Problem 8 · 1996 AJHSME
Medium
Arithmetic & Operations
number-lineextremal
Points A and B are 10 units apart. Points B and C are 4 units apart. Points C and D are 3 units apart. If A and D are as close as possible, then the number of units between them is
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Answer: B — 3 units.
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Hint 1 of 2
Each distance is just a length — the points can sit anywhere on a line as long as the gaps are right. To pull A and D close, walk OUT to B, then turn around and come back toward A as far as the next steps allow.
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Hint 2 of 2
Starting 10 away (A to B), the most you can walk back toward A is C then D: 4 + 3 = 7. Subtract that from the 10 you went out.
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Approach: go out, then backtrack as much as possible
- Put A at 0 and B at 10. To shrink the A-to-D gap, point each later hop back toward A: step C back 4 (to 6), then D back another 3 (to 3). So D lands at 10 − 4 − 3 = 3.
- Could D reach A at 0? That needs the return trip 4 + 3 = 7 to undo the full 10 — but 7 < 10, so it falls 3 short. The closest possible is 3 units.
- Why this transfers: 'how close/far can the ends be' on a line is the triangle inequality in disguise — closest = |big − (sum of the rest)|, farthest = the whole sum added up. Aim the hops to cancel or pile on.
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