Difference between revisions of "2023 AMC 12A Problems/Problem 21"
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==Problem== | ==Problem== | ||
| − | If <math>A</math> and <math>B</math> are vertices of a polyhedron, define the <math> | + | If <math>A</math> and <math>B</math> are vertices of a polyhedron, define the distance <math>d(A,B)</math> |
C to be the minimum number of edges of the polyhedron one must traverse in order to connect <math>A</math> and <math>B</math>. For example, if AB is an edge of the polyhedron, then <math>d(A, B)</math> = 1, but if AC and CB are edges and AB is not an edge, then <math>d(A, B)</math> = 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 <math>d(Q, R) > d(R, S)</math>? | C to be the minimum number of edges of the polyhedron one must traverse in order to connect <math>A</math> and <math>B</math>. For example, if AB is an edge of the polyhedron, then <math>d(A, B)</math> = 1, but if AC and CB are edges and AB is not an edge, then <math>d(A, B)</math> = 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 <math>d(Q, R) > d(R, S)</math>? | ||
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==Solution 1== | ==Solution 1== | ||
Revision as of 21:53, 9 November 2023
Problem
If 
and 
are vertices of a polyhedron, define the distance 
C to be the minimum number of edges of the polyhedron one must traverse in order to connect and . For example, if AB is an edge of the polyhedron, then 
= 1, but if AC and CB are edges and AB is not an edge, then = 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 
?
Solution 1
See also
| 2023 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by First Problem |
Followed by Problem 22 |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
| All AMC 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
