Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (→‎Solution 2(Cheese + Actual way))
m (→‎Problem)
Line 1: Line 1:
 
==Problem==
 
==Problem==
−
If <math>A</math> and <math>B</math> are vertices of a polyhedron, define the distance <math>d(A,B)</math> 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 <math>\overline{AB}</math> is an edge of the polyhedron, then <math>d(A, B) = 1</math>, but if <math>\overline{AC}</math> and <math>\overline{CB}</math> are edges and <math>\overline{AB}</math> is not an edge, then <math>d(A, B) = 2</math>. 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>?
+
If <math>A</math> and <math>B</math> are vertices of a polyhedron, define the distance <math>d(A,B)</math> 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 <math>\overline{AB}</math> is an edge of the polyhedron, then <math>d(A, B) = 1</math>, but if <math>\overline{AC}</math> and <math>\overline{CB}</math> are edges and <math>\overline{AB}</math> is not an edge, then <math>d(A, B) = 2</math>. Let <math>Q</math>, <math>R</math>, and <math>S</math> 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>?
  
 
<math>\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}</math>
 
<math>\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}</math>

Revision as of 00:25, 10 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.

Video Solution by epicbird08

https://youtu.be/s_6q0C0z6Ug

~EpicBird08

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2023_AMC_12A_Problems/Problem_21&oldid=201722"