AMC 10 · 2010 · #18
Grade 12 probabilityPick an answer.
Three random directions at once is too much to hold. Tool #7 (Identify Subproblems) splits the trip: first glue jumps 1 and 2 into a single step vec S of random length R, then ask what the third jump must do. Tool #4 (Introduce a Variable) names the angle between the first two jumps, which is all R depends on, and Tool #13 (Convert to Algebra) turns "lands within 1 meter" into a plain inequality on a cosine. Tool #1 (Draw a Diagram) reads that inequality off the direction circle as an arc, turning it into a probability. The load-bearing move is Tool #15 (Organize Information in More Ways): switch from the three absolute directions to the pair (length of the first two jumps, direction of the third jump measured against vec S), because in those coordinates the two pieces are independent. Tool #5 (Look for a Pattern) finishes: the resulting conditional probability is a straight-line tent in the half-angle, so its average needs no work at all.
Turn jumps into vectors
The jumps become three unit vectors.
Jumps add head-to-tail like arrows, so where the frog ends up is one vector sum and nothing else.
12.N-VM.B.4Introduce A VariableGlue the first two jumps
Gluing the first two leaves one length.
Two random jumps behave like one jump whose length is random, and one random length is far easier to track than two random directions.
10.S-CP.A.1Identify SubproblemsLength after two jumps
That length rides on the angle between them.
Two unit arrows make a rhombus, and its diagonal is twice the cosine of half the angle between them.
Two unit arrows make a rhombus, and its diagonal is fixed by the angle between them.
▸ Why?
Both arrows have the same length, so the figure they span is symmetric about the diagonal.
▸ Why?
With two side lengths known, the third side is decided entirely by the angle between them.
The last direction is still uniform
The last direction is still uniform and independent.
Spinning a fair spinner and then measuring the result from a mark you drew earlier is still a fair spin.
10.S-CP.A.1Organize Information In More WaysWhen does the frog land home
Landing home becomes one cosine condition.
To return inside the unit circle the last jump has to point far enough backwards, and "far enough" is set by how far out the first two jumps carried the frog.
11.G-SRT.D.10Convert To AlgebraRead the condition as an arc
That condition reads as an arc of the circle.
A uniform random direction hits a target set of headings with probability equal to that set's share of the full turn.
11.F-TF.A.1Draw A DiagramAverage the tent function
Averaging a tent-shaped function gives a quarter turn.
In the half-angle coordinate the chance of getting home is a straight ramp up and back down, so its average is just the midpoint height.
12.S-MD.A.2Look For A PatternAssemble the probability
The probability is 1/4, choice (C).
Splitting into "how far the first two jumps went" and "where the last one pointed" makes the whole probability one clean average.
12.S-MD.A.2Identify SubproblemsFold the first two jumps into one step: the chance the last jump brings the frog home is a straight ramp up and back down as that folded step shrinks, and the average of a ramp is just its midpoint, giving 1/4.
- Turn jumps into vectors
- Glue the first two jumps
- Length after two jumps
- The last direction is still uniform
- When does the frog land home
- Read the condition as an arc
- Average the tent function
- Assemble the probability
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