AMC 10 · 2012 · #10
Grade 7 algebraPick an answer.
AMC 10 2012 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
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.
Name the two unknowns
Call the smallest angle a and the constant gap d, so the 12 angles are a, a+d, a+2d, up to a+11d.
An arithmetic sequence is fully pinned down by its start and its step, so two letters capture all 12 angles.
6.EE.B.6Introduce A VariableAdd them to 360
The 12 angles fill the whole circle, so 12a + 66d = 360, which reduces to 2a + 11d = 60.
Collecting like terms turns twelve separate angles into a single equation between the start and the step.
7.EE.A.1Convert To AlgebraMake the start a whole number
Solving gives a = (60 - 11d)/2, and since 60 is even while 11 is odd, a is a whole number only when d 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.
An odd number times the step is even only when the step itself is even.
▸ Why?
An odd times an odd is odd, so only an even factor can make the product even.
▸ Why?
Otherwise the halving leaves a remainder, so the starting angle would not be a whole number.
Push the gap to its extreme
A bigger d makes a smaller, but a must stay positive, so d is at most 5 — the largest even value 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.
6.EE.B.5Extreme PrincipleCompute and check
With d = 4, a = (60 - 44)/2 = 8, and 8, 12, 16, ..., 52 really do add to 360 — answer (C).
Plugging the extreme gap back in and confirming the angles still sum to a full circle proves the minimum is actually reachable.
4.MD.C.7Extreme PrincipleWhen a fixed total is split into evenly spaced parts, making the gap as big as allowed makes the smallest part as tiny as possible.
- Name the two unknowns
- Add them to 360
- Make the start a whole number
- Push the gap to its extreme
- Compute and check