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I'm given the heights of a triangle. Find the inradius. I know that inradius is area/semiperimeter. But then?

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Let the area of triangle be $T$, the heights $h_a, h_b, h_c$, and $s$ the semi-perimeter. Then: $$2T = ah_a = bh_b=ch_c$$ $$a = \frac{2T}{h_a}, b = \frac{2T}{h_b}, c = \frac{2T}{h_c} $$ $$a+b+c= 2T(\frac{1}{h_a}+\frac{1}{h_b}+\frac{1}{h_c})$$ $$\frac{s}{T}=\frac{1}{h_a}+\frac{1}{h_b}+\frac{1}{h_c}$$

I hope you get it from here, the sum of the reciprocals of the heights is the reciprocal of the inradius itself. Nice little theorem.

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  • $\begingroup$ Wow! That was easy. Thanks for such a simple explanation. I'll accept (check) it, but I would like to wait a few minutes to see if there's any better answer. $\endgroup$ – user103197 Oct 25 '13 at 14:01

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