4. Let $A B C$ be an acute triangle, $\mathcal{K}$ a circle with diameter $A B$, $E$ and $F$ the intersections of the circle with sides $A C$ and $B C$, and $P$ the intersection of the tangents to the circle $\mathcal{K}$ at points $E$ and $F$. Calculate the ratio of the radii of the circumcircles of triangles $A B C$ and $E F P$.
IV/4. Let's denote the lengths of the sides and the sizes of the angles of the triangle as usual. The radius of the circumcircle of triangle $ABC$ is $\frac{c}{2 \sin \gamma}$, and the radius of the other circle is $\frac{|EF|}{2 \sin \angle EPF}$.

Triangle $AOE$ is isosceles with vertex $O$, so $\angle AEO = \alpha$, and the exterior angle $\angle BOE$ is $2\alpha$. Triangle $OBF$ is isosceles with vertex $O$, so $\angle OFB = \beta$, and the exterior angle $\angle FOA$ is $2\beta$. Thus, we get $\angle FOE = 2\alpha + 2\beta - \pi$ and $\angle EPF = \pi - (2\alpha + 2\beta - \pi) = 2(\pi - \alpha - \beta) = 2\gamma$.

Triangle $OEF$ is isosceles, with $|OF| = |OE| = \frac{c}{2}$ and the angle at the vertex $2\alpha + 2\beta - \pi$. Thus, $\frac{|EF|}{2} = \frac{c}{2} \sin \left(\alpha + \beta - \frac{\pi}{2}\right)$, from which we get $|EF| = c \sin \left(\alpha + \beta - \frac{\pi}{2}\right) = -c \cos (\alpha + \beta) = c \cos \gamma$.

Finally, we can calculate the radius of the circumcircle of triangle $EPF$: $\frac{|EF|}{2 \sin \angle EPF} = \frac{c \cos \gamma}{2 \sin (2\gamma)} = \frac{c \cos \gamma}{4 \sin \gamma \cos \gamma} = \frac{c}{4 \sin \gamma}$. The desired ratio is therefore 2.
Writing the radii $\frac{c}{2 \sin \gamma}$ and $\frac{|EF|}{2 \sin \angle EPF}$ ..... 1 point
Angles $\angle FOE = 2\alpha + 2\beta - \pi$ and $\angle EPF = 2\gamma$ ..... 2 points
Writing $|EF| = c \cos \gamma$ ..... 2 points
Radius of the circumcircle of triangle $EPF$ and the calculated ratio ..... 2 points