Problem 1057

$\begin{array}{l} \text { Proof 1 Draw perpendiculars from } A, D \text { to } B C, \text { only considering the segment between } B C \\ E F \text {, denote its length as } h \text {, and let the six sides of the hexagon } A B C D E F \text { be } a, b, c, d, e, f, \text { and the six interior angles be } A, B, \cdots, F. \text { By assumption, } \angle A=\angle D, \angle B=\angle E, \angle C=\angle F. \text { Clearly, } B F \geqslant h. \text { Thus, } 2 B F \geqslant 2 h. \text { When calculating } h, \text { the two segments between } B C \text { and } E F \text { are divided into two parts by points } A \text { and } D, \text { resulting in four segments, so, } \qquad +d \sin \angle E \end{array}$

In the right-hand side of the above equation, the six parentheses, because $\angle A=\angle D, \angle B=\angle E, \angle C=\angle F$, are each the sum of two positive reciprocals, thus each is greater than or equal to 2. Therefore, $R_{A}+R_{C}+R_{E} \qquad =\frac{1}{2} p$. Proof 2: Construct a parallelogram for $\triangle F A B, \triangle B C D, \triangle D E F$, with $F B, B D, D F$ as diagonals, and the other vertices are denoted as $A^{\prime}, C^{\prime}, E^{\prime}$, i.e., draw a line parallel to $A F$ through $B$, a line parallel to $B C$ through $D$, and a line parallel to $D E$ through $F$. These three lines intersect at $E^{\prime}$. When one pair of parallel sides of the original hexagon is equal, all three pairs must be equal, and the three points $A^{\prime}, C^{\prime}, E^{\prime}$ coincide at a point $Q$. In this case, $Q B, Q D, Q F$ are the lengths of the three sides of the hexagon, so the perimeter of the hexagon is $2(Q B+Q D+Q F)$. On the other hand, the circumradius of $\triangle F A B$ is equal to the circumradius of $\triangle F A^{\prime} B$, which is $\frac{1}{2} A^{\prime} A^{\prime \prime}$. Thus, the three radii are $\frac{1}{2} Q A^{\prime \prime}, \frac{1}{2} Q C^{\prime \prime}, \frac{1}{2} Q E^{\prime \prime}$. The inequality to be proven is the Mordell inequality.

Let the lengths of the shorter sides of the three pairs of parallel sides of the hexagon be denoted as $x, y, z$ for $\triangle A^{\prime} C^{\prime} E^{\prime}$ (in the figure, $A^{\prime} B=x, C^{\prime} D=y, E^{\prime} F=z$). The sides of $\triangle A^{\prime} C^{\prime} E^{\prime}$ are denoted as $a, c, e$. Thus, the other three sides of the hexagon are $x+e, y+a, z+c$. And $p=2 x+2 y+2 z+a+c+2$. By the cosine rule, $\qquad E^{n}$. Therefore, $\qquad \cdot \cos \left(\angle C^{\prime \prime}+\angle E^{\prime \prime}\right)$

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