AMC 10 · 2008 · #21
Grade 10 probabilitygeometry-2dPick an answer.
Tool #4 (Introduce a Variable) strips the picture down to two numbers, a and b, since the only freedom is horizontal. Tool #14 (Extreme Principle) settles the real question: how far apart may the centers be and still leave the circles touching? The tangent position is the boundary case, and it must be shown that the boundary is actually reached, not just that it cannot be passed. Tool #1 (Draw a Diagram) is the engine: plot the pair (a,b) as a point in a 2 × 2 square, so the event becomes a region and its probability becomes an area. Tool #16 (Change Focus / Count the Complement) makes the area trivial, because the failing region is two small corner triangles while the winning region is an awkward band. Tool #3 (Eliminate Possibilities) finishes with a decimal check against the five choices, which separates the correct value from the one wrong choice built on the same picture.
Reduce the picture to two numbers
Only two numbers vary, so the picture collapses.
Two random circles sound complicated, but only their horizontal positions can change, so two numbers hold all the randomness.
9.A-CED.A.2Introduce A VariablePin down what "intersect" means
Meeting means the centre distance is small enough, and the other case cannot happen.
Tangency is the breaking point, and the perpendicular-bisector construction shows a shared point really exists all the way up to it — not just that it cannot exist beyond it.
10.G-CO.A.1Extreme PrincipleTurn the criterion into a condition on a and b
That becomes one inequality in the two positions.
The vertical gap of 1 is a fixed leg of a right triangle, so the whole geometric condition collapses to how far apart the two x-values are.
8.G.B.8Introduce A VariablePlot the sample space as a square
The sample space is a square, so probability is area.
Two independent uniform picks fill a square evenly, so any question about them is a question about what fraction of the square works.
Two independent uniform picks fill a square evenly, so the question becomes what fraction of that square works.
▸ Why?
When every outcome carries the same weight, a chance is the share of the outcomes that give it.
▸ Why?
The two picks are made without regard to each other, so every pairing of values is possible and equally weighted.
Switch to the failing region
The failing region is far easier to measure.
When the good region is a wide messy strip, the bad region is usually two clean corners — measure those instead.
10.S-CP.A.1Change Focus Count The ComplementMeasure the two corner triangles
It is two corner triangles.
The line b = a - √(3) cuts off a right triangle whose legs are both the leftover length 2 - √(3).
10.G-GPE.B.7Draw A DiagramSubtract from the whole square
Subtracting from the whole square gives (4√(3)-3)/4.
Everything that is not one of the two corner triangles is a success, so one subtraction ends it.
10.S-CP.A.1Change Focus Count The ComplementCheck the value against the choices
Comparing with the choices confirms (4√(3)-3)/4, choice (E).
A quick decimal turns five look-alike radical expressions into five clearly different numbers.
8.NS.A.2Eliminate PossibilitiesTwo circles of the same size touch exactly when their centers are no farther apart than two radii, so this whole problem becomes "pick two numbers in [0,2] — how often are they within √(3) of each other?", and that is just an area in a square.
- Reduce the picture to two numbers
- Pin down what "intersect" means
- Turn the criterion into a condition on a and b
- Plot the sample space as a square
- Switch to the failing region
- Measure the two corner triangles
- Subtract from the whole square
- Check the value against the choices
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