AMC 10 Weekly Practice Round 3

Complete problem set with solutions and individual problem pages

Problem 19 Medium

If three vertices are chosen at random from the eight vertices of a cube and connected to form a triangle, what is the probability that the triangle is equilateral?

  • A.

    \frac{1}{7}

  • B.

    \frac{1}{14}

  • C.

    \frac{2}{7}

  • D.

    \frac{4}{35}

  • E.

    \frac{5}{14}

Answer:A

The number of ways to choose any three vertices from the eight vertices to form a triangle is _{8}C_{3} = 56.

Among them, there are 8 outcomes that form an equilateral triangle:

\triangle AC{{D}_{1}},\ \triangle BD{{C}_{1}},\ \triangle AC{{B}_{1}},\ \triangle BD{{A}_{1}},\ \triangle {{A}_{1}}{{C}_{1}}B,\ \triangle {{B}_{1}}{{D}_{1}}A,\ \triangle {{B}_{1}}{{D}_{1}}C,\ \triangle {{A}_{1}}{{C}_{1}}D.

Therefore, the probability is \dfrac{8}{56}=\dfrac{1}{7}, hence the answer is \text{A}.

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