4. The diagonals of the circumscribed trapezoid $A B C D$ with bases $A D$ and $B C$ intersect at point O. The radii of the inscribed circles of triangles $A O D, A O B, B O C$ are 6, 2, and $3 / 2$ respectively. Find the radius of the inscribed circle of triangle $C O D$.
Answer: 3

Solution. We will prove a more general statement, that $\frac{1}{r_{1}}+\frac{1}{r_{3}}=\frac{1}{r_{2}}+\frac{1}{r_{4}}$, where $r_{1}, r_{2}, r_{3}$ and $r_{4}$ are the radii of the inscribed circles of triangles $A O D, A O B, B O C$ and $C O D$ respectively.

Let $A B=a, B C=b, C D=c, A D=d, O A=x, O D=y, S_{\triangle A O D}=S$. Let $\frac{b}{d}=k$.

Triangle $C O B$ is similar to triangle $A O D$ with a similarity coefficient $k$, so $S_{\triangle C O B}=k^{2} S$, $O B=k y, O C=k x, S_{\triangle A O B}=S_{\triangle C O D}=k S$. Then

$$
r_{1}=\frac{2 S}{x+y+d}, \quad r_{3}=\frac{2 k^{2} S}{k x+k y+b}, \quad r_{2}=\frac{2 k S}{k y+x+a}, \quad r_{4}=\frac{2 k S}{k x+y+c}
$$

Thus,

$$
2\left(\frac{1}{r_{1}}+\frac{1}{r_{3}}\right)=\frac{x+y+d}{S}+\frac{k x+k y+b}{k^{2} S}=\frac{x+y}{S}+\frac{x+y}{k S}+\frac{d k+\frac{b}{k}}{k S}=\frac{x+y}{S}+\frac{x+y}{k S}+\frac{b+d}{k S}
$$

Here we used that $d k=b$ and $\frac{b}{k}=d$ from the similarity of triangles $A O D$ and $B O C$. Further,

$$
2\left(\frac{1}{r_{2}}+\frac{1}{r_{4}}\right)=\frac{k x+y+c}{k S}+\frac{x+k y+a}{k S}=\frac{x+y}{S}+\frac{x+y}{k S}+\frac{a+c}{k S}=\frac{x+y}{S}+\frac{x+y}{k S}+\frac{b+d}{k S}
$$

where $a+c=b+d$, since the trapezoid is circumscribed. Thus, we have proved that $\frac{1}{r_{1}}+\frac{1}{r_{3}}=$ $\frac{1}{r_{2}}+\frac{1}{r_{4}}$.

By the condition, $\frac{1}{6}+\frac{1}{1.5}=\frac{1}{2}+\frac{1}{r_{4}}$, from which $r_{4}=3$.

Criteria. The solution contains a statement that is not common knowledge and which the reader cannot establish on their own in 5 minutes - no more than 3 points.