The altitudes of a triangle are 12,15 and 20. What is the area of the triangle?

Answer: 150.
Denote the sides of the triangle by $a, b$ and $c$ and its altitudes by $h_{a}, h_{b}$ and $h_{c}$. Then we know that $h_{a}=12, h_{b}=15$ and $h_{c}=20$. By the well known relation $a: b=h_{b}: h_{a}$ it follows $b=\frac{h_{a}}{h_{b}} a=\frac{12}{15} a=\frac{4}{5} a$. Analogously, $c=\frac{h_{a}}{h_{c}} a=\frac{12}{20} a=\frac{3}{5} a$. Thus half of the triangle's circumference is $s=\frac{1}{2}(a+b+c)=\frac{1}{2}\left(a+\frac{4}{5} a+\frac{3}{5} a\right)=\frac{6}{5} a$. For the area $\Delta$ of the triangle we have $\Delta=\frac{1}{2} a h_{a}=\frac{1}{2} a 12=6 a$, and also by the well known Heron formula $\Delta=\sqrt{s(s-a)(s-b)(s-c)}=\sqrt{\frac{6}{5} a \cdot \frac{1}{5} a \cdot \frac{2}{5} a \cdot \frac{3}{5} a}=\sqrt{\frac{6^{2}}{5^{4}} a^{4}}=\frac{6}{25} a^{2}$. Hence, $6 a=\frac{6}{25} a^{2}$, and we get $a=25(b=20, c=15)$ and consequently $\Delta=150$.