AMC 10 · 2008 · #15
Grade 10 geometry-2dPick an answer.
The region S outside R lives entirely in the gaps between the boundary of R and the boundary of S, so this is a boundary problem, not an area problem. My plan is to walk the boundary of R vertex by vertex and ask at each one how much of the full 360° turn the pieces of R actually use up. Where they use up 180° the boundary runs straight and there is no gap; where they fall short of 360° there is a wedge, and that wedge is exactly what S fills in. Tool #4 (Introduce a Variable) pins every vertex with coordinates built from the single number √(3)/2, so no claim rests on how the picture looks. Tool #1 (Draw a Diagram) supplies the two angle counts that do all the work. Tool #14 (Extreme Principle) settles which vertices are corners of S — the usual one-line "clearly the outer points" is the step this problem actually turns on, and it can be proved outright by showing the eight candidates sit on one circle with every other vertex strictly inside it. Tool #17 (Visualize Spatial Relationships) uses the quarter turn of the whole figure to say the four gaps are congruent instead of computing four times. Finally tool #16 (Change Focus / Count the Complement) reruns the whole thing the other way round — measure R, measure S, subtract — as a check that shares no step with the first route.
Pin every vertex with coordinates
Coordinates pin every vertex exactly.
Every length in this picture comes from one right triangle with hypotenuse 1 and legs 1/2 and √(3)/2, so coordinates cost nothing and remove all guessing.
8.G.B.7Introduce A VariableThree triangles fuse into one trapezoid
Three triangles turn out to be collinear, fusing into one edge.
Three 60° corners meeting at one point make a straight angle, so the dent you expect at the tip of each spike simply is not there.
Three sixty degree corners meeting at one point make a straight angle, so no dent appears there.
▸ Why?
Angles sharing a vertex add to the angle they span together, so three sixties total one hundred eighty degrees.
▸ Why?
Each equilateral corner is sixty degrees because a triangle's three equal angles share a straight angle.
Count the angles at a square corner
The angles at a square corner leave a small gap.
Add the angles that actually meet at a corner; whatever is missing from a full turn is exactly the bite the convex hull will fill in.
7.G.B.5Draw A DiagramProve which points are corners of S
A distance check proves which points are the polygon's corners.
Eight points sitting on one circle with everything else strictly inside it — the circle hands you the corners of the hull, with no appeal to how the drawing looks.
8.G.B.8Extreme PrincipleMeasure one wedge
One wedge has area 1/4.
In a 30-60-90 right triangle the short leg is half the hypotenuse, so a 30° corner between two unit sides gives a height of exactly 1/2 — and that is where all the square roots drop out.
10.G-SRT.C.8Identify SubproblemsFour congruent wedges
Four congruent wedges give 1.
The whole picture looks the same after a quarter turn, so one corner's worth of work is all four corners' worth of work.
8.G.A.3Visualize Spatial RelationshipsCheck by measuring both regions
Measuring both regions separately confirms 1, choice (A).
Two areas that are each ugly with √(3) but differ by a clean 1 is a strong signal, because the 3√(3) has to cancel exactly.
10.G-GPE.B.7Change Focus Count The ComplementAt each corner, add up the angles the shape actually uses; whatever is missing from a full turn is the wedge the convex hull fills in, and four 30° wedges with unit sides come to exactly 1.
- Pin every vertex with coordinates
- Three triangles fuse into one trapezoid
- Count the angles at a square corner
- Prove which points are corners of S
- Measure one wedge
- Four congruent wedges
- Check by measuring both regions
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