AMC 10 · 2013 · #17

Grade 9 algebra

Archetype Coordinate Plane Figure Computation └ Sub-archetype Figure Constraint Equations for Unknown Quantity

symmetric-polynomialsquadratic-equationsoptimizationsign-analysis convert-to-algebraextremal-construction ↑ Prerequisites: quadratic-equationssystems-of-equations

📏 Long solution 💡 3 insights

Problem

Three real numbers have a known sum and a known sum of squares. Find the range of one of them.

Pick an answer.

(A)

(B)

$\frac{10}{3}$

(C)

(D)

$\frac{16}{3}$

(E)

$\frac{20}{3}$

How to solve

Strategy Organize Information in More Ways

Two equations in three unknowns leave a whole surface of solutions, so chasing a, b, c together goes nowhere. Reorganize instead: treat c as a number that has been fixed in advance, and ask a yes-or-no question about what is left. Once c is fixed, the pair (a,b) must have a prescribed sum and a prescribed sum of squares, and the question becomes 'does such a real pair exist?'. That reframing matters because of a trap. The usual inequality route (Cauchy-Schwarz, or the mean inequality) proves only one direction: every workable c must satisfy a certain quadratic inequality. That gives an interval c cannot leave, which is not the same as an interval c actually fills, and the question asks for the genuine maximum and minimum. So the plan is to find a test on c that is an equivalence, not just a one-way bound. The identity 2(a²+b²)-(a+b)²=(a-b)² supplies exactly that, because its equality case hands over the pair (a,b) that realizes each endpoint. Then a quadratic inequality in c alone finishes the job, and two explicit triples confirm the ends are reached.

1STEP 1

Fix c and isolate the pair

Fixing one leaves a pair with known sums.

a+b = 2-c, a²+b² = 12-c²

2STEP 2

The exact test for the pair

There is an exact test for such a pair.

2(a²+b²)-(a+b)²=(a-b)² ≥ 0; a,b=(s ± √(2q-s²))/2 → a+b=s, a²+b²=q

3STEP 3

Feed c into the test

Feeding it in gives one quadratic inequality.

2(12-c²) ≥ (2-c)² ⇔ 24-2c² ≥ 4-4c+c² ⇔ 3c²-4c-20 ≤ 0

4STEP 4

Factor and read the interval

Factoring reads off an interval.

(3c-10)(c+2) ≤ 0 ⇔ -2 ≤ c ≤ 10/3

5STEP 5

Show both ends are reached

Both ends are actually reached.

(a,b,c)=(2,2,-2) and (a,b,c)=(-2/3,-2/3,10/3) both satisfy a+b+c=2, a²+b²+c²=12

6STEP 6

Subtract the two extremes

The spread is 16/3, choice (E).

c_max-c_min=10/3-(-2)=16/3

Answer

16/3

Two independent sanity checks confirm the size and the strictness. First, c² ≤ a²+b²+c² = 12 forces |c| ≤ 2√(3) ≈ 3.464, and the maximum found, 10/3 ≈ 3.333, sits just inside that ceiling. It must sit strictly inside: reaching c=2√(3) would need a²+b²=0, hence a=b=0, hence a+b+c=2√(3) ≠ 2. So the true maximum is below 2√(3), exactly as found, and any answer relying on c=± 2√(3) would be wrong. Second, spot-check an interior value: c=2 gives a+b=0 and a²+b²=8, satisfied by (a,b)=(2,-2), so c=2 is possible and the interval is not empty in the middle. The asymmetry of the answer is also believable, since the two equations are not symmetric about c=0: the constraint a+b+c=2 tilts the whole picture, and indeed the interval [-2,10/3] is centered at 2/3, not at 0. Note what the argument really turned on. A one-directional inequality would have shown only that c cannot escape [-2,10/3]; the perfect-square identity made the test an equivalence, and its equality case produced the triples (2,2,-2) and (-2/3,-2/3,10/3) that make the endpoints real rather than hypothetical.

💡Key takeaway

Two real numbers with a given sum and a given sum of squares exist exactly when twice the sum of squares is at least the square of the sum, so that one test decides which values of c are allowed and which triple sits at each end.

Fix c and isolate the pair
The exact test for the pair
Feed c into the test
Factor and read the interval
Show both ends are reached
Subtract the two extremes