AMC 8 · 2019 · #6
Grade 7 geometry-2dprobabilityProblem
There are 81 grid points (uniformly spaced) in the square shown in the diagram below, including the points on the edges. Point P is in the center of the square. Given that point Q is randomly chosen among the other 80 points, what is the probability that the line PQ is a line of symmetry for the square?
Pick an answer.
AMC 8 2019 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 square contains $81$ evenly spaced grid points arranged as a $9 \times 9$ array (including the edge points). Point $P$ sits at the very center. Another point $Q$ is chosen at random from the remaining $80$ grid points. What is the probability that the segment $PQ$ lies along a line of symmetry of the square?
Givens: $81$ grid points form a $9 \times 9$ uniformly spaced square array; $P$ is the center point of the square (row 5, column 5); $Q$ is picked uniformly at random from the other $80$ points; Answer choices: (A) $\tfrac{1}{5}$, (B) $\tfrac{1}{4}$, (C) $\tfrac{2}{5}$, (D) $\tfrac{9}{20}$, (E) $\tfrac{1}{2}$
Unknowns: The probability that line $PQ$ is a line of symmetry of the square
Understand
Restated: A square contains $81$ evenly spaced grid points arranged as a $9 \times 9$ array (including the edge points). Point $P$ sits at the very center. Another point $Q$ is chosen at random from the remaining $80$ grid points. What is the probability that the segment $PQ$ lies along a line of symmetry of the square?
Givens: $81$ grid points form a $9 \times 9$ uniformly spaced square array; $P$ is the center point of the square (row 5, column 5); $Q$ is picked uniformly at random from the other $80$ points; Answer choices: (A) $\tfrac{1}{5}$, (B) $\tfrac{1}{4}$, (C) $\tfrac{2}{5}$, (D) $\tfrac{9}{20}$, (E) $\tfrac{1}{2}$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #7 Identify Subproblems, #2 Make a Systematic List
The problem already gives us a $9 \times 9$ 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) $\div$ $80$.
Execute — Answer: C
4.G.A.3 Step 1 - Sketch the four lines of symmetry of a square through the center $P$: one horizontal, one vertical, and the two diagonals from corner to corner.
- A square has exactly these four axes of symmetry, and every one of them passes through the center, so $PQ$ is a symmetry line of the square exactly when $Q$ lies on one of these four lines.
💡 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.
4.G.A.3 Step 2 - Break the favorable count into one subproblem per symmetry axis: how many grid points (other than $P$) lie on the horizontal axis, the vertical axis, the $\nearrow$ diagonal, and the $\nwarrow$ diagonal?
- Because all four axes meet only at $P$ — and we are excluding $P$ — the four subcounts have no overlap, so we can just add them.
💡 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.
3.OA.A.3 Step 3 - Count the grid points on each axis with a systematic list.
- The horizontal axis through $P$ is one full row of the $9 \times 9$ grid, so it has $9$ grid points; remove $P$ and $8$ remain.
- By the same reasoning the vertical axis contributes $8$ more.
- Each diagonal of the square also passes through exactly $9$ grid points of the $9 \times 9$ array (from one corner straight to the opposite corner, hitting one dot per column), so each diagonal contributes $9 - 1 = 8$ points after excluding $P$.
💡 $4$ groups of $8$ is just a Grade 3 multiplication word problem: $4 \times 8 = 32$.
7.SP.C.5 Step 4 - Compute the probability as favorable points divided by the size of the sample space.
- The sample space is the $80$ non-$P$ grid points, each equally likely, so the probability is $\tfrac{32}{80}$.
- Reduce by dividing numerator and denominator by their greatest common factor $16$.
💡 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.
4.G.A.3 Sketch the four lines of symmetry of a square through the center $P$: one horizo 4.G.A.3 Break the favorable count into one subproblem per symmetry axis: how many grid p 3.OA.A.3 Count the grid points on each axis with a systematic list. The horizontal axis t 7.SP.C.5 Compute the probability as favorable points divided by the size of the sample sp Review
Reasonableness: Sanity check the number $32$ against the grid: each symmetry axis is a chord of the square that obviously hits $9$ collinear grid points (one per row or per column or per diagonal step), and the four axes share only the center $P$. So $4 \times 9 = 36$ total hits, but $P$ is counted four times and we want to exclude it entirely, giving $36 - 4 = 32$ — the same number. The probability $\tfrac{2}{5} = 0.4$ also passes the gut check: $4$ special directions out of a roughly evenly distributed cloud of $80$ points should land somewhere noticeably below $\tfrac{1}{2}$ but not as small as $\tfrac{1}{5}$, which matches (C) nicely.
Alternative: Tool #3 (Eliminate Possibilities) cuts the choices fast: the favorable count must be a whole number, so the probability written over $80$ must reduce from $\tfrac{n}{80}$ with $n$ a positive integer. Checking each option: (A) $\tfrac{1}{5} = \tfrac{16}{80}$, (B) $\tfrac{1}{4} = \tfrac{20}{80}$, (C) $\tfrac{2}{5} = \tfrac{32}{80}$, (D) $\tfrac{9}{20} = \tfrac{36}{80}$, (E) $\tfrac{1}{2} = \tfrac{40}{80}$. Only the value matching $4$ symmetry axes $\times$ $8$ non-center points each $= 32$ survives, which is (C).
CCSS standards used (min grade 7)
4.G.A.3Recognize a line of symmetry for a two-dimensional figure (Identifying that a square has exactly $4$ lines of symmetry and that each of them passes through the center point $P$, which is what makes $PQ$ a symmetry line iff $Q$ lies on one of those $4$ axes.)3.OA.A.3Solve multiplication and division word problems within 100 (Adding $4$ equal groups of $8$ favorable points — $4 \times 8 = 32$ — to get the total number of grid points on the symmetry axes (excluding $P$).)7.SP.C.5Understand that the probability of a chance event is between 0 and 1 (Forming the probability as $\tfrac{\text{favorable}}{\text{total}} = \tfrac{32}{80} = \tfrac{2}{5}$, the formal Grade 7 definition of probability for equally likely outcomes.)
⭐ This AMC 8 problem only needs Grade 7 probability — favorable outcomes over total outcomes — that you already know!
⭐ This AMC 8 problem only needs Grade 7 probability — favorable outcomes over total outcomes — that you already know!
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