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CliffSedge
 2 years ago
Best ResponseYou've already chosen the best response.0I don't think so, but I'm not sure yet what the easiest proof of that would be.

CliffSedge
 2 years ago
Best ResponseYou've already chosen the best response.0I'm sure the triangle inequality will be in there somewhere, and maybe, since altitudes form right angles, Pythagorean theorem will be useful.

AnimalAin
 2 years ago
Best ResponseYou've already chosen the best response.1Consider an extremely obtuse isosceles triangle. With sides a,a, and b, angles theta (small), theta, and pi  2 theta. The two greater altitudes are b sin theta, so to meet the specification of the problem, 2bsin theta < b implies sin theta less than 1/2. Plenty of angles meet that specification.

vf321
 2 years ago
Best ResponseYou've already chosen the best response.0@AnimalAin That works, thanks. You proved that for triangle ABC, \(a > h_b + h_c\) for some \(a\). What if I asked you to prove that the following: \(b>h_b+h_c\) for some \(b\)?

AnimalAin
 2 years ago
Best ResponseYou've already chosen the best response.1Use the same method, assume b < a/2, and work the inequality similarly. The angles will be smaller, but it can be done.
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