Difference between revisions of "2023 AMC 12A Problems/Problem 21"
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First, note that a regular icosahedron has 12 vertices. So there are <math>{}_{12}P_{3} = 1320</math> ways to choose 3 distinct points. | First, note that a regular icosahedron has 12 vertices. So there are <math>{}_{12}P_{3} = 1320</math> 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 <math>1 | + | Now, the furthest distance we can get from one point to another point in a icosahedron is 3. |
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| + | Which gives us a range of <math>1 /leq d(Q, R), d(R, S) \leq 3</math> | ||
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With some case work, we get: | With some case work, we get: | ||
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Case 1: <math>d(Q, R)=3; d(R, S)=1,2</math> | Case 1: <math>d(Q, R)=3; d(R, S)=1,2</math> | ||
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<math>12×1×10=120</math> | <math>12×1×10=120</math> | ||
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Case 2: <math>d(Q, R)=2; d(R, S)=1</math> | Case 2: <math>d(Q, R)=2; d(R, S)=1</math> | ||
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<math>12×5×5=300</math> | <math>12×5×5=300</math> | ||
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Hence, <math>P(d(Q, R)>d(R, S)) = \frac{120+300}{1320} = \frac{7}{22}</math> | Hence, <math>P(d(Q, R)>d(R, S)) = \frac{120+300}{1320} = \frac{7}{22}</math> | ||
Revision as of 21:12, 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: 
Case 2: 
Hence, 
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. 
