AMC 8 · 2008 · #19

Grade 8 geometry-2dcounting

Archetype Casework by Position / Vertex Type └ Sub-archetype Grid / Solid Cell Position

combinations-basicprobability-basiccoordinate-geometrysystematic-enumeration systematic-enumerationcasework ↑ Prerequisites: combinations-basicprobability-basic

📏 Short solution 💡 2 insights 📊 Diagram

Problem

Eight points are spaced around at intervals of one unit around a $2 \times 2$ square, as shown. Two of the $8$ points are chosen at random. What is the probability that the two points are one unit apart?

Pick an answer.

(A)

$\frac{1}{4}$

(B)

$\frac{2}{7}$

(C)

$\frac{4}{11}$

(D)

$\frac{1}{2}$

(E)

$\frac{4}{7}$

AMC 8 2008 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.

View mode:

Toolkit + CCSS Solution

Understand

Restated: Eight points sit at one-unit intervals around the perimeter of a $2 \times 2$ square — the four corners plus the midpoint of each side. Two of the eight points are picked at random. What is the probability that the two chosen points are exactly one unit apart?

Givens: A $2 \times 2$ square with $8$ points on its perimeter, spaced $1$ unit apart; Two distinct points are chosen at random from the $8$; Every pair of points is equally likely; Answer choices: (A) $\tfrac{1}{4}$, (B) $\tfrac{2}{7}$, (C) $\tfrac{4}{11}$, (D) $\tfrac{1}{2}$, (E) $\tfrac{4}{7}$

Unknowns: The probability that the two chosen points are exactly $1$ unit apart

Understand

Plan

Primary tool: #13 Count Without Listing One-by-One

Secondary: #15 Visualize / Draw a Diagram

The problem is a classic favorable-over-total probability. Tool #13 (Count Without Listing) gives the total: $\binom{8}{2} = 28$ pairs, no listing needed. Tool #15 (Visualize) handles the favorable count: by drawing the $8$ points around the square, the unit-apart pairs are exactly the $8$ short segments around the perimeter — easy to see and count once the picture is clear. Together they give $\dfrac{8}{28} = \dfrac{2}{7}$.

Execute — Answer: B

#13 Count Without Listing One-by-One 7.SP.C.8 Step 1

Count the total number of pairs.
Two points are picked from the $8$, and the order does not matter, so use the combination $\binom{8}{2}$.

$$\binom{8}{2} = \dfrac{8 \times 7}{2} = 28$$

💡 Each of the $8$ points can pair with any of the other $7$, giving $8 \times 7 = 56$ ordered picks. Each unordered pair is counted twice, so divide by $2$.

#15 Visualize / Draw a Diagram 6.G.A.3 Step 2

Visualize the figure.
The $8$ points are the $4$ corners and the $4$ side-midpoints of the $2 \times 2$ square.
Going around the perimeter they alternate corner, midpoint, corner, midpoint, $\ldots$, with consecutive points exactly $1$ unit apart.

$$\text{8 points around the square} \to \text{8 unit-length perimeter gaps}$$

💡 Each side of the $2 \times 2$ square has length $2$, and the midpoint splits it into two unit segments — so each side contributes $2$ unit-apart pairs, and four sides give $8$ pairs.

#13 Count Without Listing One-by-One 8.G.B.7 Step 3

Confirm no other pair is $1$ unit apart.
The next-shortest distances skip a point: corner-to-corner along a side is $2$, midpoint-to-midpoint of adjacent sides is $\sqrt{2}$, corner-to-opposite-midpoint is $\sqrt{5}$.
All exceed $1$, so the $8$ perimeter pairs are the only favorable ones.

$$\text{favorable pairs} = 8$$

💡 Use the Pythagorean rule to check non-adjacent distances; every other pair is at least $\sqrt{2} > 1$.

#13 Count Without Listing One-by-One 7.SP.C.7 Step 4

Form the probability and simplify by dividing numerator and denominator by $4$.

$$P = \dfrac{8}{28} = \dfrac{2}{7} \;\Rightarrow\; \textbf{(B)}$$

💡 Every one of the $28$ pairs is equally likely, so the probability is just the fraction of pairs that are favorable.

[1] #13 7.SP.C.8 Count the total number of pairs. Two points are picked from the $8$, and the ord

[2] #15 6.G.A.3 Visualize the figure. The $8$ points are the $4$ corners and the $4$ side-midpoi

[3] #13 8.G.B.7 Confirm no other pair is $1$ unit apart. The next-shortest distances skip a poin

[4] #13 7.SP.C.7 Form the probability and simplify by dividing numerator and denominator by $4$.

Review

Reasonableness: Sanity-check with the symmetry trick: pick any single point first; among the remaining $7$ points, exactly $2$ are its perimeter neighbors at distance $1$ (one on each side). So the probability that the second pick lands at distance $1$ is $\tfrac{2}{7}$, matching the answer (B). This works for any starting point — corner or midpoint — because each point has exactly two unit-distance neighbors. The other choices are easy to rule out: (A) $\tfrac{1}{4} = \tfrac{7}{28}$ would need $7$ favorable pairs; (C)–(E) would need $10$ or more, but there are only $8$.

Alternative: Tool #16 (Change Your Focus): drop the count-pairs framing and focus on a single chosen point. By the rotational symmetry of the $8$-point figure, every point has exactly $2$ neighbors at distance $1$ out of $7$ other points. So the conditional probability is $\tfrac{2}{7}$ regardless of which point came first — answer (B) — in one line, with no combination formula.

CCSS standards used (min grade 8)

6.G.A.3 Draw polygons in the coordinate plane and find side lengths (Placing the $8$ points around the $2 \times 2$ square and seeing that consecutive perimeter points are $1$ unit apart.)
7.SP.C.7 Develop a probability model and use it to find probabilities of events (Treating each of the $28$ pairs as equally likely and computing favorable $/$ total.)
7.SP.C.8 Find probabilities of compound events using organized lists, tables, tree diagrams, and simulation (Counting the total number of unordered pairs as $\binom{8}{2} = 28$ without listing them.)
8.G.B.7 Apply the Pythagorean Theorem to find the distance between points (Checking that non-adjacent pairs (across a side, across a diagonal, etc.) all have distance greater than $1$, so only the $8$ perimeter pairs are favorable.)

⭐ Total pairs: $\binom{8}{2} = 28$. Unit-apart pairs: the $8$ short hops around the square. Probability: $\tfrac{8}{28} = \tfrac{2}{7}$.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: B

Review

CCSS standards used (min grade 8)

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