AMC 10 · 2023 · #20

Grade 11 probability

Archetype Geometric Probability by Area └ Sub-archetype Uniform Draws to Favorable Region

geometric-probabilitydouble-angle-formulatrigonometric-ratiosrotation-isometry easier-related-problemsymmetry-argumentconvert-to-algebra ↑ Prerequisites: probability-basictrigonometric-ratios

📏 Long solution 💡 4 insights

Problem

A frog starts at a point, jumps two units in one direction, then two more in another. Both directions are chosen at random, independently and uniformly, from all headings. Find the probability that its final position is less than one unit from the start.

Pick an answer.

(A)

$\frac{1}{6}$

(B)

$\frac{1}{5}$

(C)

$\frac{\sqrt{3}}{8}$

(D)

$\frac{\arctan \frac{1}{2}}{\pi}$

(E)

$\frac{2\arcsin \frac{1}{4}}{\pi}$

How to solve

Strategy Introduce a Variable

Two random directions sound like two variables, but the answer only depends on one. Tool #9 (Solve an Easier Related Problem) makes that precise: distance from the start does not change if the whole picture is rotated, so the first jump can be pinned to point east and only the second direction stays random. Tool #4 (Introduce a Variable) then names the single quantity that decides everything — θ, the angle between the two jump directions — and the crucial fact is that θ is still uniform over a full 2π. Tool #13 (Convert to Algebra) turns the distance into a formula in θ, and tool #16 (Change Focus) is the finish: instead of chasing the region of favorable landing points, measure the set of favorable θ and divide by 2π. Tool #3 (Eliminate Possibilities) settles the last risk — the answer is uncomfortably close to 1/6, so evaluate the decimals before committing.

1STEP 1

Only the angle matters

Only the angle between matters.

sample space = {θ : 0 ≤ θ ＜ 2π}, θ uniform

2STEP 2

Put it in coordinates

Put both jumps in coordinates.

S = (0,0), P₁ = (2, 0), P₂ = (2 + 2cosθ, 2sinθ)

3STEP 3

Compute the squared distance

The squared distance collapses cleanly.

d² = (2 + 2cosθ)² + (2sinθ)² = 4 + 8cosθ + 4cos²θ + 4sin²θ = 8 + 8cosθ

4STEP 4

Use the half-angle

The half-angle leaves one cosine.

d² = 8 + 8cosθ = 16cos² (θ/2) → d = 4|cos (θ/2)|

5STEP 5

Rewrite the success condition

It becomes one inequality.

4|cos (θ/2)| ＜ 1 ⇔ |cos (θ/2)| ＜ 1/4

6STEP 6

Measure the winning angles

Measure the interval of winning angles.

|cos u| ＜ 1/4 ⇔ u ∈ (arccos1/4, π - arccos1/4), length = π - 2arccos1/4 = 2arcsin1/4

7STEP 7

Divide by the whole circle

Divide by the full circle.

P = 4arcsin1/4/2π = 2arcsin1/4/π

8STEP 8

Check with decimals

Decimals confirm the arcsine expression.

P = 2arcsin1/4/π ≈ 0.1609 → (E)

Answer

(2 arcsin 1/4)/π

Three independent checks agree. (1) Bounds — the final distance ranges over [0, 4], and only turns very near a full reversal land inside radius 1, so a probability well under 1/4 is expected; 0.1609 fits. (2) A crude comparison — if the distance were uniform on [0,4] the answer would be 1/4, but d = 4|cos (θ/2)| pushes outcomes toward the extremes, so the true value being somewhat below 1/4 is right. (3) A second exact route — in triangle S P₁ P₂ with SP₁ = P₁P₂ = 2, the boundary case SP₂ = 1 gives 1 = 4 + 4 - 8cosβ by the Law of Cosines, so cosβ = 7/8 and the favorable interior angles form a set of measure 2arccos7/8 out of 2π, giving P = arccos7/8/π. Since cos(2arcsin1/4) = 1 - 2(1/4)² = 7/8, that is the same number, ≈ 0.1609. The near-miss (A) 1/6 ≈ 0.1667 is precisely the trap for anyone who rounds; (D) uses arctan and lands at 0.1476, a different value entirely.

💡Key takeaway

Two random directions really hold just one fact — the angle between them — so pin the first jump down, write the distance as 4|cos (θ/2)|, and the probability becomes the share of the circle where that distance is under 1.

Only the angle between jumps matters
Put the two jumps in coordinates
Distance squared collapses cleanly
Half-angle turns it into a cosine
Rewrite the success condition
Measure the winning angles with arcsin
Favorable arc over the whole circle
Decimals rule out the near-miss