2023 AMC 10 A Problem 25

Direct Counting Method

Since the icosahedron is symmetric, we can fix vertex R. We group the other 11 vertices by their distance from R, and then systematically count the pairs (Q, S) that satisfy the condition by considering each possible distance for S.

Step 1: Identify the distance distribution from any vertex

A regular icosahedron has 12 vertices. Pick any vertex R. Imagine holding the icosahedron by R: – R connects to a ring of 5 neighbors. These are at distance 1. – Directly opposite to R is a single “antipodal” vertex. Its distance from R is 3. – The remaining 5 vertices form a ring connected to the antipodal vertex. Their distance from R is 2. Thus, the 11 available vertices are distributed as: 5 at distance 1, 5 at distance 2, and 1 at distance 3.

Step 2: Set up the total possible outcomes

Because Q, R, and S are distinct randomly chosen vertices, we can analyze the probability by assuming R is already fixed. We then choose S and Q from the remaining 11 vertices. Since order matters (Q and S serve distinct roles in the inequality), we calculate the number of permutations.

Step 3: Count the successful pairs by conditioning on the distance to S

We want to find all pairs where the distance to Q is greater than the distance to S. We break this down by looking at the three possible values for the distance between R and S: Case 1: The distance to S is 1. There are 5 choices for S. For the distance to Q to be greater than 1, Q must be at distance 2 or 3. There are 5 + 1 = 6 choices for Q. This yields 5 \times 6 = 30 pairs. Case 2: The distance to S is 2. There are 5 choices for S. For the distance to Q to be greater than 2, Q must be at distance 3. There is only 1 choice for Q. This yields 5 \times 1 = 5 pairs. Case 3: The distance to S is 3. There is 1 choice for S. For the distance to Q to be greater than 3, there are 0 choices for Q, since no vertex is at distance 4. This yields 0 pairs.

Step 4: Calculate the probability

We divide the number of successful pairs by the total number of possible pairs to find the probability.

Symmetry and Complement Method

Because Q and S are chosen randomly and interchangeably from the vertices other than R, the probability that the distance to Q is greater is exactly equal to the probability that the distance to S is greater. We can calculate the probability of the equality case first, and use the complement to find our answer.

Step 1: Set up the symmetric probability equation

Let P_{>}, P_{<}, and P_{=} be the probabilities that the distance to Q is greater than, less than, or equal to the distance to S, respectively. Because the selection of Q and S is completely symmetric, P_{>} = P_{<}. Since all possibilities must add up to 1, we can express P_{>} in terms of P_{=}.

Step 2: Find the probability of equality

We fix R. The 11 other vertices have distances 1 (5 vertices), 2 (5 vertices), and 3 (1 vertex) from R. For the distances to be equal, Q and S must be at the same distance from R. – Distance 1: We pick 2 ordered vertices out of the 5. There are 5 \times 4 = 20 pairs. – Distance 2: We pick 2 ordered vertices out of the 5. There are 5 \times 4 = 20 pairs. – Distance 3: Since there is only 1 vertex at distance 3, we cannot pick two distinct vertices (0 pairs). The total number of ordered pairs (Q,S) is 11 \times 10 = 110.

Step 3: Solve for the target probability

We substitute P_{=} back into our symmetry equation to find the probability that the distance to Q is strictly greater.