The 35th 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 $A{D}^2=AB\cdot AE$. Similarly $A{D}^2=AC\cdot AF$, 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 $EF=x+(x+4\sqrt{7})=2\sqrt{61}$.