AMC 10 · 2010 · #22

Grade 7 algebra
Archetype Arithmetic Expression Decomposition └ Sub-archetype Multi-Segment Path or Count Sum
absolute-valueoptimizationtriangular-numbersweighted-average extreme-principle ↑ Prerequisites: absolute-value
📏 Medium solution 💡 3 insights
Problem
A long sum of absolute values grows with a different weight in each term. Find its smallest possible value.

Pick an answer.

(A)
49
(B)
50
(C)
51
(D)
52
(E)
53
How to solve
Strategy Extreme Principle

Tool #15 (Organize Information in More Ways): since every multiplier is positive, |mx-1|=m|x-1/m|, which turns the sum into a total weighted distance from x to 119 marked points — that picture says the answer lives where the weight on the left balances the weight on the right. Tool #14 (Extreme Principle): a minimum is only pinned down by squeezing from below, and |t| ≥ t together with |t| ≥ -t lets me strip every absolute value in whichever direction helps, producing an honest floor. Tool #4 (Introduce a Variable): the floor still contains x unless the sign pattern is chosen just right, so I name the cut-off k and solve for the k that erases x. Tool #13 (Convert to Algebra): with x gone the bound is a plain count of +1s and -1s. Tool #11 (Work Backwards): a floor is worthless until something stands on it, so I run each inequality backwards to find exactly which x make all of them equalities, and check one such x by hand.

1STEP 1

Read the sum as weighted distance

The sum is a weighted distance to many points.

f(x)=Σ_m=1¹¹⁹|mx-1|=Σ_m=1¹¹⁹ m|x-1/m|, Σ_m=1¹¹⁹ m=(119 · 120)/2=7140
2STEP 2

Strip the bars in the helpful direction

Each bar can be stripped in the helpful direction.

|t| ≥ t and |t| ≥ -t ⟹ f(x) ≥ Σ_m=1^k(1-mx)+Σ_m=k+1¹¹⁹(mx-1) for every real x
3STEP 3

Choose the cut-off that erases x

One cut-off makes the variable vanish.

Σ_m=k+1¹¹⁹ m-Σ_m=1^k m=0 ⇔ (k(k+1))/2=7140/2=3570 ⇔ k(k+1)=7140=84 · 85 ⇔ k=84
4STEP 4

Count what survives

What survives is a plain count.

f(x) ≥ Σ_m=1⁸⁴(1-mx)+Σ_m=85¹¹⁹(mx-1) = 84-35+x (3570-3570) = 49
5STEP 5

Show 49 is actually reached

A whole interval reaches it, so 49 stands.

x ≤ 1/84 and x ≥ 1/85 ⟹ f(x)=49 for all x∈[1/85,1/84]; f (1/84)=83/2+0+15/2=49
Answer
49
Both halves of the claim were established separately — f(x) ≥ 49 for all real x in Step 4, and f (1/84)=49 in Step 5 — so nothing rests on an assumption about where the minimum ought to sit. Numerical spot checks agree and show the shape: f(0)=119 and f(1)=1+2+…+118=7021 are far above, while f (1/50)=364/5=72.8 and f (1/100)=257/5=51.4 close in as x approaches the interval. The immediate neighbours are only barely higher — f (1/83)=4069/83≈ 49.02 and f (1/86)=2108/43≈ 49.02 — which is what a flat bottom with a gentle rise on each side should look like. The other endpoint agrees exactly too: f (1/85)=(84-3570/85)+(3485/85-34)=42+7=49, confirming the whole segment [1/85,1/84] sits at the same height. Since 49 is the smallest listed choice and it is genuinely attained, no smaller candidate is being missed.
💡Key takeaway

To make a pile of absolute values as small as possible, split the terms into two groups of equal total weight so the variable cancels out — then prove it by finding an x that really makes all those sign choices come true.

  • Read the sum as weighted distance
  • Strip the bars in the helpful direction
  • Choose the cut-off that erases x
  • Count what survives
  • Show 49 is actually reached

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