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

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== Problem ==
 
== 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?
 
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>
  
=== Solution <math>1</math> ===
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== Solutions ==
 
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=== 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.
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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.
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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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.
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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To conclude, the probability of a cutting in plane is <math>(6\cdot 4+8\cdot 1)/(12\cdot 4+8\cdot 1) = 32/56 = 4/7</math> (C)
  
 
-Vader10,Oct.6 2020-
 
-Vader10,Oct.6 2020-
  
=== Solution <math>2</math> ===
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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>
  
== Note: (Cheap solution same approach as solution 1) ==
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=== 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 3 ==
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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 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.
  
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,  

Revision as of 00:38, 19 October 2020

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 $(6\cdot 4+8\cdot 1)/(12\cdot 4+8\cdot 1) = 32/56 = 4/7$ (C)

-Vader10,Oct.6 2020-

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,

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

See Also

2009 AMC 10A (ProblemsAnswer KeyResources)
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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