2017 AMC 12A Problem 6
Attempt Problem 6 of the 2017 AMC 12A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2017 AMC 12A solutions, or check the answer key.
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6.
Joy has thin rods, one each of every integer length from cm through cm. She places the rods with lengths cm, cm, and 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?
Answer: B
Small Hint:
Four lengths form a quadrilateral exactly when the longest is less than the sum of the other three
Big Hint:
This gives ; then remove the rods already on the table
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 this requires and so
The integers from to give values, but the rods of length and are already on the table, leaving choices.
Thus, the correct answer is B.
Problem 6 in Other Years
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