AMC 10 · 2010 · #5

Grade 7 algebra
optimizationlinear-equations-one-var extreme-principleconvert-to-algebra ↑ Prerequisites: linear-equations-one-var
📏 Medium solution 💡 3 insights
Problem
A contestant leads by a known margin at the halfway point and never scores below a floor. Find the fewest perfect shots that make winning certain.

Pick an answer.

(A)
38
(B)
40
(C)
42
(D)
44
(E)
46
How to solve
Strategy Extreme Principle

The word doing all the work is 'guaranteed'. It quantifies over every remaining play of the tournament at once, which looks like infinitely many cases, so Tool #14 (Extreme Principle) is the primary move: replace all of them by the single least favourable one. That replacement is legitimate here, and it is worth saying why instead of assuming it — each archer's final total is a sum of independent shot values, so Chelsea's total is smallest exactly when every free shot of hers sits at its floor, and the opponent's is largest exactly when every shot of his sits at its ceiling. Tool #4 (Introduce a Variable) then names Chelsea's unknown current score so that the comparison can be written down, and it immediately shows that name cancelling. Tool #13 (Convert to Algebra) turns the guarantee into one inequality in n. The step that the algebra alone does not supply is tightness: an inequality that fails on paper only rules out a value of n if the failing scenario is one the rules permit, so Tool #14 returns to build that scenario explicitly with legal scores. Finally Tool #3 (Eliminate Possibilities) uses the fact that extra bullseyes never hurt to show the working values of n run upward without gaps, which is what makes 'the minimum' meaningful. Tool #16 (Change Focus / Count the Complement) supplies the independent cross-check in the review by counting the shots that are not bullseyes.

1STEP 1

Turn 'guaranteed' into one worst case

Guaranteed becomes one specific worst case.

Chelsea's free shots ≥ 4 each opponent's shots ≤ 10 each
2STEP 2

Name the totals; the current score cancels

Naming the totals makes the current score cancel.

M(n) = (k + 10n + 4(50-n)) - ((k-50) + 500) = 6n - 250
3STEP 3

Solve the inequality and keep it strict

The winning condition is a strict inequality.

6n - 250 > 0 ⟹ n > 250/6 = 41 2/3 ⟹ n ≥ 42
4STEP 4

Show the losing case is really allowed

The losing case really is allowed, so smaller counts fail.

500 + 41 · 10 + 9 · 4 = 946 < 450 + 50 · 10 = 950
5STEP 5

Confirm 42 works and nothing smaller can

Checking the neighbours gives 42, choice (C).

M(38) = -22, M(40) = -10, M(41) = -4, M(42) = +2 → (C)
Answer
42
Check the size first. Chelsea's cushion is 50 points, but across the last 50 shots the opponent can gain up to 500 while a floor-scoring Chelsea gains only 200 — a 300-point swing against a 50-point lead. She therefore has to spend nearly the whole second half at maximum output, so an answer in the high thirties or low forties is expected, and 42 out of 50 sits there. Parity is a second check: every listed score is even, so both final totals are even and the margin is even. The computed margins -22, -10, -4, +2 are all even, as they must be, and a margin of exactly 1 was never on the table. Third, test whether each hypothesis is really doing work. Drop Chelsea's 'at least 4' habit and her free shots could be 0, giving 10n > 450 and n ≥ 46 — which is choice (E). So the floor is load-bearing, not decoration, and (E) is precisely the answer produced by ignoring it. Finally, 42 ≤ 50, so the promised bullseyes fit in the shots she has left.
💡Key takeaway

'Guaranteed' means you have to survive the worst case, so score your opponent at his best and yourself at your floor — here every bullseye you swap in for a 4 buys 6 points of cushion, and digging out of 250 takes 42 of them.

  • Turn 'guaranteed' into one worst case
  • Name the totals; the current score cancels
  • Solve the inequality and keep it strict
  • Show the losing case is really allowed
  • Confirm 42 works and nothing smaller can