AMC 10 · 2006 · #15
Grade 11 algebraPick an answer.
The question asks for a minimum, so Tool #14 (Extreme Principle) sets the standard the work has to meet: it is not enough to produce one z that works, because several of the answer choices are values that do work. The winner must be a value that works AND that nothing positive can undercut. To get there, Tool #9 (Solve an Easier Related Problem) does the heavy lifting: two coupled angles is hard, so the angle-addition formula is used to erase x entirely and turn the pair of equations into a single condition on z alone. Tool #15 (Organize Information in More Ways) prepares that move by rewriting cos x = 0 in the form the formula actually needs, which is a statement about sin x. Tool #6 (Guess and Check) supplies an explicit x proving the candidate is really attainable, and Tool #3 (Eliminate Possibilities) sorts the remaining choices into 'impossible' and 'possible but not smallest'.
Rewrite the condition on x
A zero cosine forces the sine to be plus or minus one, with both available.
On the unit circle the points with first coordinate 0 are the top and the bottom, so the height is +1 or -1 — never just one of them.
11.F-TF.C.8Organize Information In More WaysErase x with the addition formula
The addition formula erases the unknown angle entirely.
When cos x = 0 the shift by x is a quarter-turn, and a quarter-turn just swaps cosine for sine up to a sign — so the whole problem is really about sin z.
11.F-TF.C.9Solve An Easier Related ProblemCheck the reverse direction too
Both signs are matchable, so the condition is exactly on the shift's sine size.
Having two choices for x means either sign of sin z can be flipped into place, so no candidate is lost to a sign mismatch.
11.F-TF.A.2Guess And CheckRule out everything below pi/6
Below that value the sine is too small, so nothing smaller works.
Sine climbs steadily from 0 up to 1 over the first quarter-turn, so it cannot reach the height 1/2 before the angle reaches π/6.
Sine climbs steadily from zero over the first quarter turn, so it cannot reach one half before the angle reaches thirty degrees.
▸ Why?
A value that only rises cannot pass a level twice, so the first crossing marks the smallest angle that works.
▸ Why?
An angle is a fixed fraction of the full turn, so thirty degrees names one definite place on the circle.
Confirm pi/6 itself and conclude
An explicit pair confirms it, so the answer is π/6, choice (A).
A value nothing can undercut, backed by one explicit pair of angles that reaches it, is by definition the smallest.
11.F-TF.A.2Extreme PrincipleWhen a problem says 'smallest possible', find every value that works before picking one — here cos x = 0 allows sin x = 1 and sin x = -1, and using both is what drops the answer from 7π/6 down to π/6.
- Rewrite the condition on x
- Erase x with the addition formula
- Check the reverse direction too
- Rule out everything below pi/6
- Confirm pi/6 itself and conclude