AMC 10 Daily Practice Round 1

Complete problem set with solutions and individual problem pages

Problem 15 Medium

From all triangles with integer side lengths and a perimeter of 24, one is selected at random. What is the probability that the triangle is a right triangle?

  • A.

    \frac{1}{4}

  • B.

    \frac{1}{6}

  • C.

    \frac{1}{10}

  • D.

    \frac{1}{12}

  • E.

    \frac{1}{15}

Answer:D

Let the side lengths of the triangle be a, b, and c, where a \geq b \geq c. From the perimeter condition, we have 3a \geq a + b + c = 24 and 2a < a + (b + c) = 24, which gives 8 \leq a < 12. Therefore, the possible values for a are 8, 9, 10, and 11. The valid sets of side lengths (a, b, c) are: (8, 8, 8), (9, 9, 6), (9, 8, 7), (10, 10, 4), (10, 9, 5), (10, 8, 6), (10, 7, 7), (11, 11, 2), (11, 10, 3), (11, 9, 4), (11, 8, 5), (11, 7, 6), for a total of 12 combinations. Among these, only one set of side lengths forms a right triangle. Therefore, the probability that a randomly chosen triangle is a right triangle is \frac{1}{12}.

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