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Pirkuliyev Rovsen   1
N Oct 7, 2016 by george_54
Prove that the triangle $ABC$ is equilateral if and only if $\frac{p^2-3r^2}{4Rr}+\frac{r}{R}=\frac{7}{2}$
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Pirkuliyev Rovsen
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Pirkuliyev Rovsen
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Pirkuliyev Rovsen
#1 h  Oct 5, 2016, 5:48 PM • 1 Y
Y by Adventure10
Prove that the triangle $ABC$ is equilateral if and only if $\frac{p^2-3r^2}{4Rr}+\frac{r}{R}=\frac{7}{2}$
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#2 h  Oct 7, 2016, 6:21 AM • 3 Y
Y by Pirkuliyev Rovsen, Adventure10, Mango247
Pirkuliyev Rovsen wrote:
Prove that the triangle $ABC$ is equilateral if and only if $\frac{p^2-3r^2}{4Rr}+\frac{r}{R}=\frac{7}{2}$

If $ABC$ is equilateral then $p = \frac{{3a}}{2},R = \frac{{a\sqrt 3 }}{3},r = \frac{{a\sqrt 3 }}{6}$ and the result is obvious.

If $S$ is the area of the triangle: $\frac{{{p^2} - 3{r^2}}}{{4Rr}} + \frac{r}{R} = \frac{7}{2} \Leftrightarrow \frac{{{p^2} - \frac{{3{S^2}}}{{{p^2}}}}}{{\frac{{abc}}{S} \cdot \frac{S}{p}}} + \frac{{\frac{S}{p}}}{{\frac{{abc}}{{4S}}}} = \frac{7}{2} \Leftrightarrow \frac{{{p^4} - 3{S^2}}}{{abcp}} + \frac{{4{S^2}}}{{abcp}} = \frac{7}{2} \Leftrightarrow $

$\frac{{{p^3} + (p - a)(p - b)(p - c)}}{{abc}} = \frac{7}{2} \Leftrightarrow ...(a + b + c)(ab + bc + ca) = 9abc \Leftrightarrow (a + b + c)\left( {\frac{1}{a} + \frac{1}{b} + \frac{1}{c}} \right) = 9$

But $(a + b + c)\left( {\frac{1}{a} + \frac{1}{b} + \frac{1}{c}} \right) \geqslant 9$ and the equality holds when $a=b=c$, hence the triangle $ABC$ is equilateral.
This post has been edited 1 time. Last edited by george_54, Oct 7, 2016, 7:26 AM
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