Competition · AMC preparation · step 4 of 4
AMC 8 · 2019 · #6
Grade 7 geometry-2dprobability
Pick an answer.
AMC 8 2019 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
The problem already gives us a 9 × 9 grid with P at the center, so Tool #1 (Draw a Diagram) is the natural entry point — sketch the four lines of symmetry of the square right on top of the grid and the favorable Q points become visible as the dots those lines hit. Tool #7 (Identify Subproblems) then breaks the count into four independent pieces (one per symmetry axis), and Tool #2 (Make a Systematic List) tallies the grid points along each axis without double-counting P. The final probability is just (favorable points) ÷ 80.
Draw the four symmetry axes
A square has exactly four lines of symmetry, all through P — horizontal, vertical, two diagonals — so PQ is an axis only when Q sits on one.
Grade 4 students already learn that a square has 4 lines of symmetry — drawing them on the grid turns the probability question into a counting question.
The segment PQ lies along a line of symmetry of the square exactly when Q lands on one of the square's four symmetry axes, and every one of those axes runs through the center point P.
▸ Why?
A line of symmetry folds the square exactly onto itself, and such a fold must hold the center still, so every symmetry axis passes through the center P; since PQ already goes through P, it can be a symmetry line only when Q sits somewhere on one of those axes.
▸ Why?
Folding is a flip, and a flip that lays the square back onto itself cannot move the one point sitting the same distance from all four sides, so that center point stays on the fold line.
▸ Why?
Only four folds lay the square back onto itself — across the horizontal, the vertical, and the two corner-to-corner diagonals — so the square has exactly these four symmetry axes and no others.
▸ Why?
Testing folds, a flip matches the square onto itself only along those four lines; along any other line part of the square swings past the edge instead of landing on the copy.
Split the count by axis
Count each axis separately; the four axes meet only at P (excluded), so the four subcounts are disjoint and simply add.
Splitting one hard count into four clean, non-overlapping counts is the Tool #7 move — and it's safe because P is the only shared point.
4.G.A.3Identify SubproblemsCount the points on each axis
Each axis is a line of 9 grid points, so each gives 9 - 1 = 8 after dropping P; four disjoint axes give 32 favorable points for Q.
4 groups of 8 is just a Grade 3 multiplication word problem: 4 × 8 = 32.
3.OA.A.3Make A Systematic ListDivide to get the probability
Divide favorable by the 80 equally likely points: , which reduces (dividing by 16) to — choice (C).
The Grade 7 definition of probability — favorable outcomes over total equally likely outcomes — is the only place "probability" really enters; the rest was just counting and reducing a fraction.
7.SP.C.5Identify SubproblemsThis AMC 8 problem only needs Grade 7 probability — favorable outcomes over total outcomes — that you already know!
- Draw the four symmetry axes
- Split the count by axis
- Count the points on each axis
- Divide to get the probability
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