AMC 10 · 2011 · #16
Grade 8 geometry-2dA dart board is a regular octagon divided into regions as shown. Suppose that a dart thrown at the board is equally likely to land anywhere on the board. What is the probability that the dart lands within the center square?
Pick an answer.
AMC 10 2011 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: 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.
Givens: The board is a regular octagon (all sides equal, all angles equal).; Four segments are drawn joining opposite edges; they bound a square in the middle.; The dart is equally likely to land at any point of the board.
Unknowns: The probability that the dart lands inside the center square.
Understand
Restated: 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.
Givens: The board is a regular octagon (all sides equal, all angles equal).; Four segments are drawn joining opposite edges; they bound a square in the middle.; The dart is equally likely to land at any point of the board.
Plan
Primary tool: #7 Identify Subproblems
Secondary: #1 Draw a Diagram, #4 Introduce a Variable, #3 Eliminate Possibilities
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.
Execute — Answer: A
7.SP.C.7 Step 1 Turn the probability into an area ratio
- Since the dart is equally likely to land anywhere, the probability of landing in the center square is the square's area divided by the whole octagon's area.
- So the whole problem becomes: find these two areas.
💡 For an evenly-spread target, chance is just the fraction of the surface you are aiming at.
6.G.A.1 Step 2 Cut the octagon into simple pieces
- The four drawn lines split the octagon into nine regions: the center square, four rectangles along the sides, and four right-triangle corners.
- A ratio does not depend on scale, so I give the corner triangles legs of length 1.
- Each corner triangle is a right isosceles triangle with both legs equal to 1.
💡 A hard shape becomes easy once you slice it into squares, rectangles, and triangles you already know how to measure.
8.G.B.7 Step 3 Measure a corner triangle and the shared length
- Each corner triangle has legs 1 and 1, so by the Pythagorean theorem its hypotenuse is the square root of 1 squared plus 1 squared, which is the square root of 2.
- That hypotenuse is a side of the octagon, and it is also the side of the center square.
- Each triangle's area is one half times 1 times 1, so the four triangles together have area 2.
💡 The slanted cut of a corner is the hypotenuse of a right triangle, so the Pythagorean theorem hands you its length.
6.G.A.1 Step 4 Measure the square and the rectangles
- The center square has side square-root-of-2, so its area is the square root of 2, squared, which equals 2.
- Each rectangle has one side of length 1 (touching a triangle leg) and one side of length square-root-of-2 (the octagon side), so its area is square root of 2.
- The four rectangles together have area 4 times the square root of 2.
💡 Once you know each piece's side lengths, its area is just length times width.
8.NS.A.1 Step 5 Add up and divide
- The octagon's total area is the square plus the rectangles plus the triangles: 2 plus 4 times the square root of 2 plus 2, which is 4 plus 4 times the square root of 2.
- The probability is 2 divided by that.
- Simplify by pulling out the 4, then multiply top and bottom by (square-root-of-2 minus 1); since (square-root-of-2 plus 1)(square-root-of-2 minus 1) equals 1, the fraction cleans up to (square root of 2, minus 1) over 2.
- This is choice (A).
💡 Clearing the square root from the bottom turns the messy fraction into one of the listed answers.
7.SP.C.7 Since the dart is equally likely to land anywhere, the probability of landing in 6.G.A.1 The four drawn lines split the octagon into nine regions: the center square, fou 8.G.B.7 Each corner triangle has legs 1 and 1, so by the Pythagorean theorem its hypoten 6.G.A.1 The center square has side square-root-of-2, so its area is the square root of 2 8.NS.A.1 The octagon's total area is the square plus the rectangles plus the triangles: 2 Review
Reasonableness: 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.
Alternative: Instead of choosing triangle legs of 1, use the octagon-area formula for side length s: area equals 2(1 + square root of 2) times s squared. With s equal to the octagon side and the center square also of side s, the square's area is s squared, and dividing gives 1 over 2(1 + square root of 2), the same value (square root of 2 minus 1) over 2.
CCSS standards used (min grade 8)
7.SP.C.7Develop probability models and use them to find probabilities of events (Turning 'equally likely everywhere' into probability equals target area over total area.)6.G.A.1Find area of triangles, special quadrilaterals, and polygons by composing (Decomposing the octagon into a square, four rectangles, and four triangles and adding their areas.)8.G.B.7Apply the Pythagorean theorem to determine unknown side lengths in right triangles (Finding the hypotenuse (square root of 2) of the right isosceles corner triangles, which is the square's side.)8.NS.A.1Know that numbers that are not rational are called irrational numbers (Working with the square root of 2 and rationalizing the final fraction.)
⭐ 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.
⭐ 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.
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