92. There are 9 coins, 8 of which are genuine and of the same weight, and one is counterfeit and heavier than the others. What is the minimum number of weighings on a balance scale without weights that are needed to find the counterfeit coin?
$\triangle$ Let's number the coins with natural numbers from 1 to 9. Here, it is better to weigh the coins by placing three on each pan of the balance. Place coins numbered $1,2,3$ on the left pan, and coins numbered $4,5,6$ on the right pan. There are two possible outcomes, as in problem 90.

Suppose the balance is in equilibrium (Fig. 30). Then the counterfeit coin is in the triplet with numbers $7,8,9$, and it can be found using one more weighing (see the solution to problem 90).

Suppose the balance is not in equilibrium. In this case, the counterfeit coin is in one of the first two triplets, which is heavier, and it can also be found using one more weighing.

Thus, two weighings were needed. Could the goal have been achieved in one? To answer this question, we need to change the weighing system, placing not three, but four, or two, or even one coin on each pan of the balance. But each of the mentioned options is less advantageous than the one we discussed.

Answer: two.