engths of these segments such that their inverses also follow the ratio $1: 2: 3$:

Let the lengths of the segments be $L_1$, $L_2$, and $L_3$. The ratio of their fundamental frequencies being $1:2:3$ implies:

$$ \frac{1}{L_1} : \frac{1}{L_2} : \frac{1}{L_3} = 1 : 2 : 3 $$

Inverting the terms in each ratio, we have:

$$ L_1 : L_2 : L_3 = \frac{1}{1} : \frac{1}{2} : \frac{1}{3} $$

To simplify this ratio, find a common multiple of the denominators (2 and 3), which is 6. Therefore, multiplying each term by 6 gives:

$$ L_1 : L_2 : L_3 = 6 : 3 : 2 $$

Since the total length of the wire is 110 cm, we can write:

$$ L_1 + L_2 + L_3 = 110 $$

Substituting the ratio terms:

$$ 6x + 3x + 2x = 110 $$

Solving for $x$:

$$ 11x = 110 $$

$$ x = 10 $$

Thus, the lengths of the segments are:

$L_1 = 6x = 6 \times 10 = 60$ cm

$L_2 = 3x = 3 \times 10 = 30$ cm

$L_3 = 2x = 2 \times 10 = 20$ cm

Therefore, the positions of the bridges from point $A$ should be at:

First bridge at 60 cm (end of the first segment)

Second bridge at 60 cm + 30 cm = 90 cm (end of the second segment)

Hence, the placement of the bridges should be at 60 cm and 90 cm.

Answer: Option A: 60 cm and 90 cm

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