AMC 10 · 2023 · #25

Grade 7 probability

probability-basicspatial-visualizationpolyhedron-netssymmetry-argument complementary-countingcaseworksystematic-enumeration ↑ Prerequisites: probability-basicspatial-visualization

📏 Medium solution 💡 3 insights

Problem

If $A$ and $B$ are vertices of a polyhedron, define the distance $d(A,B)$ to be the minimum number of edges of the polyhedron one must traverse in order to connect $A$ and $B$ . For example, if $\overline{AB}$ is an edge of the polyhedron, then $d(A, B) = 1$ , but if $\overline{AC}$ and $\overline{CB}$ are edges and $\overline{AB}$ is not an edge, then $d(A, B) = 2$ . Let $Q$ , $R$ , and $S$ be randomly chosen distinct vertices of a regular icosahedron (regular polyhedron made up of 20 equilateral triangles). What is the probability that $d(Q, R) > d(R, S)$ ?

Pick an answer.

(A)

$frac{7}{22}$

(B)

$frac{1}{3}$

(C)

$frac{3}{8}$

(D)

$frac{5}{12}$

(E)

$frac{1}{2}$

AMC 10 2023 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: On a regular icosahedron ($20$ triangular faces, $12$ vertices), define the distance $d(A,B)$ as the minimum number of edges in a path from $A$ to $B$. Three distinct vertices $Q, R, S$ are chosen at random. Find the probability that $d(Q, R) > d(R, S)$.

Givens: Icosahedron: $12$ vertices, $30$ edges, $20$ triangular faces; $5$ triangles (and so $5$ edges) meet at each vertex; $d(A, B)$ is the minimum edge-count path from $A$ to $B$; $Q, R, S$ are three distinct vertices chosen uniformly at random; Answer choices: (A) $\tfrac{7}{22}$, (B) $\tfrac{1}{3}$, (C) $\tfrac{3}{8}$, (D) $\tfrac{5}{12}$, (E) $\tfrac{1}{2}$

Unknowns: The probability $P(d(Q, R) > d(R, S))$

Understand

Plan

Primary tool: #16 Change Focus / Count the Complement

Secondary: #10 Create a Physical Representation, #2 Make a Systematic List

Directly counting triples with $d(Q,R) > d(R,S)$ would require casework on three distance values for each of $Q$ and $S$. Tool #16 (Change Focus) is the smart move: by symmetry, $P(>) = P(<)$, so $1 = P(>) + P(<) + P(=) = 2 P(>) + P(=)$. We just need $P(=)$. Tool #10 (Create a Physical Representation) — build or visualize a model of the icosahedron — gives the vertex-distance distribution $(5, 5, 1)$ from any fixed vertex. Then Tool #2 (Systematic List) counts ordered pairs $(Q, S)$ with $d(R, Q) = d(R, S)$ by summing over distance classes.

Execute — Answer: A

#10 Create a Physical Representation 6.G.A.4 Step 1

Find the distance distribution from a fixed vertex.
By symmetry, fix any vertex $R$.
The $5$ vertices sharing an edge with $R$ are at distance $1$.
The antipodal vertex (directly opposite) is at distance $3$ (and only $1$ such vertex exists).
Subtracting from $12 - 1 = 11$ other vertices: $11 - 5 - 1 = 5$ vertices are at distance $2$.

$$\text{From } R: \;5 \text{ at } d=1,\;5 \text{ at } d=2,\;1 \text{ at } d=3$$

💡 Grade 6 "represent 3D figures via nets/surface" — picture or build the icosahedron and count neighbors at each ring.

#16 Change Focus / Count the Complement 7.SP.C.7 Step 2

Use symmetry between $Q$ and $S$.
Since $Q$ and $S$ are drawn the same way (just two of the $11$ non-$R$ vertices), $P(d(Q,R) > d(R,S)) = P(d(Q,R) < d(R,S))$.
Together with $P(=)$ these three events partition all outcomes: $1 = 2 P(>) + P(=)$, so $P(>) = \tfrac{1 - P(=)}{2}$.

$$P(>) = \tfrac{1 - P(=)}{2}$$

💡 Grade 7 probability — by symmetry the $>$ and $<$ events are equally likely, so we only need $P(=)$.

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

Count the total ordered pairs $(Q, S)$ with $Q \ne S$ and both different from $R$.
There are $11$ ways to pick $Q$ and $10$ remaining for $S$, so $110$ ordered pairs.

$$11 \times 10 = 110$$

💡 Grade 7 counting principle — multiply choices.

#2 Make a Systematic List 7.SP.C.8 Step 4

Count the ordered pairs with $d(R, Q) = d(R, S)$ by listing cases on the common distance value.
Case $d=1$: pick $Q$ from the $5$ neighbors, then $S$ from the remaining $4$: $5 \times 4 = 20$.
Case $d=2$: similarly $5 \times 4 = 20$.
Case $d=3$: only $1$ vertex at distance $3$, cannot pick two distinct vertices there: $0$.
Total equal-distance pairs $= 20 + 20 + 0 = 40$.

$$\#\{(Q,S): d(R,Q)=d(R,S)\} = 20 + 20 + 0 = 40$$

💡 Grade 7 "sample space via organized list" — sum over the three possible common distances.

#16 Change Focus / Count the Complement 7.SP.C.7 Step 5

Compute $P(=)$ and finish.
$P(=) = \tfrac{40}{110} = \tfrac{4}{11}$.
Then $P(>) = \tfrac{1 - \tfrac{4}{11}}{2} = \tfrac{7/11}{2} = \tfrac{7}{22}$, which is choice (A).

$$P(>) = \tfrac{1 - 4/11}{2} = \tfrac{7}{22} \;\Rightarrow\; \textbf{(A)}$$

💡 Plug $P(=) = \tfrac{4}{11}$ into the symmetry formula from step 2 — done.

[1] #10 6.G.A.4 Find the distance distribution from a fixed vertex. By symmetry, fix any vertex

[2] #16 7.SP.C.7 Use symmetry between $Q$ and $S$. Since $Q$ and $S$ are drawn the same way (just

[3] #2 7.SP.C.8 Count the total ordered pairs $(Q, S)$ with $Q \ne S$ and both different from $R

[4] #2 7.SP.C.8 Count the ordered pairs with $d(R, Q) = d(R, S)$ by listing cases on the common

[5] #16 7.SP.C.7 Compute $P(=)$ and finish. $P(=) = \tfrac{40}{110} = \tfrac{4}{11}$. Then $P(>)

Review

Reasonableness: Sanity. $P(>) = \tfrac{7}{22} \approx 0.318$, just under $\tfrac{1}{3}$, which is the right ballpark: if all $11 \cdot 10 = 110$ pairs were three-way-equally split among $>, <, =$, each would be $\tfrac{1}{3}$, but the $=$ event is slightly less likely than $\tfrac{1}{3}$ ($\tfrac{4}{11} \approx 0.364$ is just a touch above $\tfrac{1}{3}$, leaving a bit less than $\tfrac{1}{3}$ for $>$). The distance distribution $(5, 5, 1)$ accounts for all $11$ non-$R$ vertices, and the count $40$ properly excludes the single-antipodal class. The fraction $\tfrac{7}{22}$ is one of the offered choices exactly.

Alternative: Tool #3 (Eliminate Possibilities) on the choices: any answer with denominator $12$ (such as $\tfrac{5}{12}$) cannot come from $11 \times 10$ ordered pairs because $12 \nmid 110$. Similarly $\tfrac{3}{8}$ and $\tfrac{1}{3}$ would need denominators $8$ or $3$, but $110 = 2 \cdot 5 \cdot 11$ has $11$ in it — the answer must have an $11$ in the denominator after reducing. Of the choices, only $\tfrac{7}{22}$ has $11$ in the denominator, so even without the full count it must be (A).

CCSS standards used (min grade 7)

6.G.A.4 Represent three-dimensional figures using nets made up of rectangles and triangles, and use the nets to find the surface area (Visualizing the icosahedron (20 triangular faces, 12 vertices, 5 triangles meeting at each vertex) to read off the distance distribution from any fixed vertex.)
7.SP.C.7 Develop a probability model and use it to find probabilities of events; compare probabilities from a model to observed frequencies (Setting up the uniform probability model on triples of distinct vertices and using symmetry to write $P(>) = (1 - P(=))/2$.)
7.SP.C.8 Find probabilities of compound events using organized lists, tables, tree diagrams, and simulation (Counting $110$ ordered pairs $(Q, S)$ and the $40$ with equal distance from $R$ by summing over the three possible common distances.)

⭐ This AMC 10 problem only needs Grade 7 probability you already know — picture the icosahedron, see that from any vertex the other $11$ split $5 + 5 + 1$, then use $P(>) = P(<)$ to write $P(>) = (1 - P(=))/2$. Counting the $40$ equal-distance pairs out of $110$ gives $\tfrac{7}{22}$.