[ The area of a figure is equal to the sum of the areas of the figures into which it is divided ]

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A circle with its center on side $AC$ of isosceles triangle $ABC$ ($AB = BC$) touches sides $AB$ and $BC$, and divides side $AC$ into three equal parts. Find the radius of the circle if the area of triangle $ABC$ is $9 \sqrt{2}$.
Let a circle of radius $r$ with center $O$ on the base $A C$ of an isosceles triangle $A B C$ touch the lateral sides $A B$ and $A C$ at points $M$ and $N$ respectively and intersect the base $A C$ at points $P$ and $Q$, such that $A P=P Q=Q C$. Then $O M$ and $O N$ are the altitudes of triangles $A O B$ and $C O B$, $A P=P Q=2 r$, $A O=3 r$. By the Pythagorean theorem in the right triangle $A O M$, we find that

$$
A M=\sqrt{A O^{2}-O M^{2}}=\sqrt{9 r^{2}-r^{2}}=2 r \sqrt{2}
$$

$$
M B=\frac{O M^{2}}{A M}=\frac{r^{2}}{2 r \sqrt{2}}=\frac{r \sqrt{2}}{4}, A B=A M+M B=2 r \sqrt{2}+\frac{r \sqrt{2}}{4}=\frac{9 r \sqrt{2}}{4}
$$

$$
9 \sqrt{2}=S_{\triangle A B C}=S_{\triangle A O B}+S_{\triangle C O B}=\frac{1}{2} A B \cdot O M+\frac{1}{2} B C \cdot O N=A B \cdot O M=\frac{9 r \sqrt{2}}{4} \cdot r=\frac{9 r^{2} \sqrt{2}}{4} .
$$

From which we find that $r^2=4$. Therefore, $r=2$.

## Answer

2.00