Problem 15 · 2007 AMC 8
Medium
Algebra & Patterns
inequality-reasoning
Let a, b and c be numbers with 0 < a < b < c. Which of the following is impossible?
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Answer: A — a + c < b is impossible.
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Hint 1 of 2
'Impossible' means it can never happen, so look for the one choice that follows from the rules alone — no clever numbers can break it. Compare each side against b.
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Hint 2 of 2
Adding only makes things bigger: with c > b already, adding a positive a can't drag the sum below b. Multiplying by a fraction, though, can shrink — that's where the 'possible' ones live.
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Approach: prove one is forced, exhibit examples for the rest
- (A) says a + c < b. But c is already bigger than b, and a is positive, so a + c > c > b. It can never dip below b — (A) is impossible.
- The others survive with small fractions, because multiplying by something under 1 shrinks a number: take a = 1/3, b = 1/2, c = 1. Then a·c = 1/3 < 1/2 = b (D works), a+b < c and a·b < c (B, C), and b/c = a is easy to arrange too.
- Only (A) is locked out, so the answer is (A).
- Strategy that transfers: on 'which is impossible' problems, one choice is provable from the given order and the rest fall to a single well-chosen example — fractions between 0 and 1 are your best test values.
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