AMC 10 · 2012 · #7
Grade 7 geometry-2dPick an answer.
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
Two unknowns describe all twelve angles.
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
Their sum being a full turn is one equation.
Collecting like terms turns twelve separate angles into a single equation between the start and the step.
Collecting like terms turns twelve separate angles into a single equation between the start and the step.
▸ Why?
The twelve angles fill the full turn about the centre, so their total is fixed before anything else.
▸ Why?
Each angle is the previous one plus the same fixed step, so all twelve are written from two letters.
Make the start a whole number
Whole angles force the step to be 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.5Introduce A VariablePush the gap to its extreme
Pushing the step up gives 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
The first angle is then 8, choice (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