Problem 1289

We will show that the given equality is true in acute and right triangles, but not in obtuse triangles.

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First, we will show that in any triangle, the vertices of the triangle $A_{1} B_{1} C_{1}$ - the so-called medial triangle - and $A_{0}$ form the vertices of an isosceles trapezoid if $A_{0} \neq A_{1}$. $B_{1} C_{1}$ is the midline of the triangle parallel to $BC$ (Figure 1), so it is parallel to $A_{0} A_{1}$, and thus perpendicular to $A A_{0}$ and bisects it. Furthermore, $A_{1} B_{1}$ is the midline parallel to $AB$, so $A_{1} B_{1} = AB / 2 = C_{1} A = C_{1} A_{0}$, and $\angle A_{1} B_{1} C_{1} = \angle B_{1} C_{1} A = \angle B_{1} C_{1} A_{0}$; from these, it follows that $A_{0}, A_{1}, B_{1}, C_{1}$ are indeed the vertices of an isosceles trapezoid if $A_{0} \neq A_{1}$, and if $A_{0} = A_{1}$, they form an isosceles triangle.

Accordingly, a circle can be circumscribed around the trapezoid $A_{0} A_{1} B_{1} C_{1}$, which means the circle circumscribed around the medial triangle passes through the foot of the altitude $A A_{0}$. This is also true for the other two feet of the altitudes, so the three feet of the altitudes lie on the circumference of the circle $k$ circumscribed around the medial triangle. 11

If the triangle $ABC$ is acute, the first angle in (1) is $\angle B_{0} A_{1} C_{0} = \angle B_{0} A_{0} C_{0}$, because they are inscribed angles subtending the same arc of $k$, the arc $B_{0} C_{0}$ containing $C_{1}$. By similar reasoning, the left side of (1) can be transformed into:

$$\angle B_{0} A_{0} C_{0} + \angle C_{0} B_{0} A_{0} + \angle A_{0} C_{0} B_{0}$$

which is the sum of the angles of the orthic triangle $A_{0} B_{0} C_{0}$, which is $180^\circ$, so (1) is indeed true.

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If the triangle $ABC$ has a right angle at $C$ (Figure 2), then $A_{0}$ and $B_{0}$ coincide at $C$, so the last angle on the left side of (1) is $0$. On the other hand, $k$ passes through $C$. The first two angles are the interior angles of the cyclic quadrilateral $A_{1} C B_{1} C_{0}$ at the opposite vertices $A_{1}$ and $B_{1}$, so their sum is $180^\circ$, hence (1) is true.

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Now let $ABC$ be an obtuse triangle (Figure 3, $\angle BAC > 90^\circ$):

$$\begin{gathered} \angle B_{0} A_{1} C_{0} + \angle C_{0} B_{1} A_{0} + \angle A_{0} C_{1} B_{0} = \\ \angle B_{0} A_{0} C_{0} + \angle C_{0} B_{0} A_{0} + \angle A_{0} C_{0} B_{0} - 180^\circ \end{gathered}$$

(again using the sum of the angles of the orthic triangle), so (1) is indeed not true.

János Havas (Budapest, Berzsenyi D. g. I. o. t.)

Remark. Equation (1) is also valid for obtuse triangles if every angle appearing in it is considered as a directed angle, a rotation angle less than $180^\circ$, e.g., the angle $\angle B_{0} A_{1} C_{0}$ is the rotation angle less than $180^\circ$ that maps the ray $A_{1} B_{0}$ to the ray $A_{1} C_{0}$ (i.e., we distinguish the first and second sides of the angle). On Figure 3, the rotation $\angle B_{0} A_{1} C_{0}$ is negative, in the direction of the clock's hands, and all other rotations are positive.

If, however, as we have done so far, we consider the angle to be the absolute value of the rotation, then (1) is valid for the triangle in Figure 3, provided we put a minus sign in front of the first term, as evident from the transformation (2).

[^0]: ${ }^{1}$ This circle is known as the Feuerbach circle. It passes through the midpoints of the segments from the vertices to the orthocenter as well. Its center is the midpoint of the segment between the orthocenter and the circumcenter.