AMC 10 · 2009 · #19
Grade 7 geometry-2dCircle A has radius 100. Circle B has an integer radius r<100 and remains internally tangent to circle A as it rolls once around the circumference of circle A. The two circles have the same points of tangency at the beginning and end of circle B's trip. How many possible values can r have?
Pick an answer.
AMC 10 2009 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: 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.
Givens: Circle $A$ has radius $100$, so its circumference is $2\pi(100) = 200\pi$.; Circle $B$ has an integer radius $r$ with $r < 100$ and stays internally tangent while rolling once around circle $A$.; Circle $B$ begins and ends its trip at the same point of tangency.
Unknowns: The number of integer radii $r$ for which the starting and ending points of tangency are the same.
Understand
Restated: 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.
Givens: Circle $A$ has radius $100$, so its circumference is $2\pi(100) = 200\pi$.; Circle $B$ has an integer radius $r$ with $r < 100$ and stays internally tangent while rolling once around circle $A$.; Circle $B$ begins and ends its trip at the same point of tangency.
Plan
Primary tool: #4 Introduce a Variable
Secondary: #17 Visualize Spatial Relationships, #2 Make a Systematic List, #13 Convert to Algebra, #3 Eliminate Possibilities
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.
Execute — Answer: B
7.G.B.4 Step 1 Picture the rolling trip
- Imagine circle $B$ pressed against the inside of circle $A$ and rolling once all the way around.
- Its point of contact sweeps along the entire inside edge of circle $A$, a distance equal to circle $A$'s circumference, $200\pi$.
- Circle $B$ can only meet at the same spot on itself again if it rolls a whole number of its own circumferences during the trip.
💡 The contact point goes once around circle $A$, so the trip length is exactly circle $A$'s circumference.
7.G.B.4 Step 2 Name the radius, compare circumferences
- Let $r$ be circle $B$'s radius.
- Circle $A$'s circumference is $2\pi(100)$ and circle $B$'s is $2\pi r$.
- The number of times $B$ rolls during one full trip is the big circumference divided by the small one.
💡 How many small circles' worth of edge fit around the big one is just the big circumference divided by the small.
6.RP.A.3 Step 3 Turn the condition into divisibility
- The tangency lines up again only when $B$ rolls a whole number of times, so $\tfrac{100}{r}$ must be a whole number.
- The $2\pi$ already cancelled, so this simply says that $r$ divides $100$ with no remainder.
💡 A whole number of turns means the small radius must divide $100$ evenly.
4.OA.B.4 Step 4 List every divisor of 100
- Find all whole numbers that divide $100$ by pairing factors: $1\times100$, $2\times50$, $4\times25$, $5\times20$, and $10\times10$.
- Collecting them gives $1, 2, 4, 5, 10, 20, 25, 50, 100$ — nine divisors in all.
💡 Pairing factors that multiply to $100$ catches every divisor without missing any.
4.OA.B.4 Step 5 Apply r < 100 and count
- The radius must satisfy $r < 100$, so drop $100$ from the list.
- That leaves $1, 2, 4, 5, 10, 20, 25, 50$ — eight values.
- So $r$ can take $8$ possible values, which is answer $\textbf{(B)}$.
💡 Every divisor of $100$ works except $100$ itself, which the rule $r < 100$ rules out.
7.G.B.4 Imagine circle $B$ pressed against the inside of circle $A$ and rolling once all 7.G.B.4 Let $r$ be circle $B$'s radius. Circle $A$'s circumference is $2\pi(100)$ and ci 6.RP.A.3 The tangency lines up again only when $B$ rolls a whole number of times, so $\tf 4.OA.B.4 Find all whole numbers that divide $100$ by pairing factors: $1\times100$, $2\ti 4.OA.B.4 The radius must satisfy $r < 100$, so drop $100$ from the list. That leaves $1, Review
Reasonableness: 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) $= 8$.
Alternative: Count divisors straight from the prime factorization instead of listing them. Since $100 = 2^2\cdot 5^2$, the number of divisors is $(2+1)(2+1) = 9$. Removing $r = 100$ leaves $9 - 1 = 8$, the same answer.
CCSS standards used (min grade 7)
7.G.B.4Know the formulas for the area and circumference of a circle and use them to solve problems (Writing circle $A$'s circumference as $200\pi$ and circle $B$'s as $2\pi r$ to compare the trip length to $B$'s size.)6.RP.A.3Use ratio and rate reasoning to solve real-world and mathematical problems (Reading the ratio $\tfrac{100}{r}$ as the number of rolls and requiring it to be a whole number, i.e. $r \mid 100$.)4.OA.B.4Find all factor pairs for a whole number in the range 1 to 100 (Listing all divisors of $100$ by factor pairs and counting how many are less than $100$.)
⭐ 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.
⭐ 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.
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