10. Let $A B C$ be an acute triangle and let $D$ be the foot of the angle bisector from $A$. Let $\omega$ be the circle passing through $A$ and tangent to $B C$ at $D$, and let $E, F$ be the intersections of $\omega$ with $A B, A C$ respectively. The tangents to $\omega$ at $E$ and $F$ intersect at $P$. Given that $P E=3$ and the radius of $\omega$ is 4, what is the length of the segment $P D$?
(A) $\frac{3}{4}$
(B) 1
(C) $\frac{1+\sqrt{2}}{2}$
(D) $\sqrt{2}$
(E) The data is insufficient to determine it
10. The answer is (B). Let $O$ be the center of $\omega$; by hypothesis, $\widehat{E A D}=\widehat{F A D}$, so $D$ is the midpoint of the arc $E F$ and in particular $O D$ passes through the midpoint of the segment $E F$. It follows that $O D$ is the perpendicular bisector of $E F$, because it is perpendicular to it and passes through its midpoint. Now, since $P E$ and $P F$ are tangents, we have $P E=P F$ and thus $P$ also lies on the perpendicular bisector of $E F$. This implies that $P, D, O$ are collinear. The triangle $P E O$ is right-angled at $E$ by construction, so we have $P O^{2}=(P D+D O)^{2}=(P D+4)^{2}=E O^{2}+P E^{2}=4^{2}+3^{2}$, from which we obtain $P D=1$.

Note. A mistake in the preparation of the competition led to the omission of the hypothesis " $\triangle ABC$ is acute-angled" in the final version of the text, and thus the correct answer was (E): if the angle at $A$ is obtuse, the segment $P D$ measures 9. However, to avoid disadvantaging those who did not notice this configuration issue, for selection purposes we awarded full marks to both answers (B) and (E), although the former was formally incorrect with the text used on the day of the competition.