AMC 8 · 2019 · #6

Grade 7 geometry-2dprobability

Archetype Casework by Position / Vertex Type

line-symmetryprobability-basicsystematic-enumeration caseworksystematic-enumeration ↑ Prerequisites: line-symmetryprobability-basic

📏 Medium solution 💡 2 insights 📊 Diagram

Problem

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?

$\textbf{(A) }\frac{1}{5}\qquad\textbf{(B) }\frac{1}{4} \qquad\textbf{(C) }\frac{2}{5} \qquad\textbf{(D) }\frac{9}{20} \qquad\textbf{(E) }\frac{1}{2}$

Pick an answer.

(A)

$\frac{1}{5}$

(B)

$\frac{1}{4}$

(C)

$\frac{2}{5}$

(D)

$\frac{9}{20}$

(E)

$\frac{1}{2}$

View mode:

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

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

#1 Draw a Diagram 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.

$$\text{Symmetry axes through } P: \text{ horizontal, vertical, diagonal}_\nearrow, \text{ diagonal}_\nwarrow$$

💡 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.

#7 Identify Subproblems 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.

$$\text{favorable} = (\text{horiz}) + (\text{vert}) + (\text{diag}_\nearrow) + (\text{diag}_\nwarrow)$$

💡 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.

#2 Make a Systematic List 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 $P$.

$$8 + 8 + 8 + 8 = 32 \text{ favorable points for } Q$$

💡 $4$ groups of $8$ is just a Grade 3 multiplication word problem: $4 \times 8 = 32$.

#7 Identify Subproblems 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 $16$.

$$P(\text{symmetry}) = \dfrac{32}{80} = \dfrac{32 \div 16}{80 \div 16} = \dfrac{2}{5} \;\Rightarrow\; \textbf{(C)}$$

💡 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.

[1] #1 4.G.A.3 Sketch the four lines of symmetry of a square through the center $P$: one horizo

[2] #7 4.G.A.3 Break the favorable count into one subproblem per symmetry axis: how many grid p

[3] #2 3.OA.A.3 Count the grid points on each axis with a systematic list. The horizontal axis t

[4] #7 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.3 Recognize 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.3 Solve 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.5 Understand 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!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: C

Review

CCSS standards used (min grade 7)

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