AMC 10 · 2021 · #17

Grade 9 algebra
Archetype Small-Domain Constrained Search └ Sub-archetype Logic-Deduction / Elimination
quadratic-equationssign-analysisinterval-arithmeticsystematic-enumeration bound-inequality-then-enumeratecasework ↑ Prerequisites: quadratic-equations
📏 Long solution 💡 3 insights
Problem
Two quadratics are built from the same two positive integers, just with the roles swapped. Count the ordered pairs for which both quadratics fail to have two distinct real solutions.

Pick an answer.

(A)
4
(B)
6
(C)
8
(D)
12
(E)
16
How to solve
Strategy Extreme Principle

"How many ordered pairs" invites listing, but b and c range over all positive integers, so a raw list never ends. The fix is to translate first and bound second. Translating is mechanical: a quadratic has two distinct real solutions exactly when its discriminant is positive, so failing that means the discriminant is at most 0. Each quadratic gives one inequality, and the two together form a tight system. Then comes the load-bearing move: push the two inequalities into each other to squeeze b alone. Each inequality says one letter is small compared to the square of the other, and chaining them forces b³ ＜ = 64, so b can only be 1, 2, 3, or 4. Once the search box is finite, a short systematic list over those four values finishes the count.

1STEP 1

Translate the failure condition

Translate it into a discriminant condition.

p² - 4q ＞ 0 ⇔ two distinct real roots → not two distinct ⇔ p² - 4q ≤ 0
2STEP 2

Write the system for both quadratics

Write a system of inequalities.

b² - 4c ≤ 0 and c² - 4b ≤ 0 ⟺ b² ≤ 4c and c² ≤ 4b
3STEP 3

Squeeze b into a finite range

Feeding them into each other narrows the range.

b² ≤ 4c → b⁴ ≤ 16c² ≤ 64b → b³ ≤ 64 → b ≤ 4
4STEP 4

Check each value of b

Check each surviving value directly.

b=1:& c ≥ 1, c ≤ 2 &&→ c ∈ {1,2} ; b=2:& c ≥ 1, c ≤ 2 &&→ c ∈ {1,2} ; b=3:& c ≥ 3, c ≤ 3 &&→ c = 3 ; b=4:& c ≥ 4, c ≤ 4 &&→ c = 4
5STEP 5

Collect and count the pairs

Collecting the pairs gives 6.

{(1,1),(1,2),(2,1),(2,2),(3,3),(4,4)} 2 + 2 + 1 + 1 = 6
Answer
6
The condition is symmetric in b and c, so the set of winning pairs should be closed under swapping. It is: (1,2) and (2,1) both appear, and the rest are already on the diagonal. Reading the two inequalities as regions, b² ＜ = 4c is the inside of one sideways-opening parabola and c² ＜ = 4b is the inside of its mirror image, and the two parabolas meet at (0,0) and (4,4) - exactly matching the fact that b = 4 is the last value that survives. A boundary sanity check: at b = 5 the floor forces 4c ＞ = 25 so c ＞ = 7, while the ceiling forces c² ＜ = 20 so c ＜ = 4, an impossible squeeze. Six is also plausible against the options: the region is a thin lens, so the small answers are the credible ones.
💡Key takeaway

When a condition caps each unknown by the square of the other, feed the two caps into each other to trap one letter in a finite range, then just list what is left.

  • Translate the failure condition
  • Write the system for both quadratics
  • Squeeze b into a finite range
  • Check each value of b
  • Collect and count the pairs

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