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

(→‎Solution 2(Cheese+Actual way))
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Fix an arbitrary point, to select the rest <math>2</math> points, there are <math>\binom{11}{2}=55</math> ways. To make <math>d(Q, R)=d(R, S), d=1/2</math>. Which means there are in total <math>2\cdot \binom{5}{2}=20</math> ways to make the distance the same. <math>\frac{1}{2}(1-\frac{20}{55})=\frac{7}{22}\implies \boxed{A}</math>
 
Fix an arbitrary point, to select the rest <math>2</math> points, there are <math>\binom{11}{2}=55</math> ways. To make <math>d(Q, R)=d(R, S), d=1/2</math>. Which means there are in total <math>2\cdot \binom{5}{2}=20</math> ways to make the distance the same. <math>\frac{1}{2}(1-\frac{20}{55})=\frac{7}{22}\implies \boxed{A}</math>
 
~bluesoul
 
~bluesoul
 +
 +
== Solution 3 ==
 +
We can imagine the icosahedron as having 3 layers. 1 vertex at the top, 5 vertices below connected to the top vertex, 5 vertices below that which are 2 edges away from the top vertex, and one vertex at the bottom that is 3 edges away. WLOG because the icosahedron is symmetric around all vertices, we can say that R is the vertex at the top. So now, we just need to find the probability that S is on a layer closer to the top than Q. We can do casework on the layer S is on to get
 +
<cmath>\frac{5}{11} \cdot \frac{6}{10} + \frac{5}{11} \cdot \frac{1}{10} = \frac{35}{110} = \frac{7}{22}</cmath>
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So the answer is <math>\boxed{\textbf{(A) }\frac{7}{22}}</math>. -awesomeparrot
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== Solution 4 ==
 +
We can actually see that the probability that <math>d(Q, R) > d(R, S)</math> is the exact same as <math>d(Q, R) < d(R, S)</math> because <math>d(Q, R)</math> and <math>d(R, S)</math> have no difference. (In other words, we can just swap Q and S, meaning that can be called the same.) Therefore, we want to find the probability that <math>d(Q, R) = d(R, S)</math>.
 +
 +
WLOG, we can rotate the icosahedron so that R is the top of the icosahedron. Then we can divide this into 2 cases:
 +
 +
1. They are on the second layer
 +
 +
There are 5 ways to put one point, and 4 ways to put the other point such that <math>d(Q, R) = d(R, S) = 1</math>. So, there are <math>5 \cdot 4 = 20</math> ways to put them on the second layer.
 +
 +
2. They are on the third layer
 +
 +
There are 5 ways to put one point, and 4 ways to put the other point such that <math>d(Q, R) = d(R, S) = 2</math>. So, there are <math>5 \cdot 4 = 20</math> ways to put them on the third layer.
 +
 +
The total number of ways to choose P and S are <math>11 \cdot 10 = 110</math> (because there are 12 vertices), so the probability that <math>d(Q, R) = d(R, S)</math> is <math>\frac{20+20}{110} = \frac{4}{11}</math>.
 +
 +
Therefore, the probability that <math>d(Q, R) > d(R, S)</math> is <math>\frac{1 - \frac{4}{11}}{2} = \boxed{\textbf{(A) }\frac{7}{22}}</math>
 +
 +
~Ethanzhang1001
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 +
==Solution 5==
 +
 +
We know that there are <math>20</math> faces. Each of those faces has <math>3</math> borders (since each is a triangle), and each edge is used as a border twice (for each face on either side). Thus, there are <math>\dfrac{20\cdot3}2=30</math> edges.
 +
 +
By Euler's formula, which states that <math>v-e+f=2</math> for all convex polyhedra, we know that there are <math>2-f+e=12</math> vertices.
 +
 +
The answer can be counted by first counting the number of possible paths that will yield <math>d(Q, R) > d(R, S)</math> and dividing it by <math>12\cdot11\cdot10</math> (or <math>\dbinom{12}3</math>, depending on the approach). In either case, one will end up dividing by <math>11</math> somewhere in the denominator. We can then hope that there will be no factor of <math>11</math> in the numerator (which would cancel the <math>11</math> in the denominator out), and answer the only option that has an <math>11</math> in the denominator: <math>\boxed{\textbf{(A) }\frac{7}{22}}</math>.
 +
 +
~Technodoggo
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 +
Additional note by "Fruitz": Note that one can eliminate <math>1/2</math> by symmetry if you swap the ineq sign.
  
 
==See also==
 
==See also==
−
 
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{{AMC10 box|year=2023|ab=A|num-b=24|after=Last Problem}}
 
{{AMC12 box|year=2023|ab=A|num-b=20|num-a=22}}
 
{{AMC12 box|year=2023|ab=A|num-b=20|num-a=22}}
  
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 22:57, 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 $d(Q, R) > d(R, S)$?

$\textbf{(A) } \frac{7}{22} \qquad \textbf{(B) } \frac{1}{3} \qquad \textbf{(C) } \frac{3}{8} \qquad \textbf{(D) } \frac{5}{12} \qquad \textbf{(E) } \frac{1}{2}$

Solution 1

First, note that a regular icosahedron has 12 vertices. So there are $P_{3}^{12} = 1320$ 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 $1 \leq d(Q, R), d(R, S) \leq 3$

With some case work, we get:

Case 1: $d(Q, R) = 3; d(R, S) = 1, 2$

$12 \times 1 \times 10 = 120$ (ways to choose R × ways to choose Q × ways to choose S)

Case 2: $d(Q, R) = 2; d(R, S) = 1$

$12 \times 5 \times 5 = 300$ (ways to choose R × ways to choose Q × ways to choose S)

Hence, $P(d(Q, R)>d(R, S)) = \frac{120+300}{1320} = \boxed{\textbf{(A) } \frac{7}{22}}$

~lptoggled

Solution 2(Cheese+Actual way)

In total, there are $\binom{12}{3}=220$ ways to select the points. However, if we look at the denominators of $B,C,D$, they are $3,8,12$ which are not divisors of $220$. Also $\frac{1}{2}$ is impossible as cases like $d(Q, R) = d(R, S)$ exist. The only answer choice left is $\boxed{A}$

(Actual way)

Fix an arbitrary point, to select the rest $2$ points, there are $\binom{11}{2}=55$ ways. To make $d(Q, R)=d(R, S), d=1/2$. Which means there are in total $2\cdot \binom{5}{2}=20$ ways to make the distance the same. $\frac{1}{2}(1-\frac{20}{55})=\frac{7}{22}\implies \boxed{A}$ ~bluesoul

Solution 3

We can imagine the icosahedron as having 3 layers. 1 vertex at the top, 5 vertices below connected to the top vertex, 5 vertices below that which are 2 edges away from the top vertex, and one vertex at the bottom that is 3 edges away. WLOG because the icosahedron is symmetric around all vertices, we can say that R is the vertex at the top. So now, we just need to find the probability that S is on a layer closer to the top than Q. We can do casework on the layer S is on to get \[\frac{5}{11} \cdot \frac{6}{10} + \frac{5}{11} \cdot \frac{1}{10} = \frac{35}{110} = \frac{7}{22}\] So the answer is $\boxed{\textbf{(A) }\frac{7}{22}}$. -awesomeparrot

Solution 4

We can actually see that the probability that $d(Q, R) > d(R, S)$ is the exact same as $d(Q, R) < d(R, S)$ because $d(Q, R)$ and $d(R, S)$ have no difference. (In other words, we can just swap Q and S, meaning that can be called the same.) Therefore, we want to find the probability that $d(Q, R) = d(R, S)$.

WLOG, we can rotate the icosahedron so that R is the top of the icosahedron. Then we can divide this into 2 cases:

1. They are on the second layer

There are 5 ways to put one point, and 4 ways to put the other point such that $d(Q, R) = d(R, S) = 1$. So, there are $5 \cdot 4 = 20$ ways to put them on the second layer.

2. They are on the third layer

There are 5 ways to put one point, and 4 ways to put the other point such that $d(Q, R) = d(R, S) = 2$. So, there are $5 \cdot 4 = 20$ ways to put them on the third layer.

The total number of ways to choose P and S are $11 \cdot 10 = 110$ (because there are 12 vertices), so the probability that $d(Q, R) = d(R, S)$ is $\frac{20+20}{110} = \frac{4}{11}$.

Therefore, the probability that $d(Q, R) > d(R, S)$ is $\frac{1 - \frac{4}{11}}{2} = \boxed{\textbf{(A) }\frac{7}{22}}$

~Ethanzhang1001

Solution 5

We know that there are $20$ faces. Each of those faces has $3$ borders (since each is a triangle), and each edge is used as a border twice (for each face on either side). Thus, there are $\dfrac{20\cdot3}2=30$ edges.

By Euler's formula, which states that $v-e+f=2$ for all convex polyhedra, we know that there are $2-f+e=12$ vertices.

The answer can be counted by first counting the number of possible paths that will yield $d(Q, R) > d(R, S)$ and dividing it by $12\cdot11\cdot10$ (or $\dbinom{12}3$, depending on the approach). In either case, one will end up dividing by $11$ somewhere in the denominator. We can then hope that there will be no factor of $11$ in the numerator (which would cancel the $11$ in the denominator out), and answer the only option that has an $11$ in the denominator: $\boxed{\textbf{(A) }\frac{7}{22}}$.

~Technodoggo

Additional note by "Fruitz": Note that one can eliminate $1/2$ by symmetry if you swap the ineq sign.

See also

2023 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 24 		Followed by
Last Problem
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 10 Problems and Solutions
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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