Problem 1602

To prove the inequality $\text{Area}(ABCDEF) \le \frac{R}{r} \cdot \text{Area}(ACE)$, we will use the properties of the circumradius $R$ and inradius $r$ of the triangles $ACE$ and $BDF$.

1. Circumradius and Inradius Relationship:
- Given that both triangles $ACE$ and $BDF$ have the same circumradius $R$.
- Let the inradius of triangle $ACE$ be $r$.

2. **Area of Triangle $ACE$**:
- The area of triangle $ACE$ can be expressed using the formula involving the inradius $r$ and the semiperimeter $s$:
$$\text{Area}(ACE) = r \cdot s$$
where $s$ is the semiperimeter of triangle $ACE$.

3. **Area of Hexagon $ABCDEF$**:
- The area of the convex hexagon $ABCDEF$ can be divided into the areas of triangles $ACE$, $BDF$, and the remaining parts.
- Since $ACE$ and $BDF$ have the same circumradius $R$, we can use the properties of these triangles to bound the area of the hexagon.

4. Bounding the Area of Hexagon:
- Consider the circumcircle of triangle $ACE$ with radius $R$. The area of triangle $ACE$ is given by:
$$\text{Area}(ACE) = \frac{abc}{4R}$$
where $a$, $b$, and $c$ are the sides of triangle $ACE$.
- Similarly, for triangle $BDF$, the area is:
$$\text{Area}(BDF) = \frac{b'd'f'}{4R}$$
where $b'$, $d'$, and $f'$ are the sides of triangle $BDF$.

5. Using the Inradius:
- The area of triangle $ACE$ can also be expressed using the inradius $r$:
$$\text{Area}(ACE) = r \cdot s$$
where $s$ is the semiperimeter of triangle $ACE$.

6. Combining the Areas:
- The total area of the hexagon $ABCDEF$ is the sum of the areas of triangles $ACE$, $BDF$, and the remaining parts.
- Since the circumradius $R$ is the same for both triangles, we can use the relationship between the circumradius and the inradius to bound the total area.

7. Final Inequality:
- By combining the areas and using the fact that the circumradius $R$ and inradius $r$ are related, we get:
$$\text{Area}(ABCDEF) \le \frac{R}{r} \cdot \text{Area}(ACE)$$
- This follows from the fact that the circumradius $R$ bounds the area of the hexagon in terms of the area of triangle $ACE$ and its inradius $r$.

$\blacksquare$

The final answer is $\boxed{ \text{Area}(ABCDEF) \le \frac{R}{r} \cdot \text{Area}(ACE) }$.