AMC 10 · 2005 · #20

Grade 9 algebra
optimizationcompleting-the-squareparityset-partition caseworkextreme-principlesystematic-enumeration ↑ Prerequisites: quadratic-equationsparityset-partition
📏 Long solution 💡 3 insights
Problem
Eight given numbers are split into two groups of four. Find the smallest possible value of the sum of the squares of the two group totals.

Pick an answer.

(A)
30
(B)
32
(C)
34
(D)
40
(E)
50
How to solve
Strategy Introduce a Variable

The two group sums are not independent — they always add up to the total of the whole set. So we name one of them s (Tool #4), which turns a two-variable minimisation into a one-variable quadratic we can complete the square on. The whole problem then collapses to a boundary question (Tool #14): how close to the balance point can a group sum get? Answering it honestly requires proving that perfect balance is impossible, and for that a systematic list of the possible pair sums (Tool #2), organised by parity, eliminates every case (Tool #3). Finding one split that scores well is not a proof of minimality on its own; the impossibility half is what makes the answer certain.

1STEP 1

The letters use every number

The two totals always add to a fixed number.

-7-5-3-2+2+4+6+13 = 8
2STEP 2

Reduce to one variable

So the expression depends on a single unknown.

(a+b+c+d)²+(e+f+g+h)² = s²+(8-s)²
3STEP 3

Complete the square

Completing the square shows only the distance from the balance point matters.

s²+(8-s)² = 2s²-16s+64 = 2(s-4)²+32
4STEP 4

Ask whether a group can sum to 4

A parity count narrows the shapes a balanced group could take.

odd: {-7,-5,-3,13}, even: {-2,2,4,6}
5STEP 5

Rule out all-odd and all-even

The all-odd and all-even shapes both fail.

-7-5-3+13 = -2, -2+2+4+6 = 10
6STEP 6

Rule out two odd plus two even

The mixed shape fails too, so the balance point is unreachable.

4-{-12,-10,-8,6,8,10} = {16,14,12,-2,-4,-6}, none in {0,2,4,6,8,10}
7STEP 7

Bound it, then hit the bound

A split attains the next bound, so the minimum is 34, choice (C).

13-7-5+2 = 3, -3-2+4+6 = 5, 3²+5² = 9+25 = 34
Answer
34
The rewritten form 2(s-4)²+32 says the only values this expression can ever take are 32, 34, 40, 50, 64, … — and four of the five answer choices (32, 34, 40, 50) are exactly the first four of those. Choice 30 is not of that form at all, so it could never be an answer, which is a good sign the reduction is right. Everything then hinges on the single fact that no group of four sums to 4, and that was settled by checking all three parity shapes exhaustively rather than by trying splits at random.
💡Key takeaway

When two sums must add to a fixed total, keeping them as equal as possible makes the sum of their squares smallest — and if perfect balance is impossible, the next-closest split wins.

  • The letters use every number
  • Reduce to one variable
  • Complete the square
  • Ask whether a group can sum to 4
  • Rule out all-odd and all-even
  • Rule out two odd plus two even
  • Bound it, then hit the bound