250. An application of the Euler's relation.

by Virgil Nicula, Mar 10, 2011, 12:17 PM

PP. Prove that for any acute or right-angled triangle $ABC$ exists the relation $\frac{m_a}{h_a}+\frac{m_b}{h_b}+\frac{m_c}{h_c} \le 1+\frac{R}{r}$ .

Proof. Suppose that $\triangle ABC$ is acute or right-angled. Denote its circumcircle $w=C(O,R)$ and the midpoints $M$ , $N$ , $P$ of the sides $[BC]$ , $[CA]$ , $[AB]$ respectively.

Thus, $m_a\le R+OM$ , $m_b\le R+ON$ , $m_c\le R+OP$ . I"ll use the well-known relations $\left\|\begin{array}{c} OM+ON+OP=R+r\ (\mathrm{\underline{Euler's\ relation}})\\\\ a\cdot OM+b\cdot ON+c\cdot OP=2S\end{array}\right\|$ $\implies$ $\sum\frac {m_a}{h_a}=$ $\sum\frac {am_a}{ah_a}=$

$\sum\frac {am_a}{2S}\le$ $\frac {1}{2S}\cdot \sum a(R+OM)=$ $\frac {2sR+2S}{2S}=\frac Rr+1$ $\implies$ $\sum\frac {m_a}{h_a}\le 1+\frac Rr$ , where $2s=a+b+c$ . Our inequality is equivalent with $\frac {am_a+bm_b+cm_c}{a+b+c}\ \le\ R+r$ .

Since $\min\left\{m_a,m_b,m_c\right\}\le \frac {am_a+bm_b+cm_c}{a+b+c}$ obtain $\min\left\{m_a,m_b,m_c\right\}\ \le \ R+r$ . Observe that $\sum OM=R\sum\cos A=R+r\iff$ $\boxed{\sum \cos A=1+\frac rR}$ .

Remark. $\cos A+(\cos B+\cos C)=\left(1-2\sin^2\frac A2\right)+2\sin\frac A2\cos\frac {B-C}{2}=$ $1+2\sin\frac A2\left(\cos\frac {B-C}2-\cos\frac {B+C}2\right)=$ $1+4\sin\frac A2\sin\frac B2\sin\frac C2=$

$1+4\sqrt{\frac {(s-b)(s-c)}{bc}\cdot\frac {(s-c)(c-a)}{ac}\cdot\frac {(s-a)(s-b)}{ab}}=$ $1+\frac {4(s-a)(s-b)(s-c)}{abc}=$ $1+\frac {4sr^2}{4Rsr}=1+\frac rR$ . Now I"ll prove the Euler's relation.

Apply the Ptolemy's relation in $:\ \left\{\begin{array}{ccc} \mathrm{APON} & : & b\cdot OP+c\cdot ON=aR\\\\ \mathrm{BMOP} & : & c\cdot OM+a\cdot OP=bR\\\\ \mathrm{CNOM} & : & a\cdot ON+b\cdot OM=cR\\\\ \mathrm{<\ and\ >} & : & a\cdot OM+b\cdot ON+c\cdot OP=(a+b+c)r\end{array}\right\|\ \bigoplus\ \implies$ $\sum a\cdot\sum OM=(R+r)\cdot \sum a\implies$ $\sum OM=R+r$ .

Otherwise. Let incircle $C(I,r)$ , circumcircle $w=C(O,R)$ and $S\in II_a\cap w$ . Is well-known or prove easily that $OM=OS-MS=R-\frac {r_a-r}{2}$ . Thus,

$OM+ON+OP=\sum\left(R-\frac {r_a-r}{2}\right)=$ $3R-\frac {\left(r_a+r_b+r_c\right)-3r}2=$ $3R-\frac {(4R+r)-3r}{2}=R+r\implies$ $OM+ON+OP=R+r$ . I used identity

$\boxed{r_a+r_b+r_c=4R+r}$ . Indeed, $(s-a)(s-b)(s-c)=$ $s^3-s^2(a+b+c)+s(ab+bc+ca)-abc\iff$ $sr^2=s^3-2s^3+s(ab+bc+ca)-4Rsr\iff$

$r^2=-s^2+(ab+bc+ca)-4Rr\iff$ $\boxed{ab+bc+ca=s^2+r^2+r(4R+r)}$ . Thus, $rr_a=\frac Ss\cdot\frac S{s-a}=\frac {S^2}{s(s-a)}=(s-b)(s-c)$ a.s.o.

$\implies$ $r\sum r_a=\sum (s-b)(s-c)=-s^2+(ab+bc+ca)=$ $r(4R+r)\implies$ $r_a+r_b+r_c=4R+r$ . Observe that $4(ab+bc+ca)=$

$(a+b+c)^2+\sum\left[a^2-(b-c)^2\right]\iff$ $4\sum bc=4s^2+4(s-b)(s-c)\iff$ $\sum bc=s^2+\sum(s-b)(s-c)$ .

Another identities: $\boxed{II_a^2=(r_a+r)^2+(b-c)^2=(r_a-r)^2+a^2}$ and $\boxed{r_a-r=\frac {ar_a}{s}=4R\sin^2\frac A2}$ .

Remark.In the case of the obtuse triangle $ABC$ , for example $A>\frac {\pi}{2}$ , obtain analogously that $OM=-R\cos A$ and $\boxed{\begin{array}{ccc} OM+ON+OP & = & r_a-R\\\\ \cos B+\cos C-\cos A & = & \frac {r_a}{R}-1\end{array}}\ (*)$ .

Indeed, $\left\{\begin{array}{ccc} \mathrm{APON} & : & b\cdot OP+c\cdot ON=aR\\\\ \mathrm{BMOP} & : & c\cdot OM-a\cdot OP=-bR\\\\ \mathrm{CNOM} & : & -a\cdot ON+b\cdot OM=-cR\\\\ \mathrm{<\ and\ >} & : & -a\cdot OM+b\cdot ON+c\cdot OP=(-a+b+c)r_a\end{array}\right\|\ \bigoplus\ \implies$ $(b+c-a)\cdot\sum OM=(b+c-a)(r_a-R)\implies$ $\sum OM=r_a-R$ .

Otherwise. $OM+ON+OP=R(-cos A+\cos B+\cos C)$ and $(\cos B+\cos C)-\cos A=2\sin\frac A2\cos\frac {B-C}{2}-\left(1-2\sin^2\frac A2\right)=$

$-1+2\sin\frac A2\left(\cos\frac {B+C}{2}+\cos\frac {B-C}{2}\right)=$ $-1+4\sin\frac A2\cos\frac B2\cos\frac C2=$ $-1+4\sqrt {\frac {(s-b)(s-c)}{bc}\cdot\frac {s(s-b)}{ac}\cdot\frac {s(s-c)}{ab}}=$

$-1+\frac {4s(s-b)(s-c)}{abc}=$ $-1+\frac {4S^2}{4RS(s-a)}=$ $-1+\frac {S}{R(s-a)}=-1+\frac {r_a}{R}$ $\implies\ (*)$ .
This post has been edited 57 times. Last edited by Virgil Nicula, Nov 22, 2015, 11:31 AM

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27.Problem of Darij Grinberg (I - centroid of triangle DEF).
26. Outside of a triangle.
25. Geometric equations.
24. A fine and cool inequalities.
23. A very nice exponential inequality (own - from CRUX).
22. Some interesting limits (sequence/function).
21 bis. Some problems with complex numbers.
21. Probleme de geometrie (Yetty & I).
20. O ineg.cu suma de AG/AI .
19. Own proposed problems in CRUX Mathematicorum - Canada.
18. O problema a lui Toshio Seimiya.
17. O inegalitate "tare" intr-un triunghi.
16. Length of the (interior) angle bisector (10 proofs).
15. Inequalities stronger than Panaitopol's.
14. Opinii si comentarii relativ la inv. rom.
13. Inegalitatea de la Sorin Borodi.
12. Inegalitatea I. Maftei & S. Radulescu (2).
11. Inegalitatea I. Maftei & S. Radulescu (1).
10. Walker's inequality and its extension.
9. Inegalitatea HO+HA+HB+HC >= 3R .
8. The first problem Shortlist ONM 2010 Romania.
7. Some interesting "slicing" problems.
6. Interesting problems from JBMO.
5. Theoretical preliminary for inequalities.
4. Remarkable geometrical inequalities (2).
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    by ryanbear, May 9, 2023, 6:10 AM

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    by tigerzhang, Aug 1, 2021, 12:02 AM

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

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

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    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

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  • Total entries: 456
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