7. In triangle $A B C$, with an area of $180 \sqrt{3}$, the angle bisector $A D$ and the altitude $A H$ are drawn. A circle with radius $\frac{105 \sqrt{3}}{4}$ and center lying on line $B C$ passes through points $A$ and $D$. Find the radius of the circumcircle of triangle $A B C$, if $B H^{2}-H C^{2}=768$.

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# Solution.

![](https://cdn.mathpix.com/cropped/2024_05_06_555865d131a60f67fb06g-5.jpg?height=860&width=1655&top_left_y=775&top_left_x=135)

Let $B C=a, A C=b, A B=c, \angle B A D=\angle C A D=\alpha, \angle A D C=\beta$, and the radius of the given circle is $r$. Note that $\beta>\alpha$ (as the external angle of triangle $A D C$). From the condition, it follows that $c>b$, so $\angle B\alpha$, then point $C$ lies between $D$ and $O$. Therefore, $\angle C A O=\beta-\alpha$. From this, it follows that triangles $A B O$ and $C A O$ are similar by two angles. Thus, $B O: A O=A O: C O$, or

$$
\frac{r+B D}{r}=\frac{r}{r-C D}
$$

Since $A D$ is the angle bisector of triangle $A B C$, we have

$$
B D=\frac{a c}{b+c}, \quad C D=\frac{a b}{b+c}
$$

Substituting the found expressions into equation (1), we get

$\left(r+\frac{a c}{b+c}\right)\left(r-\frac{a b}{b+c}\right)=r^{2}$, or $r\left(c^{2}-b^{2}\right)=a b c$.

By the cosine theorem, we have

$$
\begin{gathered}
b^{2}=c^{2}+a^{2}-2 a c \cos (A B C), c^{2}=b^{2}+a^{2}-2 a b \cos (A C B) \\
c^{2}-b^{2}=a(c \cos (A B C)-b \cos (A C B))=a(B H-H C)=B C(B H-H C)=768 \\
R=\frac{a b c}{4 S}=\frac{105 \sqrt{3} \cdot 768}{16 \cdot 180 \sqrt{3}}=28
\end{gathered}
$$

## Answer: 28