AMC 10 · 2006 · #24
Grade 11 algebrageometry-2dPick an answer.
The expression sin² x - sin x sin y + sin² y is a quadratic in sin x once sin y is treated as a constant, so naming a = sin x and b = sin y turns an unfamiliar curve into a familiar upward parabola, and 'is at most 3/4' becomes 'lies between the two roots'. The quadratic formula then produces the boundary in closed form, and the Pythagorean identity plus the angle-addition formula collapse it into sin x = sin(y ± π/3). That is where the real work starts and where this problem punishes speed. Turning sin x ≤ sin θ into x ≤ θ is legal only while θ stays in the quarter turn where sine climbs, and y + π/3 leaves that quarter turn as soon as y ＞ π/6. So I split into cases and fold the stray angle back with sin θ = sin(π - θ). Because a case split is exactly the kind of step that quietly loses a piece of the region, I then verify the result by a completely different, case-free route: an algebraic identity that rewrites the whole condition as a product of two factors. A product is at most zero precisely when its factors disagree in sign, which gives two branches, and one of those branches is invisible to any argument that only draws the boundary lines and shades whichever side contains a convenient test point. Only after both routes agree do I read off the polygon and measure it, subtracting three corner triangles from the square rather than adding up five-sided pieces.
Read it as a quadratic
Reading it as a quadratic in one sine makes it solvable.
An upward parabola dips below a height exactly between its two roots, so finding the roots finds the whole region at once.
9.A-SSE.A.2Introduce A VariableTake the square root honestly
The square root simplifies into a plain cosine.
The Pythagorean identity turns the awkward 3 - 3sin² y into a perfect square, and staying inside the first quadrant is what lets the square root come out with no sign case.
11.F-TF.C.8Convert To AlgebraRecognize the shifted sines
The bounds turn out to be shifted sines.
1/2 and √(3)/2 are the cosine and sine of 60°, so that combination is one sine of a shifted angle in disguise.
One half and root three over two are the cosine and sine of sixty degrees, so that combination is one shifted sine in disguise.
▸ Why?
A thirty-sixty-ninety triangle fixes those two ratios exactly, so no measuring is needed.
▸ Why?
Adding a fixed angle to another is what the shifted sine records, so the combination and the shift say the same thing.
Strip sines only where legal
Sines may be stripped only where they are increasing.
Sine takes the same value on both sides of π/2, so an inequality between sines becomes one between angles only after every angle is folded back into the quarter turn where sine climbs.
11.F-TF.A.2Identify SubproblemsOne identity replaces the casework
A factoring identity replaces all the casework.
Once a tangled inequality becomes one factor times another, the whole question is only which factor is positive and which is negative.
9.A-SSE.B.3Organize Information In More WaysThe stray branch is one point
The escaping branch is a single point, so it costs no area.
Two inequalities that both push x upward can only be satisfied at the very top of its range, which pins a single point instead of leaving room for a region.
9.A-CED.A.3Extreme PrincipleFind the pentagon's five corners
The region is a pentagon with five findable corners.
When straight lines slice a square, every corner of what is left is a crossing point, so listing the crossings lists the corners.
10.G-GPE.B.7Draw A DiagramSubtract three corner triangles
Subtracting three corner triangles gives π²/6, choice (B).
Trimming corners off a square is three easy triangles instead of one awkward five-sided shape, provided the trimmed pieces do not overlap.
6.G.A.1Change Focus Count The ComplementWhen an inequality looks tangled, rewrite it as one thing times another: a product is negative only when its two factors disagree in sign, and that turns a curvy-looking condition into a few straight lines you can actually measure.
- Read it as a quadratic
- Take the square root honestly
- Recognize the shifted sines
- Strip sines only where legal
- One identity replaces the casework
- The stray branch is one point
- Find the pentagon's five corners
- Subtract three corner triangles