TITLE: Question on triangle with heights
QUESTION [8 upvotes]: Prove that there exists no triangle with heights 4,7, and 10 units. I am completely puzzled.

REPLY [17 votes]: The area of the triangle is half the product of a side of the triangle and the height of the triangle on that side. So calling the area of the triangle $A$, the sides would have to be
$$\frac{2\cdot A}{10}, \; \frac{2\cdot A}{7},\; \frac{2\cdot A}{4}.$$
But
$$\frac{1}{4} > \frac{1}{7} + \frac{1}{10},$$
so these "sides" could not make a triangle, since the longest is longer than the sum of the two shorter.

REPLY [5 votes]: Hint: Well let the sides of the triangle be $a, b, c$ - ordered so that twice the area of the triangle is $4a=7b=10c$ (base times height). Can you use this information to express all the sides of the triangle as multiples of $a$?