2009 AMC 10A Problem 24

Below is the professionally curated solution for Problem 24 of the 2009 AMC 10A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2009 AMC 10A solutions, or check the answer key.

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Concepts:complementary countingcube geometrycombinations

Difficulty rating: 1860

24.

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?

14\dfrac{1}{4}

38\dfrac{3}{8}

47\dfrac{4}{7}

57\dfrac{5}{7}

34\dfrac{3}{4}

Solution:

Three vertices determine a plane that cuts through the interior unless all three lie on a single face.

Each of the 66 faces gives (43)=4\binom{4}{3} = 4 triples, so 64=246 \cdot 4 = 24 triples lie on a face out of (83)=56\binom{8}{3} = 56 total.

The probability of hitting the interior is 12456=47.1 - \dfrac{24}{56} = \dfrac{4}{7}.

Thus, the correct answer is C.

Problem 24 in Other Years