AMC 10 · 2015 · #11

Grade 10 geometry-2d

Archetype Casework by Position / Vertex Type └ Sub-archetype Count of Type

tangent-circlescircle-equationcoordinate-geometry caseworksystematic-enumeration ↑ Prerequisites: tangent-circlescircle-equation

📏 Long solution 💡 3 insights

Problem

Two circles of fixed sizes may sit anywhere, and their common tangents are counted. Count the possible tallies.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

How to solve

Strategy Introduce a Variable

Placing two circles looks like four degrees of freedom, two center coordinates each, but sliding or turning the whole sheet cannot change how many common tangents there are. So one number carries the entire configuration: d, the distance between the centers. Name it, and k becomes a function of the single variable d. Next, rewrite "tangent to both circles" as two distance equations, which turns that function into algebra that can be solved exactly rather than eyeballed from pictures. The solution count of that algebra changes only at a couple of boundary values of d, and those boundaries cut the d-line into a short list of regimes to tabulate. This route is chosen over sketching the familiar circle pictures because sketching shows that certain counts happen but never proves that no other count can happen, and ruling out k = 5 and k = 6 is exactly what the multiple-choice list demands.

1STEP 1

Squeeze the setup to one number

The whole setup collapses to one distance.

d = |O₁O₂| ≥ 0, k = k(d)

2STEP 2

Turn tangency into a distance equation

Tangency becomes a plain distance equation.

ℓ = { x : u · x = c }, |u| = 1 → dist(O, ℓ) = |u · O - c|

3STEP 3

Set coordinates and fix the sign

Coordinates split it into two families.

|c| = 2, |d u_x - c| = 3; c = 2 → d u_x = 5 or d u_x = -1

4STEP 4

Count the normals in each family

Each family holds zero, one, or two lines.

u_x² + u_y² = 1 → u_y² = 1 - u_x², u_x = 5/d or u_x = -1/d

5STEP 5

Find where the count jumps

The count jumps at the sum and the difference of the radii.

|5/d| ≤ 1 ⇔ d ≥ 5 = 3 + 2, |1/d| ≤ 1 ⇔ d ≥ 1 = 3 - 2

6STEP 6

List the regimes and count values

That gives 5 possible tallies, choice (D).

k(d) = 0 & 0 ≤ d ＜ 1 ; 1 & d = 1 ; 2 & 1 ＜ d ＜ 5 ; 3 & d = 5 ; 4 & d ＞ 5

Answer

The five regimes match five pictures that can be sketched by hand: one circle nested strictly inside the other with no shared tangent, the two internally tangent with the single line at the touch point, the two crossing at two points with only the outer pair of tangents, the two externally tangent with the outer pair plus the line through the touch point, and the two fully apart with the outer pair plus the pair that crosses between them. The algebra also names which family is which: the -1 family, alive whenever d ＞ 3 - 2, is the outer pair, and the +5 family, alive whenever d ＞ 3 + 2, is the crossing pair. The thresholds landing exactly on the difference and the sum of the radii is the expected place for a circle configuration to change character. One correction worth making to the usual quick telling: k = 0 is not just the concentric case, since every d ＜ 1 gives 0 lines, so the real condition is that one circle lies strictly inside the other. That does not change the count of possible values. Five values is choice (D). Choices (A) 2, (B) 3 and (C) 4 all come from dropping regimes, most often by ignoring the two tangency boundaries and keeping only 0, 2, 4; choice (E) 6 is impossible because the algebra caps k at 4.

💡Key takeaway

Only the distance between the two centers matters: one pair of tangent lines switches on once that distance passes 3 - 2 and the other pair once it passes 3 + 2, so the count climbs through 0, 1, 2, 3, 4.

Squeeze the setup to one number
Turn tangency into a distance equation
Set coordinates and fix the sign
Count the normals in each family
Find where the count jumps
List the regimes and count values