177. A nice proof of one identity in a triangle.

by Virgil Nicula, Nov 24, 2010, 12:35 PM

PP1. Prove that in any triangle $ABC$ there is the identity $\sum\frac {a(s-a)}{h_a}=2\cdot (R+r)$ , where $2s=a+b+c$ .

Proof 1 (nice !). I"ll use the well-known relations $\begin{array}{cc} (1) & \boxed{\ 2Rh_a=bc\ }\\\\ (2) & \boxed{\ \sum\cos A=1+\frac rR\ }\end{array}$ . Thus, $\sum\frac {a(s-a)}{h_a}\ \stackrel{(1)}{=}\ 2R\cdot\sum\frac {a(s-a)}{bc}=$ $R\cdot\sum\frac {a(b+c-a)}{bc}=$

$R\cdot\sum\left(\frac ac+\frac ab-\frac {a^2}{bc}\right)=$ $R\cdot\sum\left(\frac bc+\frac cb-\frac {a^2}{bc}\right)=$ $R\cdot\sum\frac {b^2+c^2-a^2}{bc}=$ $2R\cdot\sum\cos A\ \stackrel{(2)}{=}\ 2R\cdot \left(1+\frac rR\right)=2\cdot (R+r)$ .

Remark. Since $2S=ah_a=2(s-a)r_a$ obtain that $2\cdot (R+r)=\sum\frac {a(s-a)}{h_a}=\sum\frac {a^2(s-a)}{2(s-a)r_a}$ In conclusion, $\boxed{\ \sum \frac {a^2}{r_a}=4\cdot (R+r)\ }$ .

Observe that $4\cdot (R+r)=\sum \frac {a^2}{r_a}\stackrel{(C.B.S.)}{\ \ge\ }\frac {\left(\sum a\right)^2}{\sum r_a}=\frac {4s^2}{4R+r}$ $\implies$ $\boxed{\ s^2\le (R+r)(4R+r)\ }\ (*)$ . Otherwise, from the Gerretsen's inequality

$s^2\le 4R^2+4Rr+3r^2$ and the equivalence $4R^2+4Rr+3r^2\le (R+r)(4R+r)\iff 2r\le R$ obtain that $s^2\le (R+r)(4R+r)$ , i.e. $(*)$ .

Proof 2. Using the well-known relations $\left\|\begin{array}{c} \sum bc=s^2+r^2+4Rr\\\\ \sum a^2=2\cdot\left(s^2-r^2-4Rr\right)\\\\ \sum a^3=\left(\sum a\right)^3-3\cdot\sum a\cdot\sum bc+3abc\end{array}\right\|$ obtain that $\sum\frac {a(s-a)}{h_a}=\frac {1}{2S}\cdot\sum a^2(s-a)=$

$\frac {1}{2sr}\cdot\left(s\cdot\sum a^2-\sum a^3\right)=$ $\frac {1}{2sr}\cdot\left[2s\left(s^2-r^2-4Rr\right)-8s^3+6s\left(s^2+r^2+4Rr\right)-12Rsr\right]=$ $2(R+r)$ .

Proof 3. Using the well-known relations $\tan\frac A2=\frac {r}{s-a}$ and $S=2R^2\cdot \prod\sin A$ obtain that $\sum\frac {a(s-a)}{h_a}=$ $\sum \frac {2R\cdot\sin A}{2R\cdot\sin B\sin C}\cdot\frac {r}{\tan\frac A2}=$

$r\cdot\sum \frac {2\sin\frac A2\cos\frac A2}{\sin B\sin C}\cdot\frac {\cos\frac A2}{\sin\frac A2}=$ $r\cdot\sum\frac {1+\cos A}{\sin B\sin C}=$ $\frac {r}{\prod\sin A}\cdot\sum \left(\sin A+\frac 12\cdot\sin 2A\right)=$ $\frac {r}{\prod\sin A}\cdot \left(\frac sR+2\cdot\prod \sin A\right)=$ $2\cdot (R+r)$ .

Proof 4. Denote $\left\{\begin{array}{c} x=s-a\\\ y=s-b\\\ z=s-c\end{array}\right\|$ . Observe that $\left\{\begin{array}{c} a=y+z\\\ b=z+x\\\ c=x+y\\\ x+y+z=s\end{array}\right\|$ and $xy+yz+zx=r(4R+r)\ ,\ xyz=sr^2$ .

Therefore, $\sum\frac {a(s-a)}{h_a}=$ $\frac {1}{2sr}\cdot\sum a^2(s-a)=$ $\frac {1}{2sr}\cdot \sum x(y+z)^2=$ $\frac {1}{2sr}\cdot \sum\left[x\left(y^2+z^2\right)+2xyz\right]=$

$\frac {1}{2sr}\cdot \left[6xyz+\sum yz(y+z)\right]=$ $\frac {1}{2sr}\cdot \left(3xyz+\sum x\cdot\sum yz\right)=$ $\frac {1}{2sr}\cdot \left[3sr^2+sr(4R+r)\right]=$ $2\cdot (R+r)$ .
This post has been edited 39 times. Last edited by Virgil Nicula, Dec 1, 2015, 11:26 AM

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    by samrocksnature, Apr 14, 2021, 5:16 AM

  • Holy- Darn this is good. shame it's inactive now

    by the_mathmagician, Jan 17, 2021, 7:43 PM

  • godly blog. opopop

    by OlympusHero, Dec 30, 2020, 6:08 PM

  • long blog

    by MrMustache, Nov 11, 2020, 4:52 PM

  • 372554 views!

    by mrmath0720, Sep 28, 2020, 1:11 AM

  • wow... i am lost.

    369302 views!

    -piphi

    by piphi, Jun 10, 2020, 11:44 PM

  • That was a lot! But, really good solutions and format! Nice blog!!!! :)

    by CSPAL, May 27, 2020, 4:17 PM

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

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

72 shouts
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About Owner
  • Posts: 7054
  • Joined: Jun 22, 2005
Blog Stats
  • Blog created: Apr 20, 2010
  • Total entries: 456
  • Total visits: 404395
  • Total comments: 37
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