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2017 AMC 12A Problem 6

Problem 6 of 25EasierGeometryCombinatorics

Joy has 3030 thin rods, one each of every integer length from 11 cm through 3030 cm. She places the rods with lengths 33 cm, 77 cm, and 1515 cm on a table. She then wants to choose a fourth rod that she can put with these three to form a quadrilateral with positive area. How many of the remaining rods can she choose as the fourth rod?

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Solution

Four lengths form a quadrilateral with positive area if and only if the longest is strictly less than the sum of the other three. With a fourth rod of length n,n, this requires 15<3+7+n15\lt 3+7+n and n<3+7+15,n\lt 3+7+15, so 5<n<25. 5\lt n\lt 25. The integers from 66 to 2424 give 1919 values, but the rods of length 77 and 1515 are already on the table, leaving 19−2=1719-2=17 choices. Thus, the correct answer is B.
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Tagged: triangle inequality · counting integers in a range

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