Geometric Inequality

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Source: Estonia IMO TST 1996 Day 1 P2

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IstekOlympiadTeam

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IstekOlympiadTeam

#1 h Nov 27, 2015, 7:48 PM • 2 Y

Y by Adventure10, Mango247

Let $a,b,c$ be the sides of a triangle, $\alpha ,\beta ,\gamma$ the corresponding angles and $r$ the inradius. Prove that $a\cdot sin\alpha+b\cdot sin\beta+c\cdot sin\gamma\geq 9r$

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xzlbq

#3 h Nov 28, 2015, 12:49 AM • 2 Y

Y by Adventure10, Mango247

IstekOlympiadTeam wrote:

Let $a,b,c$ be the sides of a triangle, $\alpha ,\beta ,\gamma$ the corresponding angles and $r$ the inradius. Prove that $a\cdot sin\alpha+b\cdot sin\beta+c\cdot sin\gamma\geq 9r$

$\sin \left( A \right) a+\sin \left( B \right) b+\sin \left( C \right) c-{\it h_a}-{\it h_b}-{\it h_c}=1/2\,{\frac {{a}^{2}+{b}^{2}+{c}^{2}-bc-a c-ab}{R}}$

This post has been edited 2 times. Last edited by xzlbq, Nov 28, 2015, 12:51 AM

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#4 h Nov 28, 2015, 3:53 AM • 3 Y

Y by Abhinandan18, cr1, Adventure10

From the Law of Sines, we obtain that $\frac{a}{sin\alpha}=2R$ , so the inequality is equivalent to ${a}^{2}+{b}^{2}+{c}^{2}\geq18Rr$ But $R=\frac{abc}{4P}$ and $r=\frac{P}{s}$ , where $R$ is the circumradius, $P$ is the area and $s$ is the semiperimeter. Therefore, $18Rr=\frac{18abc}{4s}=\frac{9abc}{a+b+c}$ , so it is only left to prove that $({a}^{2}+{b}^{2}+{c}^{2})(a+b+c)\geq9abc$ which is a trivial application of AM-GM.

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#6 h Jan 16, 2016, 6:09 PM • 2 Y

Y by Adventure10, Mango247

$\frac {a}{\sin\alpha}=2R$ $sin\alpha=\frac {a}{2R}$ so our inequality transforms to $\frac{a^2}{2R}+\frac {b^2}{2R}+\frac {c^2}{2R}\geq9r$ then $a^{2}+b^{2}+c^{2}=(a+b+c)^{2}-2(ab+bc+ac)\geq 18Rr$ we know that intriangle $a+b+c=2p$ , $ab+bc+ac=p^{2}+r^{2}+4Rr$ so $4p^{2}-2p^{2}-2r^{2}-8Rr\geq 18Rr$ then $p^{2}-r^{2}\geq 13Rr$ then we also know thet in triangle $p^{2}\geq \frac {27Rr}{2}$ . (it isn't hard to prove) it means $\frac{27Rr}{2}-13Rr\geq r^{2}$
$\frac {27Rr-26Rr}{2}\geq r^{2}$ , $\frac {R}{2}\geq r$ and it's true because it's Euler's theorem $d^{2}=R^{2}-2Rr$ $d^{2}\geq 0$ $R\geq 2r$

This post has been edited 1 time. Last edited by geometryshota, Jan 16, 2016, 7:01 PM

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#7 h Jan 16, 2016, 6:23 PM • 2 Y

Y by Adventure10, Mango247

xzlbq wrote:

IstekOlympiadTeam wrote:

Let $a,b,c$ be the sides of a triangle, $\alpha ,\beta ,\gamma$ the corresponding angles and $r$ the inradius. Prove that $a\cdot sin\alpha+b\cdot sin\beta+c\cdot sin\gamma\geq 9r$

$\sin \left( A \right) a+\sin \left( B \right) b+\sin \left( C \right) c-{\it h_a}-{\it h_b}-{\it h_c}=1/2\,{\frac {{a}^{2}+{b}^{2}+{c}^{2}-bc-a c-ab}{R}}$

Nice solution BQ! Greetings!

I was thinking to write the sum of $a^2+b^2+c^2=2(s^2-4Rr-r^2) \ge 18Rr$ and after this to see...comparing with Gerretsen inequality.

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