9. (40 points) Among nine visually identical gold coins, one weighs $a$, seven weigh $b$, and the last one weighs $c$, and $a < b < c$. Using a balance scale, find the minimum number of weighings required to identify the coin weighing $a$ and the coin weighing $c$.

To solve this problem, we need to devise a strategy to identify the coins with weights $a$ and $c$ using the fewest number of weighings. Here is a step-by-step approach:

1. **First Weighing:**
   - Divide the nine coins into three groups of three coins each: Group 1, Group 2, and Group 3.
   - Weigh Group 1 against Group 2.

   **Case 1:** If Group 1 and Group 2 balance, then the coin weighing $a$ and the coin weighing $c$ must be in Group 3.
   - Weigh two coins from Group 3 against each other.
     - If they balance, the remaining coin in Group 3 is the one weighing $c$.
     - If they do not balance, the lighter coin is the one weighing $a$ and the heavier coin is the one weighing $c$.

   **Case 2:** If Group 1 and Group 2 do not balance, then the coin weighing $a$ and the coin weighing $c$ are in the lighter and heavier groups, respectively.
   - Identify the lighter group (Group 1 or Group 2) and the heavier group (Group 1 or Group 2).
   - Weigh two coins from the lighter group against each other.
     - If they balance, the remaining coin in the lighter group is the one weighing $a$.
     - If they do not balance, the lighter coin is the one weighing $a$.
   - Weigh two coins from the heavier group against each other.
     - If they balance, the remaining coin in the heavier group is the one weighing $c$.
     - If they do not balance, the heavier coin is the one weighing $c$.

In both cases, we can identify the coins weighing $a$ and $c$ in a minimum of 2 weighings.

Thus, the minimum number of weighings required is \(\boxed{2}\).
9. Solution 1: First, label the nine coins as
$A, B, C, D, E, F, G, H, I$.
Weigh $(A, B, C, D)$ against $(E, F, G, H)$. If they balance, then $a+c=2b$. If they do not balance, assume without loss of generality that $(A, B, C, D) > (E, F, G, H)$.

For the second weighing, weigh $(A, B)$ against $(C, D)$ and $(E, F)$ against $(G, H)$. At this point, there are four possible scenarios:
(1) $(A, B) > (C, D)$ and $(E, F) > (G, H)$. Then the heavy coin is in $(A, B)$, and the light coin can only be in $(G, H)$. For the fourth weighing, place $(A, B, G, H)$ on one side and $(C, D, E, F)$ on the other side. The lightness or heaviness or balance of $(A, B, G, H)$ side indicates whether $a+c$ is lighter, heavier, or equal to $2b$.
(2) $(A, B) > (C, D)$ and $(E, F) = (G, H)$. Then the heavy coin is in $(A, B)$, and the light coin can only be in $(C, D, I)$.

If the light coin is in $(C, D)$, then it is already known that $a+c > 2b$. For the fourth weighing, place $(A, B, I)$ on one side and $(E, F, G)$ on the other side. The lightness or heaviness of the $(A, B, I)$ side indicates whether $a+c$ is lighter or heavier than $2b$.
(3) $(A, B) = (C, D)$ and $(E, F) > (G, H)$. Then the light coin is in $(G, H)$, and the heavy coin can only be in $(E, F, I)$. If the heavy coin is in $(E, F)$, then it is already known that $a+c < 2b$. For the fourth weighing, place $(A, B, C)$ on one side and $(G, H, I)$ on the other side. The lightness or heaviness or balance of the $(G, H, I)$ side indicates whether $a+c$ is lighter, heavier, or equal to $2b$.
(4) If both weighings are unbalanced, assume $(A, B) > (C, D)$ and $(E, F) > (G, H)$. For the third weighing, compare $(A, B)$ with $(G, H)$. If they balance, then the fake coins are in $(C, D, E, F)$. For the fourth weighing, compare $(A, B, G, H)$ with $(C, D, E, F)$. The lightness or heaviness or balance of the $(C, D, E, F)$ side indicates whether $a+c$ is lighter, heavier, or equal to $2b$; if they do not balance, then the fake coins are in $(A, B, G, H)$. For the fourth weighing, compare $(A, B, G, H)$ with $(C, D, E, F)$. The lightness or heaviness or balance of the $(A, B, G, H)$ side indicates whether $a+c$ is lighter, heavier, or equal to $2b$. (For the third weighing, compare $(A, B)$ with $(E, F)$ (or symmetrically $(C, D)$ with $(G, H))$. The heavier side contains the fake heavy coin, and the other side contains the real coins, thus identifying the location of the light coin.)
In summary, four weighings are sufficient to determine the result.