Problem 16. The answer is (B).

Since the angles at the circumference $\widehat{E F X}$ and $\widehat{G F X}$ are equal, it follows that the arcs $\widehat{E X}$ and $\widehat{G X}$ are congruent. Moreover, $\widehat{Y E}$ and $\widehat{Y G}$ are also congruent by hypothesis; thus $\widehat{Y E}+\overparen{E X}=\widehat{Y G}+\widehat{G X}$ is equal to a semicircle, and $X Y$ is a diameter. It follows that the triangle $X F Y$ is right-angled at $F$, therefore $X Y^{2}=F Y^{2}+F X^{2}=5^{2}+12^{2}=169$; the value of the radius is then $\frac{1}{2} X Y=\frac{13}{2}$.

(Problem proposed by V. Ricciuti)
16. Given a triangle $E F G$, let $\mathcal{C}$ be the circumcircle of the triangle. Denote by $X$ the intersection point (different from $F$) of the angle bisector of $\angle E F G$ with the circle $\mathcal{C}$, and by $Y$ the midpoint of the arc $\widehat{E G}$ containing $F$. Knowing that $\overline{F X}=12$ and $\overline{F Y}=5$, what is the radius of the circle $\mathcal{C}$?
(A) 9
(B) $13 / 2$
(C) 7
(D) 8
(E) $11 / 2$