Problem 1141

Solution. Let the bases of the trapezoid be $AD$ and $BC$, and the diagonals intersect at point $O$. First, consider the more complex case where $\angle COD = 60^\circ$.

Lemma. Suppose a regular triangle $ABK$ is constructed outside a side of an arbitrary triangle $ABC$. Then for any point $P$, the inequality $PA + PB + PC \geq CK$ holds.

Proof of the lemma: Construct a regular triangle $APM$ oriented like $ABK$. Then triangles $AKM$ and $ABP$ are equal by two sides and the angle between them, and $PA + PB + PC = KM + MP + PC \geq CK$.

!

!

Now construct parallelograms $BCED$ and $ABDK$. Triangle $KDE$ is obtained from $ABC$ by translating by the vector $\overrightarrow{BD}$, so $AC = KE$. Due to the parallelism, $\angle CEK = \angle AOB = 60^\circ$, and $\angle ACE = \angle AOD = 120^\circ$.

To apply the lemma, construct a regular triangle $CNE$ outside $CKE$. Triangles $KEN$ and $ACE$ are equal by two sides and the angle $120^\circ$ between them, so $KN = AE$. Using the lemma, we have

$$AB + CD = (DC + DK + DE) - DE \geq KN - DE = AE - DE = AD$$

which is what we needed to prove.

Now consider the simpler case where $\angle AOD = 60^\circ$. We will reduce this case to the previous one using compressions. Specifically, we will shift $BC$ towards $AD$ in the direction perpendicular to $AD$ (downwards in the diagram). As the angles $CAD$ and $CDA$ decrease, $\angle AOD$ increases. We can bring $BC$ to a position where $\angle AOD = 120^\circ$. By the already proven part of the problem, in this case, the sum of the lateral sides will be no less than the base. But the lateral sides decrease during the compression (by the Pythagorean theorem), while the base remains unchanged. Therefore, before the compression, this inequality was even more satisfied.