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Difference between revisions of "2023 AMC 12A Problems/Problem 21"
(→‎Solution 2(Cheese+Actual way))
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In total, there are <math>\binom{12}{3}=220</math> ways to select the points. However, if we look at the denominators of <math>B,C,D</math>, they are <math>3,8,12</math> which are not divisors of <math>220</math>. Also <math>\frac{1}{2}</math> is impossible as cases like <math>d(Q, R) = d(R, S)</math> exist. The only answer choice left is <math>\boxed{A}</math>
 
In total, there are <math>\binom{12}{3}=220</math> ways to select the points. However, if we look at the denominators of <math>B,C,D</math>, they are <math>3,8,12</math> which are not divisors of <math>220</math>. Also <math>\frac{1}{2}</math> is impossible as cases like <math>d(Q, R) = d(R, S)</math> exist. The only answer choice left is <math>\boxed{A}</math>
 +
 
(Actual way)
 
(Actual way)
 +
 
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

Revision as of 22:16, 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

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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