AMC 10 · 2009 · #19

Grade 7 geometry-2d
Archetype Small-Domain Constrained Search └ Sub-archetype Factor / Divisibility Constraint
factorsdivisor-countratio-proportion convert-to-algebrasystematic-enumeration ↑ Prerequisites: factors
📏 Medium solution 💡 3 insights
Problem
Circle A has radius 100. A smaller circle B with a whole-number radius r ＜ 100 stays touching the inside of circle A and rolls once all the way around it. Circle B starts and finishes at the same point of tangency. Find how many whole-number values r can have.

Pick an answer.

(A)
$\ 4\$
(B)
$\ 8\$
(C)
$\ 9\$
(D)
$\ 50\$
(E)
$\ 90\$

AMC 10 2009 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Introduce a Variable

The unknown that ties the whole picture together is circle B's radius, so tool #4 (Introduce a Variable) names it r and lets us compare the two circumferences as one clean ratio. Tool #17 (Visualize Spatial Relationships) turns the rolling picture into a countable condition: the tangency lines up again only after a whole number of B's turns. That condition becomes "r divides 100," and tool #2 (Make a Systematic List) then enumerates every divisor of 100 so none is missed before the r ＜ 100 rule trims the count.

1STEP 1

Picture the rolling trip

Roll B once around the inside of A: its contact point sweeps A's whole edge, a distance of 200π.

C_A = 2π(100) = 200π
2STEP 2

Name the radius, compare circumferences

Let r be B's radius: A's circumference is 2π(100), B's is 2π r, and B's turns are the big one divided by the small.

C_A/C_B = 2π(100)/(2π r) = 100/r
3STEP 3

Turn the condition into divisibility

The 2π cancels, so the turn count is a whole number exactly when r divides 100 with no remainder.

100/r ∈ Z ⇔ r ∣ 100
4STEP 4

List every divisor of 100

Pair up factors — 1×100, 2×50, 4×25, 5×20, 10×10 — to catch all 9 divisors of 100.

100 = 2² · 5² → 1, 2, 4, 5, 10, 20, 25, 50, 100 (9 divisors)
5STEP 5

Apply r ＜ 100 and count

The rule r ＜ 100 drops 100, leaving 1, 2, 4, 5, 10, 20, 25, 50 — 8 values, choice (B).

9 - 1 = 8 → (B)
Answer
8\
Test a few radii directly. If r = 1, then 100/1 = 100 whole turns; if r = 50, then 100/50 = 2 whole turns — both bring the tangency back to the same point. A non-divisor like r = 3 gives 100/3, not a whole number, so it fails. Exactly the eight divisors of 100 that are below 100 work, matching choice (B) = 8.
💡Key takeaway

A small circle lands back on the same touching point only if its radius divides the big radius exactly, so the answer is just how many divisors fit the rule.

  • Picture the rolling trip
  • Name the radius, compare circumferences
  • Turn the condition into divisibility
  • List every divisor of 100
  • Apply r ＜ 100 and count

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