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Difference between revisions of "2023 AMC 12A Problems/Problem 21"

(→‎Solution 1)
(→‎Solution 1)
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==Solution 1==
 
==Solution 1==
 
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 ≤ d(Q, R), d(R, S) ≤ 3</math>
 
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 ≤ d(Q, R), d(R, S) ≤ 3</math>
 
With some case work, we get:
 
With some case work, we get:
−
''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>
 
<math>12×1×10=120</math>
 
<math>12×1×10=120</math>
−
''Case 2: <math>d(Q, R)=2; d(R, S)=1</math>''
+
Case 2: <math>d(Q, R)=2; d(R, S)=1</math>
 
<math>12×5×5=300</math>
 
<math>12×5×5=300</math>
 
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 21:11, 9 November 2023

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 $d(Q, R) > d(R, S)$?

Solution 1

First, note that a regular icosahedron has 12 vertices. So there are ${}_{12}P_{3} = 1320$ 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 $1 ≤ d(Q, R), d(R, S) ≤ 3$ (Error compiling LaTeX. Unknown error_msg) With some case work, we get: Case 1: $d(Q, R)=3; d(R, S)=1,2$ $12×1×10=120$ Case 2: $d(Q, R)=2; d(R, S)=1$ $12×5×5=300$ Hence, $P(d(Q, R)>d(R, S)) = \frac{120+300}{1320} = \frac{7}{22}$

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

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