Problem 1317

We arrive at a contradiction from the assumption that the diagonals $A D, B E, C F$ are all greater than 2.

Let $A D > 2$, and draw a unit circle $k$ around $A$. The adjacent vertices $B, F$ are inside or on the circumference of $k$, so the angle subtended by the segment $B F$ from $D$ cannot be greater than the angle between the tangents $T_{1} D T_{2}$ drawn from $D$ to $k$, where $T_{1}, T_{2}$ are the points of tangency.

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Figure 1

We will show that

$$\angle B D F \leq \angle T_{1} D T_{2}$$

and $A A' \leq A T_{2} + T_{2} A' = A T_{2} + T_{1} A = 2 < A D$, so the angle at $D$ in the triangle is smaller than the other two, which are equal, and thus smaller than $60^\circ$. However,

$$\angle A D A' = \angle A D T_{2} + \angle T_{2} D A' = \angle A D T_{2} + \angle T_{1} D A = \angle T_{1} D T_{2},$$

which proves our statement.

If now $A D, E B$, and $C F$ were all greater than 2 - or twice the length of any side of our convex hexagon - then by assigning the role of $D$ to $B$ and $F$ in turn - and simultaneously the role of $A$ to the opposite vertex - we would find that the sum of the angles in the triangle $B D F$ is less than $180^\circ$, since each angle is individually less than $60^\circ$.

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Figure 2

This contradiction shows the correctness of the problem statement.

Gábor Hetyei (Pécs, Leőwey K. Gymnasium, 2nd grade)

Remarks. 1. A pleasant idea is to extend every second side of the hexagon as shown in Figure 3, thus embedding the figure in a triangle, which many solutions started from.

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By selecting the smallest angle of the triangle - or one of them if there are more, in the diagram $\angle A H F = \alpha$ is such an angle - we have $\alpha \leq 60^\circ$. Then the diagonal $B E$ that subtends both sides of the angle $\alpha$ is not longer than 2 units.

Reflect $F$ over the midpoint of segment $A E$, let the image be $G$. Thus, on the one hand, the quadrilateral $A F E G$ is a parallelogram, and on the other hand, $\angle G A B = \angle F H A = \alpha \leq 60^\circ$, so $E G = F A \leq 1$, and among the sides opposite to $\alpha$, the one that is larger than $\alpha$ is $G B$; possibly $A G = A B = G B$. Therefore, $\bar{G} B \leq 1$, and from the triangle $G B E$, $E B \leq E G + G B \leq 2$. Equality can occur if the triangle $G B E$ is degenerate, and simultaneously the two sides of the hexagon used are exactly 1 unit long. - This proves the statement for the hexagon in Figure 3.

The reader can easily extend the above to hexagons where neither the extensions in Figure 3 nor the use of the side triplet $B C, D E, F A$ form an enclosing triangle. If, for example, the angle of rotation that maps the direction of the ray $A B$ to the direction of the ray $C D$ (considered through the direction of $B C$) is exactly $180^\circ$ or more, then two sides of the above triangle are parallel, or the side $E F$ lies outside the hexagon, touching one side of the resulting triangle. And simultaneously, it can happen that the direction of the ray $B C$ is rotated at least $180^\circ$ to the direction of the ray $D E$ (Figure 4). - From these two "failures," the statement can be proven for one of the diagonals in question. (That is, to "patch" the previously incomplete proof.)

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2. In competition committees, it is often said: incomplete, but easily completable to completeness. However, this little bit - as exemplified by the previous remark - is only added to the contestant's work if there is a sign that the contestant felt: something is still missing. In other words, if they did not consider the proof complete in its incomplete state.
3. Another lesson from Figure 3: don't just draw one diagram, and not just a slightly different one from something regular, because such is not humility, but a lack of thoroughness!