Track / Stage 4 / 5 of 340 #265 of 2000

Problem 265

AMC 12 late, AIME early

Geometry Difficulty 4.3 Prove it 1st Annual Harvard-MIT November Tournament · United States

Problem:
A triangle has altitudes of length $15$ , $21$ , and $35$ . Find its area.

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

Solution:
Answer: $245 \sqrt{3}$

If $A$ is the area of the triangle, the sides are $\frac{2A}{15}$ , $\frac{2A}{21}$ , and $\frac{2A}{35}$ . So the triangle is similar to a $\frac{1}{15}$ , $\frac{1}{21}$ , $\frac{1}{35}$ triangle, which is similar to a $3, 5, 7$ triangle. Let the sides be $3k$ , $5k$ , and $7k$ . Then the angle between the sides of length $3k$ and $7k$ is $120^{\circ}$ , so the area is $\frac{15 \sqrt{3}}{4} k^2$ . But the area can also be calculated as $\frac{(3k)(35)}{2} = \frac{105k}{2}$ . Setting these values equal, $k = \frac{14 \sqrt{3}}{3}$ and the area is $245 \sqrt{3}$ .

Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.