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vf321

  • 2 years ago

Can the sum of any two altitudes of a triangle be smaller than one of its legs?

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  1. CliffSedge
    • 2 years ago
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    I don't think so, but I'm not sure yet what the easiest proof of that would be.

  2. CliffSedge
    • 2 years ago
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    I'm sure the triangle inequality will be in there somewhere, and maybe, since altitudes form right angles, Pythagorean theorem will be useful.

  3. AnimalAin
    • 2 years ago
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    Consider 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.

  4. vf321
    • 2 years ago
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    @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 \(b\)?

  5. AnimalAin
    • 2 years ago
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    Use 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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