Three girls play three rounds of a game. On each round
there are two winners and one loser. The girl who loses
on a round has to double the number of chips that each
of the other girls has by giving up some of her own
chips. Each girl loses one round. At the end of three
rounds, each girl has 40 chips. How many chips did
each girl have at the beginning of the game?

To solve this problem, we need to backtrack from the end of the game to the beginning. Here's how we can approach it:

### Initial Setup:
- Let's denote the number of chips that the three girls (A, B, and C) had at the beginning of the game by \( a \), \( b \), and \( c \), respectively.

### Round Analysis:
- Each girl loses exactly one round. 
- When a girl loses a round, she doubles the chips of the other two girls by giving them some of her own chips.

### End Condition:
- After all three rounds, each girl has 40 chips.

Let's consider the reverse process, where we imagine that the girls end up with 40 chips, and we work backward to determine the initial chips they had.

### Working Backwards:
1. **Final State:**
   \[
   \text{At the end, each girl has 40 chips.}
   \]

2. **Before the last round:**
   Suppose \( C \) lost the last round. 
   - Before losing, \( C \) would have had \( 40 - x \) chips.
   - \( A \) and \( B \) would have had \( 40/2 = 20 \) chips each.

   Thus, before the last round, we had:
   \[
   A = 20, \quad B = 20, \quad C = 80 - 40 = 40 - x + 40 = 40 + x
   \]

3. **Before the second round:**
   Suppose \( B \) lost the second round.
   - Before losing, \( B \) would have had \( 20 - y \) chips.
   - \( A \) and \( C \) would have had \( 20/2 = 10 \) and \( 40 + x \) chips respectively.

   Thus, before the second round, we had:
   \[
   A = 10, \quad B = 80 - 40 + y = 40 - y, \quad C = 40 + x
   \]

4. **Before the first round:**
   Suppose \( A \) lost the first round.
   - Before losing, \( A \) would have had \( 10 - z \) chips.
   - \( B \) and \( C \) would have had \( 10/2 = 5 \) and \( 40 + x - y = x - y + 40 \) chips respectively.

   Thus, before the first round, we had:
   \[
   A = 5, \quad B = 5 + y, \quad C = x - y + 40
   \]

Given the initial chips before any round:
   \[
   a = 5, \quad b = 5 + y, \quad c = x - y + 40
   \]

### Solving Backwards for \( a = b = c = x = y = z = \text{initial chips} \):

This analysis shows that we can simplify this by assuming equal distribution backward until reaching a constant value:
- \( a = b = c = 35 \)
- The \( x, y, z \) values will add consistently to maintain the balance.

### Final Answer:
- Each girl had 35 chips at the beginning.

Would you like to see the detailed equations or any further questions on this topic? Here are some related questions you might find interesting:

1. What if one girl had started with more chips than the others?
2. How does the distribution change if we consider different initial conditions?
3. What if the number of rounds changes? How does it affect the final distribution?
4. Can you find a general formula for any number of rounds and players?
5. How would the outcome change if the loser gave away double the chips from both girls combined?

**Tip:** Working backward from the known final condition is a powerful strategy in problems where actions undo previous states.

Explore how to solve a problem involving the distribution of chips among three girls over three rounds of a game. Each girl ends with 40 chips after a sequence of losses and redistributions. Detailed step-by-step analysis provided.

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