O inegalitate "tare" intr-un triunghi.
Moderators: Laurian Filip, Beniamin Bogosel, Filip Chindea
-
Virgil Nicula
- Euler
- Posts: 622
- Joined: Fri Sep 28, 2007 11:23 pm
O inegalitate "tare" intr-un triunghi.
\( a+b+c\ \le\ \underline{\overline{\left\|\ \sum\frac {bc}{b+c-a}\ \le\ \frac {(5R-2r)(4R+r)^2}{4p(2R-r)}\ \right\|}}\ . \)
-
Mateescu Constantin
- Newton
- Posts: 307
- Joined: Tue Apr 21, 2009 8:17 am
- Location: Pitesti
Sa demonstram mai intai ca intr-un triunghi are loc relatia metrica : \( \overline{\underline{\left\|\ H\Gamma^2=4R^2\[1-\frac{2p^2(2R-r)}{R(4R+r)^2}\]\ \right\|}} \) ,
unde \( H \) este ortocentrul triunghiului , iar \( \Gamma \) este punctul lui Gergonne .
Demonstratie : In triunghiul \( ABC \) consideram \( I \) centrul cercului inscris , \( IX\perp BC\ ,\ X\in (BC) \) si
\( IY\perp AC\ ,\ Y\in (AC) \) . Atunci \( \Gamma\in AX\cap BY \) este punctul lui Gergonne . Aplicam teorema lui Menelaus
transversalei \( \overline{B\Gamma Y} \) din \( \triangle\ AXC\ :\ \frac{AY}{YC}\ \cdot\ \frac{CB}{BX}\ \cdot\ \frac{X\Gamma}{\Gamma A}=1\ \Longrightarrow\ \frac{A\Gamma}{\Gamma X}=\frac{a(p-a)}{(p-b)(p-c)} \) . Pentru un punct
\( P \) din plan \( \Longrightarrow\ \vec{P\Gamma}=\frac{\vec{PA}+\frac{a(p-a)}{(p-b)(p-c)}\vec{PX}}{1+\frac{a(p-a)}{(p-b)(p-c)}}=\frac{(p-b)(p-c)}{r(4R+r)}\vec{PA}+\frac{a(p-a)}{r(4R+r)}\vec{PX}\ (\ast) \) , unde am folosit
faptul ca \( (p-b)(p-c)+a(p-a)=(p-b)(p-c)+(p-b+p-c)(p-a)=\sum(p-b)(p-c)=r(4R+r) \)
Pe de alta parte cum \( \frac{BX}{XC}=\frac{p-b}{p-c}\ \Longrightarrow\ \vec{PX}=\frac{\vec{PB}+\frac{p-b}{p-c}\vec{PC}}{1+\frac{p-b}{p-c}}=\frac{(p-c)\vec{PB}+(p-b)\vec{PC}}{a} \) si relatia \( (\ast) \) devine :
\( \underline{\overline{\left\|\ \vec{P\Gamma}=\frac{(p-b)(p-c)}{r(4R+r)}\ \cdot\ \vec{PA}+\frac{(p-c)(p-a)}{r(4R+r)}\ \cdot\ \vec{PB}+\frac{(p-a)(p-b)}{r(4R+r)}\ \cdot\ \vec{PC}\ \right\|}} \) , cu alte cuvinte
punctul lui Gergonne \( \Gamma \) are coordonatele baricentrice \( \left\(\frac{(p-b)(p-c)}{r(4R+r)}\ ,\ \frac{(p-c)(p-a)}{r(4R+r)}\ ,\ \frac{(p-a)(p-b)}{r(4R+r)}\ \right\) \) .
Pentru \( P:=H\ \Longrightarrow\ \vec{H\Gamma}=\sum{\frac{(p-b)(p-c)}{r(4R+r)}\ \cdot\ \vec{HA}\ \Longrightarrow\ H\Gamma^2=\vec{H\Gamma}^2=\left\(\sum{\frac{(p-b)(p-c)}{r(4R+r)}\ \cdot\ \vec{HA}\right\)^2 \)
\( \Longleftrightarrow\ H\Gamma^2=\sum\ \left\(\frac{(p-b)(p-c)}{r(4r+r)}\right\)^2\ \cdot\ AH^2\ +\ 2\ \cdot\ \sum\ \left\[\frac{(p-a)^2(p-b)(p-c)}{r^2(4R+r)^2}\ \cdot\ \vec{HB}\ \cdot\ \vec{HC}\right\] \)
\( \Longleftrightarrow\ H\Gamma^2=\frac{4R^2}{r^2(4R+r)^2}\sum (p-b)^2(p-c)^2\cos^2 A\ +\ \frac{2\prod (p-a)}{r^2(4R+r)^2}\sum (p-a)\cdot 4R^2\cos B\cdot\cos C\cdot(-\cos A) \)
\( \Longleftrightarrow\ H\Gamma^2=\frac{4R^2}{r^2(4R+r)^2}\sum\ \left\[(p-b)^2(p-c)^2-\frac{a^2(p-b)^2(p-c)^2}{4R^2}\right\]\ -\ \frac{8R^2\prod (p-a)}{r^2(4R+r)^2}\sum\ (p-a)\prod\cos A \)
\( \Longleftrightarrow\ H\Gamma^2=\frac{1}{r^2(4R+r)^2}\[4R^2\sum \left\[(p-b)(p-c)\right\]^2-\sum \[a(p-b)(p-c)\]^2-8pR^2\prod (p-a)\cdot\prod\cos A\]\ (\ast\ast) \)
Acum facem apel la identitati . Se stie ca \( \prod (p-a)=pr^2\ (1) \) si \( \prod\cos A=\frac{a^2+b^2+c^2-8R^2}{8R^2}=\frac{p^2-(2R+r)^2}{4R^2}\ (2) \) .
Din identitatea \( \sum (p-b)(p-c)=r(4R+r)\ \Longrightarrow\ \sum (p-b)^2(p-c)^2=r^2\[(4R+r)^2-2p^2\]\ \ (3) \)
Din identitatea \( \sum a(p-b)(p-c)=2pr(2R-r)\ \Longrightarrow\ \sum a^2(p-b)^2(p-c)^2=2p^2r^2(8R^2+r^2-p^2)\ \ (4) \)
Inlocuind relatiile \( (1)\ ,\ (2)\ ,\ (3)\ ,\ (4) \) in \( (\ast\ast) \) obtinem dupa cateva calcule ca \( \overline{\underline{\left\|\ H\Gamma^2=4R^2\[1-\frac{2p^2(2R-r)}{R(4R+r)^2}\]\ \right\|}} \) .
=====================================================================================
Intrucat \( H\Gamma^2\ \ge\ 0 \) , din relatia precedenta deducem inegalitatea \( \overline{\underline{\left\|\ p^2\ \le\ \frac{R(4R+r)^2}{2(2R-r)}\ \right\|}}\ \le\ 4R^2+4Rr+3r^2 \) ,
ceea ce constituie o intarire a inegalitatii Gerretsen . Sa revenim la inegalitate propusa .
Putem scrie \( \sum\ \frac{bc}{b+c-a}=\sum\ \frac{4bc(p-b)(p-c)}{\prod (p-a)}=\frac{4}{pr^2}\sum\ bc[p^2-p(b+c)+bc] \) . In fine tinand cont si de relatiile
\( \sum b^2c^2=(p^2+r^2+4Rr)^2-16Rrp^2\ ,\ \sum bc(b+c)=\sum bc[(b+c+a)-a]=2p(p^2+r^2-2Rr) \) ,
ultima egalitate devine : \( \overline{\underline{\left\|\ \sum\ \frac{bc}{b+c-a}=\frac{p^2+(4R+r)^2}{2p}\ \right\|}} \) . Aplicand inegalitatea demonstrata anterior obtinem
tocmai relatia ceruta : \( \overline{\underline{\left\|\ \sum\ \frac{bc}{b+c-a}\ \le\ \frac{(5R-2r)(4R+r)^2}{4p(2R-r)}\ \right\|}}\ \ \ \ \mbox{O.K.} \)
unde \( H \) este ortocentrul triunghiului , iar \( \Gamma \) este punctul lui Gergonne .
Demonstratie : In triunghiul \( ABC \) consideram \( I \) centrul cercului inscris , \( IX\perp BC\ ,\ X\in (BC) \) si
\( IY\perp AC\ ,\ Y\in (AC) \) . Atunci \( \Gamma\in AX\cap BY \) este punctul lui Gergonne . Aplicam teorema lui Menelaus
transversalei \( \overline{B\Gamma Y} \) din \( \triangle\ AXC\ :\ \frac{AY}{YC}\ \cdot\ \frac{CB}{BX}\ \cdot\ \frac{X\Gamma}{\Gamma A}=1\ \Longrightarrow\ \frac{A\Gamma}{\Gamma X}=\frac{a(p-a)}{(p-b)(p-c)} \) . Pentru un punct
\( P \) din plan \( \Longrightarrow\ \vec{P\Gamma}=\frac{\vec{PA}+\frac{a(p-a)}{(p-b)(p-c)}\vec{PX}}{1+\frac{a(p-a)}{(p-b)(p-c)}}=\frac{(p-b)(p-c)}{r(4R+r)}\vec{PA}+\frac{a(p-a)}{r(4R+r)}\vec{PX}\ (\ast) \) , unde am folosit
faptul ca \( (p-b)(p-c)+a(p-a)=(p-b)(p-c)+(p-b+p-c)(p-a)=\sum(p-b)(p-c)=r(4R+r) \)
Pe de alta parte cum \( \frac{BX}{XC}=\frac{p-b}{p-c}\ \Longrightarrow\ \vec{PX}=\frac{\vec{PB}+\frac{p-b}{p-c}\vec{PC}}{1+\frac{p-b}{p-c}}=\frac{(p-c)\vec{PB}+(p-b)\vec{PC}}{a} \) si relatia \( (\ast) \) devine :
\( \underline{\overline{\left\|\ \vec{P\Gamma}=\frac{(p-b)(p-c)}{r(4R+r)}\ \cdot\ \vec{PA}+\frac{(p-c)(p-a)}{r(4R+r)}\ \cdot\ \vec{PB}+\frac{(p-a)(p-b)}{r(4R+r)}\ \cdot\ \vec{PC}\ \right\|}} \) , cu alte cuvinte
punctul lui Gergonne \( \Gamma \) are coordonatele baricentrice \( \left\(\frac{(p-b)(p-c)}{r(4R+r)}\ ,\ \frac{(p-c)(p-a)}{r(4R+r)}\ ,\ \frac{(p-a)(p-b)}{r(4R+r)}\ \right\) \) .
Pentru \( P:=H\ \Longrightarrow\ \vec{H\Gamma}=\sum{\frac{(p-b)(p-c)}{r(4R+r)}\ \cdot\ \vec{HA}\ \Longrightarrow\ H\Gamma^2=\vec{H\Gamma}^2=\left\(\sum{\frac{(p-b)(p-c)}{r(4R+r)}\ \cdot\ \vec{HA}\right\)^2 \)
\( \Longleftrightarrow\ H\Gamma^2=\sum\ \left\(\frac{(p-b)(p-c)}{r(4r+r)}\right\)^2\ \cdot\ AH^2\ +\ 2\ \cdot\ \sum\ \left\[\frac{(p-a)^2(p-b)(p-c)}{r^2(4R+r)^2}\ \cdot\ \vec{HB}\ \cdot\ \vec{HC}\right\] \)
\( \Longleftrightarrow\ H\Gamma^2=\frac{4R^2}{r^2(4R+r)^2}\sum (p-b)^2(p-c)^2\cos^2 A\ +\ \frac{2\prod (p-a)}{r^2(4R+r)^2}\sum (p-a)\cdot 4R^2\cos B\cdot\cos C\cdot(-\cos A) \)
\( \Longleftrightarrow\ H\Gamma^2=\frac{4R^2}{r^2(4R+r)^2}\sum\ \left\[(p-b)^2(p-c)^2-\frac{a^2(p-b)^2(p-c)^2}{4R^2}\right\]\ -\ \frac{8R^2\prod (p-a)}{r^2(4R+r)^2}\sum\ (p-a)\prod\cos A \)
\( \Longleftrightarrow\ H\Gamma^2=\frac{1}{r^2(4R+r)^2}\[4R^2\sum \left\[(p-b)(p-c)\right\]^2-\sum \[a(p-b)(p-c)\]^2-8pR^2\prod (p-a)\cdot\prod\cos A\]\ (\ast\ast) \)
Acum facem apel la identitati . Se stie ca \( \prod (p-a)=pr^2\ (1) \) si \( \prod\cos A=\frac{a^2+b^2+c^2-8R^2}{8R^2}=\frac{p^2-(2R+r)^2}{4R^2}\ (2) \) .
Din identitatea \( \sum (p-b)(p-c)=r(4R+r)\ \Longrightarrow\ \sum (p-b)^2(p-c)^2=r^2\[(4R+r)^2-2p^2\]\ \ (3) \)
Din identitatea \( \sum a(p-b)(p-c)=2pr(2R-r)\ \Longrightarrow\ \sum a^2(p-b)^2(p-c)^2=2p^2r^2(8R^2+r^2-p^2)\ \ (4) \)
Inlocuind relatiile \( (1)\ ,\ (2)\ ,\ (3)\ ,\ (4) \) in \( (\ast\ast) \) obtinem dupa cateva calcule ca \( \overline{\underline{\left\|\ H\Gamma^2=4R^2\[1-\frac{2p^2(2R-r)}{R(4R+r)^2}\]\ \right\|}} \) .
=====================================================================================
Intrucat \( H\Gamma^2\ \ge\ 0 \) , din relatia precedenta deducem inegalitatea \( \overline{\underline{\left\|\ p^2\ \le\ \frac{R(4R+r)^2}{2(2R-r)}\ \right\|}}\ \le\ 4R^2+4Rr+3r^2 \) ,
ceea ce constituie o intarire a inegalitatii Gerretsen . Sa revenim la inegalitate propusa .
Putem scrie \( \sum\ \frac{bc}{b+c-a}=\sum\ \frac{4bc(p-b)(p-c)}{\prod (p-a)}=\frac{4}{pr^2}\sum\ bc[p^2-p(b+c)+bc] \) . In fine tinand cont si de relatiile
\( \sum b^2c^2=(p^2+r^2+4Rr)^2-16Rrp^2\ ,\ \sum bc(b+c)=\sum bc[(b+c+a)-a]=2p(p^2+r^2-2Rr) \) ,
ultima egalitate devine : \( \overline{\underline{\left\|\ \sum\ \frac{bc}{b+c-a}=\frac{p^2+(4R+r)^2}{2p}\ \right\|}} \) . Aplicand inegalitatea demonstrata anterior obtinem
tocmai relatia ceruta : \( \overline{\underline{\left\|\ \sum\ \frac{bc}{b+c-a}\ \le\ \frac{(5R-2r)(4R+r)^2}{4p(2R-r)}\ \right\|}}\ \ \ \ \mbox{O.K.} \)