When the orthocenter of a triangle is reflected over its sides, we obtain a convex hexagon whose angles are $130^{\circ}, 140^{\circ}, 110^{\circ}, 80^{\circ}, 120^{\circ}$, and $140^{\circ}$. Determine the largest angle of the original triangle, given that the area of the hexagon is more than twice the area of the triangle.
It is known that the reflections of the orthocenter of a triangle over its sides lie on the circumcircle. Therefore, the convex hexagon mentioned in the problem is a cyclic hexagon. It is easy to see that if the triangle is acute, then the area of the cyclic hexagon formed by the reflections is exactly twice the area of the triangle. If the triangle is right, then its orthocenter is the right-angle vertex, whose reflections over the legs are itself, so the hexagon mentioned in the problem does not form. Therefore, the original triangle is obtuse. Let the vertices of this triangle be \(A, B, C\), and its orthocenter be \(M\). First, assume that the reflections of \(M\) over the sides forming the obtuse angle are \(M_1\) and \(M_2\), which lie on the segments \(MA\) and \(MB\) respectively (Figure 1). It is clear that the orthocenter of \(\triangle ABM\) is \(C\), and therefore this triangle is acute. In this case, the hexagon formed by the reflections is \(M_3 A M_1 C M_2 B\), and from our previous observation, \(\angle AMB = \angle AM_3B\) is an acute angle, so the 80° angle of the hexagon is at \(M_3\). From the order of the angles, we can see that the 140° angles are at \(M_1\) and \(M_2\). From the cyclic quadrilateral \(AM_3BC\), \(\angle ACB = 100^\circ\), and thus the triangle is isosceles. However, the hexagon, considering that the angles at \(A\) and \(B\) are different, is not symmetric about \(MC\), so there is no such solution.

Next, consider the case where \(M_1\) and \(M_2\) lie on the extensions of \(MA\) beyond \(A\) and \(MB\) beyond \(B\) respectively (Figure 2). The vertices of the hexagon are now in the order \(M_3 M_1 A C B M_2\). According to the diagram, \(\angle M_1AB = 90^\circ + \beta\), so in the hexagon, \(\angle M_1AC = 90^\circ + \alpha + \beta\). Similarly, \(\angle CBM_2 = 90^\circ + \alpha + \beta\). In this hexagon, the angles at \(A\) and \(B\) are equal and are \(140^\circ\), so \(\alpha + \beta = 50^\circ\), and thus \(\angle ACB = 130^\circ\). This is indeed a solution because the broken line \(M_1ACBM_2\) can be completed to the desired hexagon. Reflect \(C\) over \(AB\), and let the reflection be \(C'\). Let \(CC'\) intersect the circle at \(M_3\). From the cyclic quadrilateral \(ACM_3M_1\), \(\angle AM_1M_2 = 90^\circ + \alpha\), so when \(\alpha = 20^\circ\), this angle is \(110^\circ\), and similarly, the angle at \(M_2\) is \(120^\circ\). Then the angle at \(M_3\) can be calculated, which is \(80^\circ\). Since the reflection of \(C\) over \(AB\) is an internal point of \(CM_3\), the area of the hexagon is greater than twice the area of the triangle.

The only remaining case is when \(M_1\) is an internal point of \(MA\) and \(M_2\) is on the extension of \(MB\) beyond \(B\), or vice versa. (Neither \(M_1\) nor \(M_2\) can coincide with \(A\) or \(B\) because then the hexagon would not form.) In this case, \(\angle ABM_2 = 90^\circ + \alpha\), which is an obtuse angle, and thus in the cyclic quadrilateral \(ABM_2M_3\), the angle at \(M_3\) is acute (Figure 3). This angle is therefore the 80° angle of the hexagon, and from the conditions, \(\angle M_1BC = 130^\circ\) and \(\angle AM_1C = 140^\circ\). From the cyclic quadrilateral \(ABCM_1\), \(\beta = 40^\circ\), so \(\angle M_1AM_3 = 100^\circ\), which is a contradiction because the hexagon does not have a 100° angle. Therefore, this configuration does not satisfy our conditions. The answer to the problem is that the largest angle of the triangle is \(130^\circ\).

Based on the work of Koblinger Egmont (Fazekas M. Főv. Gyak. Gimn., III. o.t.)

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