7.3. Given a triangle $A B C$, side $A B$ is divided into 4 equal segments $A B_{1}=B_{1} B_{2}=B_{2} B_{3}=B_{3} B$, and side $A C$ into 5 equal segments $A C_{1}=C_{1} C_{2}=C_{2} C_{3}=C_{3} C_{4}=C_{4} C$. How many times larger is the area of triangle $A B C$ compared to the sum of the areas of triangles $C_{1} B_{1} C_{2}, C_{2} B_{2} C_{3}, C_{3} B_{3} C_{4}, C_{4} B C$?
Answer: 2 times.

Solution: Let the area of $A B_{1} C_{1}$ be $S$. Then the area of $B_{1} C_{1} C_{2}$ is also $S$, since $B_{1} C_{1}$ is the median in $A B_{1} C_{2}$. Similarly, the area of $B_{1} B_{2} C_{2}$ is $2S$, as $C_{2} B_{1}$ is the median in $A B_{2} C_{2}$. The area of $B_{2} C_{2} C_{3}$ is half the area of $A B_{2} C_{2}$, because these triangles share the same height, and the base $C_{2} C_{3}$ is half the length of the base $A C_{2}$. Therefore, the area of $B_{2} C_{2} C_{3}$ is $2S$. From similar reasoning, the area of $B_{2} C_{3} B_{3}$ is half the area of $A B_{2} C_{3}$, which is $3S$. Since $C_{3} C_{4}$ is one-third of $A C_{3}$, the area of $B_{3} C_{3} C_{4}$ is one-third of $A B_{3} C_{3}$ and equals $3S$. Similarly, the areas of $B_{3} C_{4} B$ and $C_{4} B C$ are each $4S$. Therefore, the total area of the shaded part is $S + 2S + 3S + 4S = 10S$, and the unshaded part is $S + 2S + 3S + 4S = 10S$, meaning the shaded area is half of the entire triangle.

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