Problem 4 · AMC 8 Stretch
Stretch
Algebra & Patterns
symmetrypattern-recognitionreduce-and-expand
Three friends sit in a circle, each holding some money. Going around, each says, 'If I doubled my money by taking what the next person has, I could buy the prize.' If the friends hold \(x_1, x_2, x_3\) and the prize costs \(h\), the three statements are \(x_1 + x_2 = h\), \(x_2 + x_3 = h\), \(x_3 + x_1 = h\). How much does each friend hold, in terms of \(h\)?
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Answer: each holds h/2
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Hint 1 of 4
All three equations look the same as you go around the circle — that's a symmetry. Try comparing two equations directly instead of grinding through substitution.
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Hint 2 of 4
Subtract the first equation from the second: \((x_2 + x_3) - (x_1 + x_2) = h - h\). The \(x_2\) cancels, leaving \(x_3 - x_1 = 0\), so \(x_3 = x_1\). Do this with another pair.
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Hint 3 of 4
Comparing pairs shows \(x_1 = x_2 = x_3\). Call it \(x\). Then any equation becomes \(x + x = h\).
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Approach: Exploit the rotational symmetry by subtracting equations
- The three equations are identical in form around the circle, a strong hint the three amounts are equal.
- Subtract the first from the second: \((x_2 + x_3) - (x_1 + x_2) = 0 \Rightarrow x_3 = x_1\). Subtract the second from the third: \((x_3 + x_1) - (x_2 + x_3) = 0 \Rightarrow x_1 = x_2\).
- So \(x_1 = x_2 = x_3\). Call the common amount \(x\); any equation reads \(x + x = h\), so \(2x = h\) and \(x = \tfrac{h}{2}\).
- Each friend holds exactly half the prize price, \(\tfrac{h}{2}\).
Mark:
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