AMC 10 · 2022 · #22

Grade 8 arithmetic

tangent-circlescoordinate-geometryarea-circlescaseworksystems-of-equations caseworkidentify-subproblemsconvert-to-algebra ↑ Prerequisites: tangent-circlescoordinate-geometry

📏 Long solution 💡 4 insights

Problem

Three given circles: C₁: x²+y²=4 (center O, radius 2), C₂: x²+y²=64 (center O, radius 8), and C₃: (x-5)²+y²=3 (center (5,0), radius √(3)). Find every circle S tangent to all three. Sum the areas of those circles S.

Pick an answer.

(A)

$48\pi$

(B)

$68\pi$

(C)

$96\pi$

(D)

$102\pi$

(E)

$136\pi$

AMC 10 2022 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Draw a Diagram

Tool #1 (Diagram) — sketch the two concentric circles with the small offset C₃ inside the annulus; the picture immediately shows S has to fit between C₁ and C₂ in one of two simple ways. Tool #9 (Easier) — solve the easier sub-problem first: "which circles are tangent to BOTH concentric circles?" That forces just two radii (3 or 5). Tool #7 (Subproblems) — split into two cases (radius 3 or 5) and within each, two more sign-choices for the C₃ tangency. Tool #2 (List) — systematically list each tangency configuration to be sure none are missed. Tool #13 (Algebra) — set up the distance equation for the center to confirm each case yields valid circles.

1STEP 1

C₁ (radius 2) and C₂ (radius 8) share center O, forming an annulus of width 6; C₃ at (5,0) with radius √(3) sits entirely inside it.

annulus width = 8 - 2 = 6, C₃ ⊂ annulus

2STEP 2

Tangency to both concentric circles is a distance condition that pins the solution radius to exactly r = 3 or r = 5.

r = 3, d = 5 or r = 5, d = 3

3STEP 3

So each center lies on x²+y²=25 (r=3) or x²+y²=9 (r=5); now impose tangency to C₃: (x-5)²+y²=(r±√(3))².

(x-5)² + y² = (r ± √(3))²

4STEP 4

Case r=3: substituting x²+y²=25 gives 50−10x=(3±√(3))², so two x-values, each mirrored across the x-axis — 4 circles.

50 - 10x = (3 ± √(3))² → 4 circles

5STEP 5

Case r=5: with x²+y²=9 it becomes 34−10x=(5±√(3))²; both x-values stay inside the locus, so again 4 circles.

34 - 10x = (5 ± √(3))² → 4 circles

6STEP 6

Add the areas: four r=3 circles give 36π, four r=5 circles give 100π, for a total of 136π — choice (E).

36π + 100π = 136π → (E)

Answer

136π

Sanity. The two concentric circles alone force radius ∈ {3, 5} — this is a clean geometric pinch and easy to double-check by drawing. The third circle C₃ then permits two tangency sign-choices and the configuration is symmetric in y, multiplying by 2. So we expect a count divisible by 4 per radius — and we got exactly 4 + 4 = 8 circles. Each x-value computed lies inside the corresponding locus circle (|x| ＜ 5 for case A, |x| ＜ 3 for case B), so all y's are real and the eight circles truly exist. Sum 36π + 100π = 136π matches (E). Choices 48π, 68π, 96π, 102π are decoys for forgetting one case or one sign.

💡Key takeaway

This AMC 10 problem only needs Grade 8 coordinate-plane distance you already know — drawing the two same-center circles immediately pins the unknown radius to 3 or 5, and each of those two cases gives 4 valid circles (two sign-choices for tangency with the third circle, times mirror symmetry). Eight circles in total, area 4π(9) + 4π(25) = 136π.