[Theorem of the length of a tangent and a secant; the product of the entire secant and its external part [  Sine Theorem  The midline of a triangle

A circle with radius 3 passes through vertex $B$, the midpoints of sides $A B$ and $B C$, and is tangent to side $A C$ of triangle $A B C$. Angle $B A C$ is acute, and $\sin \angle B A C=\frac{1}{3}$. Find the area of triangle $A B C$.

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Let $M$ and $N$ be the midpoints of sides $BC$ and $AB$ respectively, $K$ be the point of tangency of the circumcircle of triangle $BMN$ with side $AC$, $O$ be the center of the circle, and $R=3$ be its radius. The segment $MN$ is the midline of triangle $ABC$, so $MN \parallel AC$, which means $\angle BNM = \angle BAC$. By the Law of Sines,

$$
BM = 2R \sin \angle BNM = 2R \sin \angle BAC = 2 \cdot 3 \cdot \frac{1}{3} = 2, \quad BC = 2BM = 4
$$

By the tangent-secant theorem,

$$
CK = \sqrt{BC \cdot CM} = \sqrt{4 \cdot 2} = 2\sqrt{2}
$$

Let $P$ be the foot of the perpendicular dropped from the center $O$ of the circle to the line $BC$. Then $P$ is the midpoint of $BM$, so $MP = BP = 1$ and $CP = CM + MP = 2 + 1 = 3$. From the right triangle $BOP$, we find that

$$
OP = \sqrt{OB^2 - BP^2} = \sqrt{3^2 - 1^2} = 2\sqrt{2}
$$

The right triangles $OKC$ and $CRO$ are equal by two legs ($OK = CP = 3$ and $CK = OP = 2\sqrt{2}$), so $\angle COK = \angle OCP$, and since $\angle KCO = 90^\circ - \angle COK$, we have $\angle OCP + \angle KCO = 90^\circ$. Therefore, triangle $ABC$ is a right triangle. Then

$$
AB = \frac{BC}{\sin \angle BAC} = \frac{4}{\frac{1}{3}} = 12, \quad AC = \sqrt{12^2 - 4^2} = 8\sqrt{2}
$$

Thus,

$$
S_{\triangle ABC} = \frac{1}{2} AC \cdot BC = \frac{1}{2} \cdot 8\sqrt{2} \cdot 4 = 16\sqrt{2}
$$

## Answer

$16\sqrt{2}$.

## Problem