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Difference between revisions of "2009 AMC 10A Problems/Problem 24"

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== Problem ==
 
== Problem ==
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Three distinct vertices of a cube are chosen at random. What is the probability that the plane determined by these three vertices contains points inside the cube?
 
Three distinct vertices of a cube are chosen at random. What is the probability that the plane determined by these three vertices contains points inside the cube?
  
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</math>
 
</math>
  
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First of all, number of planes determined by any three vertices of a cube is 20 (6 surface, 6 opposing parallel edges, 8 points cut by three remote vertices). Among these 20 planes only 6 surfaces will not cut into the cube.
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== Solutions ==
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Secondly, to choose three vertices randomly, the four vertices planes each will be chosen 4 times, while the three vertices planes each will be chosen once.  
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=== Solution 1 ===
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To conclude, the probability of a cutting in plane is (6*4+8*1)/(12*4+8*1)=32/56=4/7 (C)
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First of all, number of planes determined by any three vertices of a cube is <math>20</math> (<math>6</math> surface, <math>6</math> opposing parallel edges, <math>8</math> points cut by three remote vertices). Among these <math>20</math> planes only <math>6</math> surfaces will not cut into the cube.
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Secondly, to choose three vertices randomly, the four vertices planes each will be chosen <math>4</math> times, while the three vertices planes each will be chosen once.
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To conclude, the probability of a cutting in plane is <math>\frac {(6\cdot 4+8\cdot 1)}{(12\cdot 4+8\cdot 1)} = \frac {32}{56}</math> = <math>\boxed {\textbf{(C)}}</math>.
  
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----
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-Vader10,Oct.6 2020-
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Vader10
 
  
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=== Solution <math>2</math> ===
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Minor LaTeX edit by Arcticturn
  
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=== Solution 2 ===
 
We will try to use symmetry as much as possible.
 
We will try to use symmetry as much as possible.
  
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</cmath>
 
</cmath>
  
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== Note: (Cheap solution same approach as solution 1) ==
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=== Solution 3 (Cheap solution, same approach as Solution 2) ===
  
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This problem can be approached the same way, by picking vertices, but with a much faster and kind of cheap solution: Pick any vertex A and a face it touches. For vertex B, out of the 7 remaining vertices, 4 of them aren't on the same face as the one chosen for vertex A, so vertex C can be placed anywhere and the plane will no matter what be in the cube. Therefore, the probability of choosing a valid vertex B is 4/7.
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This problem can be approached the same way, by picking vertices, but with a much faster and kind of cheap solution: Pick any vertex A and a face it touches. For vertex B, out of the <math>7</math> remaining vertices, <math>4</math> of them aren't on the same face as the one chosen for vertex A, so vertex C can be placed anywhere and the plane will no matter what be in the cube. Therefore, the probability of choosing a valid vertex B is <math>4/7</math>.
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== Solution 3 ==
 
  
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=== Solution 4 ===
 
There are <math>\binom{8}{3}=56</math> ways to pick three vertices from eight total vertices; this is our denominator. In order to have three points inside the cube, they cannot be on the surface. Thus, we can use complementary probability.
 
There are <math>\binom{8}{3}=56</math> ways to pick three vertices from eight total vertices; this is our denominator. In order to have three points inside the cube, they cannot be on the surface. Thus, we can use complementary probability.
  
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There are <math>\binom{4}{3}=4</math> to choose three points from the vertices of a single face. Since there are six faces, <math>4 \times 6 = 24</math>.
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There are <math>\binom{4}{3}=4</math> ways to choose three points from the vertices of a single face. Since there are six faces, <math>4 \times 6 = 24</math>.
  
 
Thus, the probability of what we don't want is <math>\frac{24}{56} = \frac{3}{7}</math>. Using complementary probability,  
 
Thus, the probability of what we don't want is <math>\frac{24}{56} = \frac{3}{7}</math>. Using complementary probability,  
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1- \frac 37 = \boxed{\frac 47}
 
1- \frac 37 = \boxed{\frac 47}
 
</cmath>
 
</cmath>
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=== Solution 5 (Casework) ===
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This problem is fairly simple. Start with <math>2</math> points WLOG.
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Case <math>1</math>: You pick 2 points that are diagonally across from each other but still on the same face. This happens with probability <math>\frac {3}{7}</math>. We notice that <math>4</math> out of the <math>6</math> possible outcomes include a point inside the cube. Therefore, the probability of this happening is <math>\frac {2}{7}</math>.
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Case <math>2</math>: You pick 2 points that a directly across from each other. This happens with probability <math>\frac {3}{7}</math>. We notice that <math>2</math> out of the <math>6</math> possible outcomes include a point inside the cube. Therefore, the probability of this happening is <math>\frac {1}{7}</math>.
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Case <math>3</math>: You pick 2 points that is a space diagonal of the cube. This happens with probability <math>\frac {1}{7}</math>. Clearly, all of the points contain a point inside the cube, so our probability is <math>\frac {1}{7}</math>.
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Adding these probabilities gives us <math>\boxed {\frac {4}{7}}</math>
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~Arcticturn
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== (Video solution)==
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Video: https://youtu.be/5PiNMIxItVQ
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~DaBobWhoLikeMath1234
  
 
== See Also ==
 
== See Also ==

Latest revision as of 19:25, 31 October 2022

Problem

Three distinct vertices of a cube are chosen at random. What is the probability that the plane determined by these three vertices contains points inside the cube?

$\mathrm{(A)}\ \frac{1}{4} \qquad \mathrm{(B)}\ \frac{3}{8} \qquad \mathrm{(C)}\ \frac{4}{7} \qquad \mathrm{(D)}\ \frac{5}{7} \qquad \mathrm{(E)}\ \frac{3}{4}$

Solutions

Solution 1

First of all, number of planes determined by any three vertices of a cube is $20$ ($6$ surface, $6$ opposing parallel edges, $8$ points cut by three remote vertices). Among these $20$ planes only $6$ surfaces will not cut into the cube. Secondly, to choose three vertices randomly, the four vertices planes each will be chosen $4$ times, while the three vertices planes each will be chosen once. To conclude, the probability of a cutting in plane is $\frac {(6\cdot 4+8\cdot 1)}{(12\cdot 4+8\cdot 1)} = \frac {32}{56}$ = $\boxed {\textbf{(C)}}$.

-Vader10,Oct.6 2020-

Minor LaTeX edit by Arcticturn

Solution 2

We will try to use symmetry as much as possible.

Pick the first vertex $A$, its choice clearly does not influence anything.

Pick the second vertex $B$. With probability $3/7$ vertices $A$ and $B$ have a common edge, with probability $3/7$ they are in opposite corners of the same face, and with probability $1/7$ they are in opposite corners of the cube. We will handle each of the cases separately.

In the first case, there are $2$ faces that contain the edge $AB$. In each of these faces there are $2$ other vertices. If one of these $4$ vertices is the third vertex $C$, the entire triangle $ABC$ will be on a face. On the other hand, if $C$ is one of the two remaining vertices, the triangle will contain points inside the cube. Hence in this case the probability of choosing a good $C$ is $2/6 = 1/3$.

In the second case, the triangle $ABC$ will not intersect the cube if point $C$ is one of the two points on the side that contains $AB$. Hence the probability of $ABC$ intersecting the inside of the cube is $2/3$.

In the third case, already the diagonal $AB$ contains points inside the cube, hence this case will be good regardless of the choice of $C$.

Summing up all cases, the resulting probability is: \[\frac 37\cdot\frac 13 + \frac 37\cdot \frac 23 + \frac 17\cdot 1 = \boxed{\frac 47}\]

Solution 3 (Cheap solution, same approach as Solution 2)

This problem can be approached the same way, by picking vertices, but with a much faster and kind of cheap solution: Pick any vertex A and a face it touches. For vertex B, out of the $7$ remaining vertices, $4$ of them aren't on the same face as the one chosen for vertex A, so vertex C can be placed anywhere and the plane will no matter what be in the cube. Therefore, the probability of choosing a valid vertex B is $4/7$.

Solution 4

There are $\binom{8}{3}=56$ ways to pick three vertices from eight total vertices; this is our denominator. In order to have three points inside the cube, they cannot be on the surface. Thus, we can use complementary probability.

There are $\binom{4}{3}=4$ ways to choose three points from the vertices of a single face. Since there are six faces, $4 \times 6 = 24$.

Thus, the probability of what we don't want is $\frac{24}{56} = \frac{3}{7}$. Using complementary probability,

\[1- \frac 37 = \boxed{\frac 47}\]

Solution 5 (Casework)

This problem is fairly simple. Start with $2$ points WLOG.

Case $1$: You pick 2 points that are diagonally across from each other but still on the same face. This happens with probability $\frac {3}{7}$. We notice that $4$ out of the $6$ possible outcomes include a point inside the cube. Therefore, the probability of this happening is $\frac {2}{7}$.

Case $2$: You pick 2 points that a directly across from each other. This happens with probability $\frac {3}{7}$. We notice that $2$ out of the $6$ possible outcomes include a point inside the cube. Therefore, the probability of this happening is $\frac {1}{7}$.

Case $3$: You pick 2 points that is a space diagonal of the cube. This happens with probability $\frac {1}{7}$. Clearly, all of the points contain a point inside the cube, so our probability is $\frac {1}{7}$.

Adding these probabilities gives us $\boxed {\frac {4}{7}}$

~Arcticturn

(Video solution)

Video: https://youtu.be/5PiNMIxItVQ ~DaBobWhoLikeMath1234

See Also

2009 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 23 		Followed by
Problem 25
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

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