XXIX OM - I - Problem 12

Determine the least upper bound of such numbers $ \alpha \leq \frac{\pi}{2} $, that every acute angle $ MON $ of measure $ \alpha $ and every triangle $ T $ on the plane have the following property. There exists a triangle $ ABC $ isometric to $ T $ such that the side $ \overline{AB} $ is parallel to $ OM $ and the lines perpendicular to $ ON $ and passing through the centroids of triangles $ ABC $ and $ ABC' $ respectively intersect the segment $ \overline{AB} $; $ C' $ is the reflection of vertex $ C $ across the perpendicular bisector of side $ \overline{AB} $.
Note. Physically, the problem means: What should be the angle of inclination of an inclined plane so that for any triangle, one can choose a certain side such that the triangle, placed with this side on the inclined plane, will not tip over.
First, note that if $Q$ is the midpoint of side $\widehat{AC}$, then $\measuredangle ABQ \leq \frac{\pi}{6}$. Indeed, denoting by $D$ the projection of point $A$ onto the line $BQ$, we get $AD \leq AQ = \frac{1}{2} AC \leq \frac{1}{2} AB$ (Fig. 9).
Therefore, $\sin \measuredangle ABQ = \frac{AD}{AB} \leq \frac{1}{2}$. Since angle $ABC$ is not the largest angle of triangle $ABC$, it is acute. Hence, angle $ABQ$ is also acute. From the inequality $\sin \measuredangle ABQ \leq \frac{1}{2}$, we obtain that $\measuredangle ABQ \leq \frac{\pi}{6}$.
Let $\overline{AB}$ be the longest side of triangle $ABC$ isometric to triangle $T$, let $P$ be the centroid of triangle $ABC$, and $P$ be the centroid of triangle $ABC$. We can assume that $A = O$, as this does not change the conditions of the problem (Fig. 10).
om29_1r_img_9.jpgom29_1r_img_10.jpg
From the initial observation, it follows that each of the angles $BOP$ and $BOP$ has a measure not greater than $\frac{\pi}{6}$. Therefore, if $\measuredangle MON \leq \frac{\pi}{3}$, then $\measuredangle NOP = \measuredangle MON + \measuredangle BOP \leq \frac{\pi}{2}$ and similarly $\measuredangle NOP \leq \frac{\pi}{2}$. It follows that the lines perpendicular to $ON$ and passing through points $P$ and $P$ respectively intersect the segment $\overline{AB}$. Therefore, the sought upper bound of the numbers $\alpha$ is not less than $\frac{\pi}{3}$.
Conversely, if the line perpendicular to $ON$ passing through point $P$ intersects the segment $\overline{AB}$, then $\measuredangle NOP \leq \frac{\pi}{2}$. Therefore, $\measuredangle MON = \measuredangle NOP - \measuredangle BOP \leq \frac{\pi}{2} - \measuredangle BOP$. In particular, if $T$ is an equilateral triangle, then $\measuredangle BOP = \frac{\pi}{6}$ and hence
$\measuredangle MON \leq \frac{\pi}{2} - \frac{\pi}{6} = \frac{\pi}{3}$. It follows that the sought upper bound of the numbers $\alpha$ is the number $\frac{\pi}{3}$.