AMC 10 · 2011 · #16

Grade 8 geometry-2dprobability

Archetype Geometric Probability by Area └ Sub-archetype Area-Proportional Region Probability

geometric-probabilityarea-rectanglesarea-triangles identify-subproblems ↑ Prerequisites: area-triangles

📏 Medium solution 💡 3 insights 📊 Diagram

Problem

A dart lands somewhere on a regular octagon dartboard, with every point equally likely. The board is cut by four lines into a center square, four rectangles, and four corner triangles. Find the chance the dart lands inside the center square.

Pick an answer.

(A)

$\frac{\sqrt{2} - 1}{2}$

(B)

$\frac{1}{4}$

(C)

$\frac{2 - \sqrt{2}}{2}$

(D)

$\frac{\sqrt{2}}{4}$

(E)

$2 - \sqrt{2}$

AMC 10 2011 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Identify Subproblems

Because the dart is uniform, the probability is just one area divided by another, so I never need real measurements — only a ratio. The octagon is an awkward shape, so I break it into pieces I can measure: one center square, four rectangles, four corner triangles. I pick a convenient size for the pieces (a ratio is unaffected by scale), add up the octagon's area, and divide the square's area by it.

1STEP 1

Turn the probability into an area ratio

The dart is uniform, so the probability is the center square's area divided by the octagon's area — find those two areas.

P = (area of center square)/(area of octagon)

2STEP 2

Cut the octagon into simple pieces

The four cuts split the board into nine regions; a ratio ignores scale, so give each corner triangle legs of length $1$ .

octagon = 1 square + 4 rectangles + 4 corner triangles

3STEP 3

Measure a corner triangle and the shared length

Legs $1$ and $1$ give hypotenuse $\sqrt{2}$ , an octagon side and the square's side; the four triangles have total area $2$ .

hyp = √(1² + 1²) = √(2); 4 × 1/2(1)(1) = 2

4STEP 4

Measure the square and the rectangles

The square's side $\sqrt{2}$ makes its area $2$ ; each rectangle is $1$ by $\sqrt{2}$ , so the four total $4\sqrt{2}$ .

square = (√(2))² = 2; 4 rectangles = 4(1 · √(2)) = 4√(2)

5STEP 5

Add up and divide

Total area is $2+4\sqrt{2}+2$ , so the probability $\frac{2}{4+4\sqrt{2}}$ rationalizes to $\frac{\sqrt{2}-1}{2}$ , choice (A).

P = 2/(4 + 4√(2)) = 1/(2(1+√(2))) = (√(2)-1)/2

Answer

(√(2) - 1)/2

The center square is one of nine regions and is clearly not the biggest chunk of the board, so a probability well under one third makes sense. Numerically (square root of 2 minus 1) over 2 is about 0.207, roughly one fifth, which matches how small the middle square looks in the figure. The other choices are larger or, like 1/4, do not survive the exact area computation.

💡Key takeaway

For an even target, probability is just the target's area over the whole area — so cut the hard shape into squares, rectangles, and triangles you can measure.

Turn the probability into an area ratio
Cut the octagon into simple pieces
Measure a corner triangle and the shared length
Measure the square and the rectangles
Add up and divide

Turn the probability into an area ratio

Cut the octagon into simple pieces

Measure a corner triangle and the shared length

Measure the square and the rectangles

Add up and divide

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