AMC 10 · 2011 · #18

Grade 8 algebra
Archetype Coordinate Plane Figure Computation └ Sub-archetype Figure Constraint Equations for Unknown Quantity
absolute-valuecompleting-the-squareoptimization extreme-principlecasework ↑ Prerequisites: absolute-value
📏 Medium solution 💡 2 insights
Problem
Two absolute values of a sum and a difference add to a fixed amount. Find the largest value of a given expression.

Pick an answer.

(A)
5
(B)
6
(C)
7
(D)
8
(E)
9
How to solve
Strategy Extreme Principle

A maximum over a region lives at the edge of that region (Tool #14 Extreme Principle), so the first job is to know exactly what the region is. The constraint is a sum of two absolute values, which is unreadable as written, so rewrite it in a different form (Tool #15): comparing |x| with |y| collapses the whole left side into a single clean expression. That turns the constraint into a picture — the border of a square (Tool #1). Then bound x² - 6x + y² piece by piece on that square and test the candidate corner against the original equation (Tool #6). The test matters: stripping the absolute-value bars four ways produces four full straight lines, and on those lines the expression has no maximum at all, so an argument that stops at the four lines has not answered the question.

1STEP 1

Collapse the two absolute values

The pair collapses to twice the larger size.

|x+y| + |x-y| = 2max(|x|,|y|) for all real x, y
2STEP 2

Read off the region exactly

So the region is a square boundary.

max(|x|,|y|) = 1 ⇔ (|x| ≤ 1 and |y| ≤ 1) and (|x| = 1 or |y| = 1)
3STEP 3

Bound each piece on the square

Each term has its own easy ceiling.

x² ≤ 1, -6x ≤ 6, y² ≤ 1 ⟹ x² - 6x + y² ≤ 1 + 6 + 1 = 8
4STEP 4

Show the ceiling is reached

One corner reaches all three at once.

(x,y) = (-1,1): |-1+1| + |-1-1| = 0 + 2 = 2 ✓, (-1)² - 6(-1) + 1² = 1 + 6 + 1 = 8
5STEP 5

Confirm with the distance picture

Completing the square confirms 8, choice (D).

x² - 6x + y² = (x-3)² + y² - 9; (-1-3)² + (± 1)² - 9 = 16 + 1 - 9 = 8 → (D) 8
Answer
8
Sample the border and watch 8 hold as the top. At (1,0): |1| + |1| = 2 and the value is 1 - 6 + 0 = -5. At (0,1): |1| + |-1| = 2 and the value is 0 - 0 + 1 = 1. At (-1,0): |-1| + |-1| = 2 and the value is 1 + 6 + 0 = 7. At (-1, 1/2): |-1/2| + |-3/2| = 2 and the value is 1 + 6 + 1/4 = 7.25. At (-1, ± 1): 8. Nothing beats 8, and the choices explain themselves — (C) 7 is exactly the value at (-1,0), the trap for anyone who pushes x to -1 but forgets that y can be ± 1 at that very same corner, and (E) 9 would need y² = 2, impossible once |y| ≤ 1 is established. Reaching (D) 8 requires both extremes to happen at one point, and a corner of the square is exactly where they do.
💡Key takeaway

The equation |x+y| + |x-y| = 2 is just a disguised way of saying the bigger of |x| and |y| is 1, so the point sits on the border of the unit square; at the corner (-1, ± 1) both -6x and y² are as large as they can be at once, and x² - 6x + y² reaches (D) 8.

  • Collapse the two absolute values
  • Read off the region exactly
  • Bound each piece on the square
  • Show the ceiling is reached
  • Confirm with the distance picture

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