Japanese Mathematical Olympiad

Without loss of generality assume $AB \ge AC$. First, since $\angle AED = \angle ADB = 90^\circ$, triangles $AED$ and $ADB$ are similar, so $AE : AD = AD : AB$, that is $AD^2 = AB \cdot AE$. Similarly $AD^2 = AC \cdot AF$, hence $AB \cdot AE = AC \cdot AF$ and points $E$, $B$, $C$, $F$ are concyclic. Now,
$$\angle BAO = \frac{1}{2}(180^\circ - \angle AOB) = 90^\circ - \angle ACB = \angle DAC,$$

so $\angle BAO + \angle AEF = \angle DAC + \angle ACB = 90^\circ$ and lines $AO$ and $EF$ are perpendicular.
Let $M$ be the midpoint of $AD$. Since $\angle AED = \angle AFD = 90^\circ$, $M$ is the circumcenter of triangle $AFE$. Therefore in the similarity between triangles $ABC$ and $AFE$, the points $(A, B, C, O, D)$ correspond to $(A, F, E, M, P)$. In particular triangles $AOD$ and $AMP$ are similar, and since $\angle AOD = 90^\circ$, we have $\angle AMP = 90^\circ$. As $M$ is the midpoint of $AD$, we have $PD = PA = 11$.
From $\angle POD = 90^\circ$ and $OD = 4\sqrt{7}$, the Pythagorean theorem gives $OP = 3$. Also $\angle AOD = \angle AED = \angle AFD = 90^\circ$, so points $A$, $E$, $O$, $D$, $F$ are concyclic. Moreover $EF$ and $OD$ are both perpendicular to $AO$, hence parallel, making $EODF$ an isosceles trapezoid. Let $EP = x$, so that $FP = EP + OD = x + 4\sqrt{7}$. By the power of the point $P$, we have $EP \cdot FP = AP \cdot OP$, i.e. $x(x + 4\sqrt{7}) = 33$, which yields $x = \sqrt{61} - 2\sqrt{7}$. Hence $EF = x + (x + 4\sqrt{7}) = 2\sqrt{61}$.