AMC 10 · 2012 · #10
Grade 7 algebraMary divides a circle into 12 sectors. The central angles of these sectors, measured in degrees, are all integers and they form an arithmetic sequence. What is the degree measure of the smallest possible sector angle?
Pick an answer.
AMC 10 2012 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: A circle is cut into 12 sectors whose central angles are whole numbers of degrees that form an arithmetic sequence. Find the smallest angle that is possible for the smallest sector.
Givens: There are 12 sectors filling one whole circle.; The 12 central angles form an arithmetic sequence (a constant gap between consecutive angles).; Every angle is a positive integer number of degrees.; The 12 angles add up to 360 degrees.
Unknowns: The smallest possible value of the smallest sector angle.
Understand
Restated: A circle is cut into 12 sectors whose central angles are whole numbers of degrees that form an arithmetic sequence. Find the smallest angle that is possible for the smallest sector.
Givens: There are 12 sectors filling one whole circle.; The 12 central angles form an arithmetic sequence (a constant gap between consecutive angles).; Every angle is a positive integer number of degrees.; The 12 angles add up to 360 degrees.
Plan
Primary tool: #14 Extreme Principle
Secondary: #4 Introduce a Variable, #13 Convert to Algebra, #3 Eliminate Possibilities
The question asks for the smallest possible first angle, so this is a min/max problem: I name the first angle and the common difference, turn 'the angles fill a circle' into one equation, and then push the common difference to its largest allowed value to force the first angle as small as it can go.
Execute — Answer: C
6.EE.B.6 Step 1 Name the two unknowns
- Call the smallest angle a and the constant gap between consecutive angles d.
- Then the 12 angles are a, a+d, a+2d, up to a+11d.
- Both a and d are whole numbers.
💡 An arithmetic sequence is fully pinned down by its start and its step, so two letters capture all 12 angles.
7.EE.A.1 Step 2 Add them to 360
- The sectors fill the whole circle, so the 12 angles add to 360.
- Adding the terms gives 12a plus (0+1+2+...+11)d, and 0+1+...+11 = 66.
- So 12a + 66d = 360.
- Dividing every term by 6 gives the cleaner equation 2a + 11d = 60.
💡 Collecting like terms turns twelve separate angles into a single equation between the start and the step.
6.EE.B.5 Step 3 Make the start a whole number
- Solve for a: 2a = 60 - 11d, so a = (60 - 11d)/2.
- For a to be a whole number, 60 - 11d must be even.
- Since 60 is even, 11d must be even, and because 11 is odd this happens only when d itself is even.
💡 An odd number times d is even only when d is even, so the parity of d controls whether a lands on a whole number.
6.EE.B.5 Step 4 Push the gap to its extreme
- To make a as small as possible, make d as large as possible.
- But a must stay positive, so 60 - 11d > 0, which means d is less than about 5.45, so d is at most 5.
- The largest even value allowed is d = 4.
💡 A fixed total shared among 12 angles means a bigger step forces the first angle lower, so the biggest legal step gives the smallest first angle.
4.MD.C.7 Step 5 Compute and check
- With d = 4, a = (60 - 44)/2 = 8.
- The angles run 8, 12, 16, ..., 52, and they add to 12 x (8 + 52)/2 = 360, so this really is a valid circle.
- The smallest possible sector angle is 8 degrees, which is answer (C).
💡 Plugging the extreme gap back in and confirming the angles still sum to a full circle proves the minimum is actually reachable.
6.EE.B.6 Call the smallest angle a and the constant gap between consecutive angles d. The 7.EE.A.1 The sectors fill the whole circle, so the 12 angles add to 360. Adding the terms 6.EE.B.5 Solve for a: 2a = 60 - 11d, so a = (60 - 11d)/2. For a to be a whole number, 60 6.EE.B.5 To make a as small as possible, make d as large as possible. But a must stay pos 4.MD.C.7 With d = 4, a = (60 - 44)/2 = 8. The angles run 8, 12, 16, ..., 52, and they add Review
Reasonableness: The angles 8, 12, 16, ..., 52 are all positive whole numbers, evenly spaced, and add to exactly 360, so the arrangement is legal. Trying the next even step d = 6 needs 66 degrees of increase in one term, which already breaks a > 0, so 8 cannot be beaten and answer (C) holds.
Alternative: Instead of the smallest term, center on the average. The 12 angles average 360/12 = 30, and for 12 terms the mean sits between the 6th and 7th terms, so the 6th term is 30 - d/2 and the smallest is 30 - 11d/2. Maximizing even d under a positive constraint again gives d = 4 and smallest = 30 - 22 = 8.
CCSS standards used (min grade 7)
6.EE.B.6Use variables to represent numbers and write expressions (Naming the smallest angle a and the common difference d, and writing all 12 angles from them.)7.EE.A.1Apply properties of operations to add, subtract, factor, and expand linear expressions (Summing the twelve terms to 12a + 66d and reducing the equation to 2a + 11d = 60.)6.EE.B.5Understand solving an equation or inequality as a process of finding values (Solving for a, forcing it to be a whole number, and searching for the value of d that minimizes it.)4.MD.C.7Recognize angle measure as additive and solve addition problems (Using the fact that the sector angles add to the full 360-degree circle, and verifying the final angles sum to 360.)
⭐ When a fixed total is split into evenly spaced parts, making the gap as big as allowed makes the smallest part as tiny as possible.
⭐ When a fixed total is split into evenly spaced parts, making the gap as big as allowed makes the smallest part as tiny as possible.
More like this
Same archetype — closest grade level first.