AMC 10 · 2006 · #22

Grade 11 geometry-2dprobability

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

geometric-probabilitytrigonometric-ratiosangle-addition-formula symmetry-argumenteasier-related-problem ↑ Prerequisites: geometric-probabilitytrigonometric-ratios

📏 Long solution 💡 4 insights

Problem

A regular hexagon of side 2 sits at the centre of a circle that lies entirely outside it. From a uniformly random point of the circle, the chance of seeing three whole sides is exactly one half. Find the radius.

Pick an answer.

(A)

$2\sqrt{2}+2\sqrt{3}$

(B)

$3\sqrt{3}+\sqrt{2}$

(C)

$2\sqrt{6}+\sqrt{3}$

(D)

$3\sqrt{2}+\sqrt{6}$

(E)

$6\sqrt{2}-\sqrt{3}$

How to solve

Strategy Organize Information in More Ways

The tempting move is to stare at the picture, decide by eye where the "good" arcs are, and start doing trigonometry inside some triangle. That skips the only genuinely hard part: nothing so far says which viewpoints see three whole sides, and an argument that assumes the arcs are where they look is proving nothing. So the real work is a change of bookkeeping — Tool #15 (Organize Information in More Ways): stop describing the hexagon as six segments and describe it as six inequalities, one per side line. Then "the entire side S_k is visible from P" becomes the single test "P lies strictly beyond the line carrying S_k", proved in both directions, and counting visible sides becomes counting satisfied inequalities. Tool #1 (Draw a Diagram) fixes coordinates and supplies the two numbers the hexagon contributes, its circumradius and its apothem. Tool #4 (Introduce a Variable) compresses all six inequalities into one angle α=arccos√3/r, a field-of-view half-width that carries every bit of the dependence on r. Tool #9 (Solve an Easier Related Problem) then uses the twelve symmetries to shrink the whole circle to a single 30° wedge. Tool #14 (Extreme Principle) finishes the counting inside that wedge by sorting the six angular gaps: only the third-smallest is ever in doubt, and the fourth-smallest can never qualify — which also settles that "three sides" and "exactly three sides" are the same event here. Tool #7 (Identify Subproblems) is the last step, an isolated exact-value computation of cos 75°.

1STEP 1

The hexagon as six inequalities

The hexagon becomes six inequalities, one per side.

R=2, a=√(2²-1²)=√3, H={X: X · n_k ≤ √3, k=0,…,5}, n_k=(cos 60k°,sin 60k°)

2STEP 2

When is a whole side visible

A whole side is visible exactly when the viewpoint is outside that side's line.

Z=λ P+(1-λ)Y, 0 ＜ λ ＜ 1 → Z · n_k=λ (P · n_k)+(1-λ)√3; #visible=#{k: P · n_k ＞ √3}

3STEP 3

One angle runs the whole count

So a single angle runs the whole count.

P · n_k=rcos(θ-60k°) ＞ √3⇔ d(θ,60k°) ＜ α, α=arccos√3/r∈(0°,90°)

4STEP 4

Twelve copies of one wedge

Twelve symmetries reduce the circle to one wedge.

12 symmetries → Pr=(measure of good t∈[0°,30°])/30°

5STEP 5

Three visible, never four

Three sides can show but never four.

t ≤ 60°-t ≤ 60°+t ≤ 120°-t ≤ 120°+t ≤ 180°-t; good⇔ 60°+t ＜ α; 120°-t ≥ 90° ＞ α

6STEP 6

Read off the probability

The given probability pins that angle at 75 degrees.

Pr=(α-60°)/30°=(12(α-60°))/360°=1/2 → α-60°=15° → α=75°

7STEP 7

Turn the angle into a length

Converting the angle to a length gives 3√(2)+√(6), choice (D).

r=√3/(cos 75°)=4√3/(√6-√2)=(4√3(√6+√2))/4=√(18)+√6=3√2+√6 → (D)

Answer

3√(2)+√(6)

Put the number straight back into the criterion. With r=3√2+√6≈ 6.6921 and apothem √3≈ 1.73205, the ratio is √3/r≈ 0.258819, and cos 75°≈ 0.258819 — agreement to six places. Better, check it exactly and without any inverse cosine. The claim behind the answer is that the good arc around the direction 0° stops exactly at θ=15°, which by Step 2 means the point r(cos 15°,sin 15°) lies exactly on the line carrying the side whose outward perpendicular points at -60°. Test it: that dot product is rcos(15°-(-60°))=rcos 75°=(3√2+√6)(√6-√2)/4=(3√(12)-6+6-√(12))/4=2√(12)/4=4√3/4=√3, which is precisely the apothem. The boundary lands on the side line exactly, so the six good arcs really are 30° wide and really do total 180°. Two structural checks. First, size: r≈ 6.69 is more than three side lengths out from the centre, and it has to be more than 2√3≈ 3.46 by Step 5, so the scale is right; it also cannot be enormous, since the probability climbs to 1 as r→∞. Second, the choices are deliberately close — 6.293, 6.610, 6.631, 6.692, 6.753 — so no amount of eyeballing separates them, but the formula (arccos(√3/r)-60°)/30° does: it gives about 0.4674 for (A), 0.4937 for (B), 0.4953 for (C), 0.5000 for (D), 0.5046 for (E). Only (D) is exactly one half, and the strict monotonicity from Step 6 says no other radius can be. One trap worth naming: at probability exactly 1/2 the good arcs and the bad arcs have equal width, so a solver who put the good arcs around the vertex directions by mistake would solve 180°-2α=30° and still get α=75° — the error is invisible at this one probability. That is why Step 6 tested t=0 and t=30 explicitly instead of trusting the picture.

💡Key takeaway

You see a whole side of a shape exactly when you have stepped past the flat wall that side lies in, so counting what is visible turns into counting how many of the six walls you are standing beyond — and then the whole problem is about angles.

The hexagon as six inequalities
When is a whole side visible
One angle runs the whole count
Twelve copies of one wedge
Three visible, never four
Read off the probability
Turn the angle into a length