2} C_{3} A_{3} C_{2}$ is a rectangle, so the Thales circles over the segments $A_{2} A_{3}$ and $C_{2} C_{3}$ are identical. Similarly, we can show that the Thales circles over the segments $A_{2} A_{3}$ and $B_{2} B_{3}$ are identical, so there is a circle $k_{0}$ that passes through all the points $A_{2}, B_{2}, C_{2}, A_{3}, B_{3}, C_{3}$, and let its center be $F_{0}$.

Since the segment $A_{2} A_{3}$ is seen from $A_{1}$ at a right angle, $A_{1}$ is also on $k_{0}$, and similarly, $B_{1}$ and $C_{1}$ are also on it. Since $F_{0}$ is the midpoint of the segment $C_{2} C_{3}$, and the lines $C_{0} C_{1}$ and $C_{2} L_{0}$ are parallel, the reflection of the point $L_{0}$ over $F_{0}$ lies on the line $C_{0} C_{1}$. Similarly, we find that this reflection lies on the altitudes $A_{0} A_{1}$ and $B_{0} B_{1}$, so this reflection is the orthocenter $M_{0}$. Therefore, $F_{0}$ indeed bisects the segment $L_{0} M_{0}$.

II. The central similarity with center $S_{0}$ and ratio $(-1 / 2)$ maps the triangle $A_{0} B_{0} C_{0}$ to the triangle $A_{2} B_{2} C_{2}$, so it maps the circumcircle of $A_{0} B_{0} C_{0}$ to the circumcircle of $A_{2} B_{2} C_{2}$ and the center $L_{0}$ of the first to the center $F_{0}$ of the second. Therefore, $S_{0}$ is the point of the segment $F_{0} L_{0}$ that is closer to $F_{0}$ and divides it in the ratio 1:2. Since $F_{0}$ bisects the segment $L_{0} M_{0}$, it follows that $S_{0}$ is the point of the segment $L_{0} M_{0}$ that is closer to $L_{0}$ and divides it in the ratio 1:2.

We now proceed to prove the statements a), b), and c) mentioned earlier. Let the original triangle be denoted by $H$, the triangle formed by the centers of the excircles of $H$ by $H_{1}$, and the medial triangle of $H$ by $H_{2}$. Since the centers of the excircles of $H$ are the intersections of the external and internal angle bisectors of $H$, the sides of $H_{1}$ are the external angle bisectors of $H$, and the internal angle bisectors of $H$ are the altitudes of $H_{1}$. Therefore, $K^{*}$ is the orthocenter of $H_{1}$, $H$ is the orthic triangle of $H_{1}$, and the circumcircle of $H$ is the Feuerbach circle of $H_{1}$. According to statement I, $L^{*}$ bisects the segment $K^{*} L$, as stated in a). Since the points $M, S, L^{*}$ are the orthocenter, centroid, and circumcenter of $H$, respectively, applying statement II to $H$ gives us statement b). The central similarity with center $S$ and ratio $(-1 / 2)$ maps $H$ to $H_{2}$ and $K^{*}$ to $K$, so $S$ is the point of the segment $K K^{*}$ that is closer to $K$ and divides it in the ratio 1:2. This completes the proof of statement c), and we have reached the end of our solution.