Joy has 30 thin rods, one each of every integer length from 1 cm through 30 cm. She places the rods with lengths 3, 7 and 15 on the table. She then wants to choose a fourth rod that she can put with these 3 such that they form a quadrilateral with positive area. How many of the remaining rods can she choose as the 4th rod
In the AOPS solution, they included the 6cm rod and the 24 cm rod. But if you draw a diagonal that creates a $(3,7,x)$ $x$ has a maximum of 9, and 9+6 = 15 and 15+9=24, violating the triangle inequality. Is there something im overlooking?
