Track / Stage 4 / 177 of 340 #437 of 1964

Problem 437

AMC 12 late, AIME early

Geometry Difficulty 4.8 Multiple choice

2. Given that the lengths of two altitudes of $\triangle A B C$ are 5 and 20. If the length of the third altitude is also an integer, then the maximum length of the third altitude is:

Pick one

Official solution

2. B.

Let the area of $\triangle A B C$ be $S$ , and the length of the third altitude be $h$ . Then the lengths of the three sides are $\frac{2 S}{5}, \frac{2 S}{20}, \frac{2 S}{h}$ .
Thus, $\frac{2 S}{5}-\frac{2 S}{20}<\frac{2 S}{h}<\frac{2 S}{5}+\frac{2 S}{20}$ .
Solving this, we get $4<h<\frac{20}{3}$ .
Therefore, the maximum integer value of $h$ is 6.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.