Difference between revisions of "2023 AMC 12A Problems/Problem 21"
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<math>12 \times 5 \times 5 = 300</math> (ways to choose R × ways to choose Q × ways to choose S) | <math>12 \times 5 \times 5 = 300</math> (ways to choose R × ways to choose Q × ways to choose S) | ||
| − | Hence, <math>P(d(Q, R)>d(R, S)) = \frac{120+300}{1320} = | + | Hence, <math>P(d(Q, R)>d(R, S)) = \frac{120+300}{1320} = \boxed{\textbf{(A) } \frac{7}{22}}</math> |
~lptoggled | ~lptoggled | ||
Revision as of 21:46, 9 November 2023
Problem
If 
and 
are vertices of a polyhedron, define the distance 
to be the minimum number of edges of the polyhedron one must traverse in order to connect and . For example, if 
is an edge of the polyhedron, then 
, but if 
and 
are edges and is not an edge, then 
. 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
First, note that a regular icosahedron has 12 vertices. So there are 
ways to choose 3 distinct points.
Now, the furthest distance we can get from one point to another point in a icosahedron is 3. Which gives us a range of 
With some case work, we get:
Case 1: 
(ways to choose R × ways to choose Q × ways to choose S)
Case 2: 
(ways to choose R × ways to choose Q × ways to choose S)
Hence, 
~lptoggled
See also
| 2023 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 20 |
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. 
