I have a triangle, the length of heights are $i,h,g$.
Prove that we can build a new triangle so that the lengths of the sides are: $i^{-1}, g^{-1}, h^{-1}$ (see picture)
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@mvw For the same reason that $2,3,8$ can't be the side lengths of a triangle. The inequality is not satisfied for all triples. – Deepak Mar 19 '16 at 01:54
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@Deepak Thanks. Time to go to bed. :-) – mvw Mar 19 '16 at 01:57
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Each side of the original teiangle is twice the area divided by the altitude to that side. So your proposed triangle is similar to the original one.
Oscar Lanzi
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QUESTION: if the aria of the original triangle is 20 and the sides of the proposed triangle is : 10*i^-1 ... what is the aria of the new triangle – Omar Abu Asbe Mar 20 '16 at 22:22
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Let the respective bases for the heights $\displaystyle h,g,i$ be $\displaystyle H, G, I$, and the latter three are the sides of the original triangle.
By considering the area of the triangle, establish that $\displaystyle hH = gG = iI$.
Rearrange (for example) to $\displaystyle \frac{1}{h} = (\frac{H}{G})\frac{1}{g}$ and similarly show that $\displaystyle \frac{1}{i} = (\frac{I}{G})\frac{1}{g}$
Now let $\displaystyle \frac 1h + \frac 1i - \frac 1g = \alpha$
This gives $\displaystyle H + I - G = \alpha Gg$ and by using the triangle inequality on the original triangle, show that $\displaystyle \alpha > 0$.
Proceed similarly for the other sides (or conclude by symmetry). You have now established the triangle inequalities for the new triangle, and you're done.
Deepak
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