We show that a such figure phi must be a disk (in advance sorry, I don’t know how to use LaTeX and math symbols so I hope my solution will still be clear haha) First of all, any disk clearly satisfies the condition (for any two points A and B we can choose the semicircle lying in the same side of AB as the center). Now let’s prove that it’s a necessary condition. Since phi is supposed to be bounded and closed, we can take the longest distance AB between two points in phi. By the hypothesis we know that phi contains the semicircle gamma whose diameter is AB. Moreover, if l and m are the lines perpendicular to AB passing through A and B respectively, we know that phi is contained between l and m (if a point P of phi was WLOG on the other side of l, then we would have that ABP has an obtuse angle in A, so BP > BA, contradiction). Now we prove the following claim : Claim 1. The whole disk formed by gamma (of diameter AB) is fully contained in phi. Proof : First we show the claim for the points strictly inside the disk. Suppose FTSOC that such a point Q is not in phi. We can clearly see that the line perpendicular to BQ through Q intersects gamma at least once. If Q’ (that is different from A) is the searched intersection, then B and Q’ should be connectable by a semicircle in phi. But the semicircle not containing Q has a part which is on the other side of m (checkable by some angle properties) ! Thus, this is a contradiction. Finally, we can show that the boundary is also included in phi by continuity, and closeness of the figure. We can finish the problem by noticing that, if R was a point in phi outside the disk, then we could pick the line passing through R and the center of the circle, and consider the intersection S which lies the furthest from R. Then the length of RS would be strictly greater then AB, we would have again a contradiction, meaning that phi is exactly the disk. CQFD