0
$\begingroup$

Prove that for every triangle the following inequality is true $$\dfrac{1}{2r} < \dfrac{1}{h_1} + \dfrac{1}{h_2} < \dfrac{1}{r}$$

My attempt was trying to get somthing from connection between area, heights and radius, but without any effect.

$\endgroup$
  • $\begingroup$ can you please explain what is $h1,h2$? $\endgroup$ – Sujit Bhattacharyya Mar 24 '18 at 12:01
  • $\begingroup$ $h_1a=h_2b=r s$ where $a,b$ are two sides and $s$ is the perimeter $\endgroup$ – Lozenges Mar 24 '18 at 12:13
  • $\begingroup$ They are any two heights $\endgroup$ – user128409235 Mar 24 '18 at 12:36
  • $\begingroup$ @Lozenges shouldn't it be $h_1 a = h_2 b = 2rs$? $\endgroup$ – kayush Mar 24 '18 at 12:55
0
$\begingroup$

$h_1a=h_2b=2rs$ Equating area

$h_1 = \cfrac{2rs}a$

$h_2 = \cfrac{2rs}b$

$\cfrac1{h_1} + \cfrac1{h_2} = \cfrac{a+b}{2rs} = \cfrac{2(a+b)}{2r(a+b+c)}$

Using traingular inequality, $a+b > c$

$\cfrac1{h_1} + \cfrac1{h_2} > \cfrac{1}{2r}$

and since c > 0, $a+b+c > a+b \implies (a+b)/(a+b+c) < 1$

$\cfrac1{h_1} + \cfrac1{h_2} < \cfrac{1}{r}$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.