4. In an acute-angled triangle $A B C$, the altitude $A A_{1}$ is drawn. $H$ is the orthocenter of triangle $A B C$. It is known that $A H=3, A_{1} H=2$, and the radius of the circumcircle of triangle $A B C$ is 4. Find the distance from the center of this circle to $H$.
Solution. Draw the altitudes $B B_{1}$ and $C C_{1}$ in the triangle. Then the quadrilateral $A C_{1} H B_{1}$ is cyclic, since its opposite angles $C_{1}$ and $B_{1}$ are right angles. Therefore: $\angle B H C=$ $\angle C_{1} H B_{1}=180^{\circ}-\angle C_{1} A B_{1}$. Reflect point $H$ symmetrically with respect to the line $B C$ and denote the resulting point by $H_{1}$. Then $\angle B H_{1} C=\angle B H C=180^{\circ}-\angle C_{1} A B_{1}$,
so the quadrilateral $B A C H_{1}$ is cyclic and point $H_{1}$ lies on the circumcircle of triangle $A B C$. Denote the center of this circle by $O$.

By construction, we have $H_{1} A_{1}=H A_{1}=2$, therefore, in the isosceles triangle $O A H_{1}$, the base $A H_{1}$ is $3+2+2=7$, and the lateral side is equal to the radius of the circle: that is, 4. The height dropped to the base is

$$
\sqrt{4^{2}-(7 / 2)^{2}}=\sqrt{15} / 2
$$

and the distance from the base $O_{1}$ of this height to $H$ is $A H_{1} / 2-A H=7 / 2-3=1 / 2$. Therefore,

$$
O H=\sqrt{O O_{1}^{2}+O_{1} H^{2}}=\sqrt{(\sqrt{15} / 2)^{2}+(1 / 2)^{2}}=2
$$

Answer: 2.