In $\triangle ABC$, the altitudes $AA'$, $BB'$, and $CC'$ intersect at point $H$. Let $P$ be the midpoint of segment $AH$. What is the condition for the quadrilateral $A'B'PC'$ to be a parallelogram?

Dr. Feldheim Ervin, Bp.
$A^{\prime} B^{\prime} C^{\prime} \triangle$ is the orthic triangle of $A B C \triangle$. It is known that the altitudes of $A B C \triangle$ are the angle bisectors of the orthic triangle; the angles of the orthic triangle are: $180^{\circ}-2 \alpha, 180^{\circ}-2 \beta, 180^{\circ}-2 \gamma$.

Furthermore, it is known that the Feuerbach circle of the triangle passes through the vertices of the orthic triangle, as well as the midpoints of the segments $A H (B H, C H)$.

![](https://cdn.mathpix.com/cropped/2024_05_02_0c5ee149d998ec5fe84fg-1.jpg?height=499&width=458&top_left_y=344&top_left_x=823)

The requirement is that the quadrilateral $A^{\prime} B^{\prime} P C^{\prime}$ inscribed in the Feuerbach circle be a parallelogram. A cyclic quadrilateral can only be a parallelogram if it is a rectangle! Therefore, it must be that

$$
B^{\prime} A^{\prime} C \varangle=180^{\circ} \quad 2 \alpha=90^{\circ}, \quad \text { or } \quad \alpha=45^{\circ}
$$

In this rectangle, the diagonal $A^{\prime} P$ bisects the angle $B^{\prime} A^{\prime} C^{\prime} \varangle$: from this it follows that the rectangle is a square; the other diagonal $B^{\prime} C^{\prime} \perp P A$, meaning $B^{\prime}$ and $C^{\prime}$ are symmetric with respect to $A A^{\prime}$. This implies that the triangle $A B C \triangle$ is isosceles, where

$$
\beta=\gamma=\frac{180^{\circ}-45^{\circ}}{2}=67^{\circ} 30^{\prime}
$$

Tibor Hoffmann (Szent István g. VI. o. Bp.)