128. Value of IG and two inequalities in neighbourhood of 0.

by Virgil Nicula, Sep 22, 2010, 4:02 PM

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=151&t=366349

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=151&t=368232
Quote:
Prove that for every triangle it holds $9\cdot GI^2 =s^2 -16Rr +5r^2$ .
Method 1. If $T$ , $M$ are the projections of $I$ , $G$ on $BC$ and $S$ is the projection of $G$ on $IT$, then

$IG^2=SG^2+SI^2$ $\iff$ $IG^2=TM^2+\left(IT-GM\right)^2$ $\iff$ $IG^2=\frac {(b-c)^2(b+c-3a)^2}{36a^2}+$ $\left(r-\frac {h_a}{3}\right)^2$ $\iff$

$36a^2(a+b+c)\cdot IG^2=$ $(a+b+c)(b-c)^2(3a-b-c)^2+(a-b+c)(a+b-c)(b+c-a)(2a-b-c)^2$ $\iff$

$\boxed{9(a+b+c)\cdot IG^2=2(a+b)(b+c)(c+a)-13abc-\left(a^3+b^3+c^3\right)}$ - a symmetrical expression in $a$ , $b$ , $c$ .

From this identity obtain easily that $9\cdot IG^2=s^2-16Rr+5r^2$ and the remarkable inequality $\boxed{s^2+5r^2\ge 16Rr}$ .

Remark. From here can obtain two inequalities in the neighbourhood of $0\ \ :\ \ \boxed {6a\cdot IG\ge |(b-c)(3a-b-c)|}$ and $\boxed{3a\cdot IG\ge r\cdot |2a-b-c|}$ .
Quote:
Proposed problem. Prove that in a triangle $ABC$ there is the inequality $\boxed {\ 40abc\ \le\ (a+b+c)^3+13abc\ \le\ 5(a+b)(b+c)(c+a)\ }$ .

An equivanent form is $\boxed {\ 16abc\ \le\ a^3+b^3+c^3+13abc\ \le\ 2(a+b)(b+c)(c+a)\ }$ .
Quote:
Proposed problem. Prove that for $\triangle ABC$ the expression $f(a,b,c)\equiv \frac {1}{a^2}\cdot \left[(b-c)^2(b+c-3a)^2+4r^2(b+c-2a)^2\right]$ is symmetrically.

Proof. Observe that $f(a,b,c)$ is symmetrically w.r.t. $b$ , $c$ . Using the relation $sr^2=(s-a)(s-b)(s-c)$ prove easily that

$f(a,b,c)=f(b,c,a)$ , i.e. $\frac {(b-c)^2(b+c-3a)^2+4r^2(b+c-2a)^2}{a^2}=\frac {(c-a)^2(c+a-3b)^2+4r^2(c+a-2b)^2}{b^2}$ .

Method 2 (classic). For any $X$ we have $\sum a\cdot XA^2=(a+b+c)\cdot XI^2+abc$ .

For $X:=G$ and using $9\cdot GA^2=4m_a^2=2(b^2+c^2)-a^2$ obtain the required identity.

Method 3 (classic). For any $X$ we have $\sum XA^2=3\cdot XG^2+\frac {a^2+b^2+c^2}{3}$ .

For $X:=I$ and using $IA^2=\frac {bc(s-a)}{s}=bc-4Rr$ obtain the required identity.

Remark. Can use and the relations $\left\|\begin{array}{c} s_1\equiv a+b+c=2s\\\\ s_2\equiv ab+bc+ca=s^2+r^2+4Rr\\\\ s_3\equiv abc=4Rrs\\\\ S_2\equiv a^2+b^2+c^2=2(s^2-r^2-4Rr)\\\\ S_3\equiv a^3+b^3+c^3=2s\left(s^2-3r^2-6Rr\right)\end{array}\right\|$ .
This post has been edited 25 times. Last edited by Virgil Nicula, Nov 23, 2015, 7:29 AM

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  • impressive :D
    awesome. 358,000 visits?????

    by OlympusHero, May 14, 2020, 8:43 PM

72 shouts
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