AMC 10 · 2013 · #12

Grade 7 probability

Archetype Casework by Position / Vertex Type └ Sub-archetype Count of Type

probability-basiccombinations-basicregular-pentagon systematic-enumerationcomplementary-counting ↑ Prerequisites: probability-basiccombinations-basic

📏 Medium solution 💡 2 insights

Problem

Let $S$ be the set of sides and diagonals of a regular pentagon. A pair of elements of $S$ are selected at random without replacement. What is the probability that the two chosen segments have the same length?

Pick an answer.

(A)

$\frac{2}5$

(B)

$\frac{4}9$

(C)

$\frac{1}2$

(D)

$\frac{5}9$

(E)

$\frac{4}5$

AMC 10 2013 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.

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Toolkit + CCSS Solution

Understand

Restated: A regular pentagon has 5 sides and 5 diagonals, giving a set S of 10 segments. Two of these 10 segments are drawn at random, without replacement. Find the probability that the two drawn segments are the same length.

Givens: The pentagon is regular, so all 5 sides are equal in length; In a regular pentagon all 5 diagonals are also equal in length to each other; S contains all 10 segments: 5 sides plus 5 diagonals; Two distinct segments are chosen at random, each pair equally likely

Unknowns: The probability that the two chosen segments have equal length

Understand

Plan

Primary tool: #2 Make a Systematic List

Secondary: #1 Draw a Diagram, #16 Change Focus / Count the Complement

Every pair of the 10 segments is equally likely, so the probability is just (favorable pairs)/(total pairs). Counting pairs is a 'how many ways' job, so I count with combinations. Sketching the pentagon first shows why the lengths split into exactly two equal groups of 5. As a check I also change focus: fix the first pick and ask how many of the remaining 9 match it.

Execute — Answer: B

#1 Draw a Diagram 4.G.A.2 Step 1

Two equal groups of five

Draw the regular pentagon and all its segments.
Because the pentagon is regular, its 5 sides are all the same length.
Its 5 diagonals are also all the same length as one another (longer than the sides).
So the 10 segments in S fall into exactly two length classes: 5 sides of one length and 5 diagonals of another.

$$|S| = 5 \text{ sides} + 5 \text{ diagonals} = 10$$

💡 A regular shape's symmetry forces its sides equal and its diagonals equal, so only two lengths exist.

#2 Make a Systematic List 7.SP.C.8 Step 2

Count all possible pairs

Choosing 2 segments from the 10 in S, order not mattering, is a combination.
The number of ways is 10 choose 2.

$$\binom{10}{2} = \dfrac{10 \times 9}{2} = 45$$

💡 Listing every unordered pair once is exactly what 'n choose 2' counts.

#2 Make a Systematic List 7.SP.C.8 Step 3

Count the matching pairs

Two segments match in length only if both are sides or both are diagonals.
Two sides from the 5 sides can be picked in 5 choose 2 ways, and likewise two diagonals in 5 choose 2 ways.
Add these two disjoint cases.

$$\binom{5}{2} + \binom{5}{2} = 10 + 10 = 20$$

💡 A same-length pair must come entirely from one length group, so count each group separately and add.

#2 Make a Systematic List 7.SP.C.7 Step 4

Divide to get the probability

The probability is the favorable pairs over the total pairs.
That gives 20/45, which reduces to 4/9.
So the answer is (B).

$$P = \dfrac{20}{45} = \dfrac{4}{9}$$

💡 With every pair equally likely, probability is simply the share of pairs that match.

[1] #1 4.G.A.2 Draw the regular pentagon and all its segments. Because the pentagon is regular,

[2] #2 7.SP.C.8 Choosing 2 segments from the 10 in S, order not mattering, is a combination. The

[3] #2 7.SP.C.8 Two segments match in length only if both are sides or both are diagonals. Two s

[4] #2 7.SP.C.7 The probability is the favorable pairs over the total pairs. That gives 20/45, w

Review

Reasonableness: 4/9 is just under 1/2, which fits: after your first segment is fixed, only 4 of the other 9 segments share its length, and 4 out of 9 is a bit below half. The tempting wrong answer (C) 1/2 comes from mistakenly assuming exactly half the leftovers match. Since 4/9 < 1/2, choice (C) is the trap and (B) is right.

Alternative: Change focus and skip the combinations entirely: pick any first segment. Whatever it is, its length group has 5 members, so 4 other segments share its length, out of 9 segments remaining. The probability is 4/9 immediately.

CCSS standards used (min grade 7)

4.G.A.2 Classify two-dimensional figures based on presence of parallel or perpendicular lines (Using the regular pentagon's symmetry to see its 5 sides are equal and its 5 diagonals are equal, giving two length groups of 5)
7.SP.C.8 Find probabilities of compound events using organized lists, tables, and simulation (Counting the 45 total unordered pairs and the 20 same-length pairs with combinations)
7.SP.C.7 Develop probability models and use them to find probabilities of events (Forming the equally-likely model and dividing favorable pairs by total pairs to get 4/9)

⭐ The 10 segments split into two equal groups of 5, so a match means both come from the same group — count those pairs over all pairs and you get 4/9.

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: B

Two equal groups of five

Count all possible pairs

Count the matching pairs

Divide to get the probability

Review

CCSS standards used (min grade 7)

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