187. Distancies.

by Virgil Nicula, Dec 10, 2010, 8:02 PM

Distanta intre doua puncte remarcabile ale unui triunghi.


Consideram un triunghi fix $ ABC$ si vom caracteriza pozitia unui punct oarecare $ P$ din planul triunghiului dat in raport cu acesta.

Notam produsul scalar $ v_1\circ v_2$ al vectorilor $ v_1$ , $ v_2\ .$ Reamintim doua relatii metrice utile din geometria sintetica in exprimare vectoriala :

Relatia lui Pitagora generalizata. $ \triangle\ ABC\ \implies\ BC^2=AB^2+AC^2-2\cdot\overrightarrow {AB}\circ\overrightarrow {AC}$ , unde $ \overrightarrow {AB}\circ\overrightarrow {AC}=bc\cdot\cos A\ .$

Relatia lui Stewart. $ A\in BC\ \implies\ (\forall )\ P\ ,\ PA^2\cdot \overline {BC}+PB^2\cdot \overline {CA}+PC^2\cdot \overline {AB}+\overline {BC}\cdot\overline {CA}\cdot\overline {AB}=0\ .$

$ \bigodot$ Lema. Pentru orice $ P$ exista si este unic tripletul ordonat $ (x,y,z)\in\mathbb R^3$ astfel incat $ \left|\begin{array}{c}
x+y+z=1\\\\
x\cdot \overrightarrow {PA}+y\cdot\overrightarrow {PB}+z\cdot\overrightarrow {PC}=0\end{array}\right|\ .$

In acest caz vom spune ca punctul $ P$ are coordonatele baricentrice $ \ (x,y,z)$ in raport cu $ \triangle\ ABC$ si vom scrie $ P\ (x,y,z)\ .$


Observatii.

$ \odot$ Daca $ P(x,y,z)$ , atunci pentru orice punct $ M$ exista relatia $ x\cdot \overrightarrow {MA}+y\cdot\overrightarrow {MB}+z\cdot\overrightarrow {MC}=\overrightarrow {MP}\ .$

$ \odot$ Un punct $ P$ are coordonatele baricentrice $ (x,y,z)$ in raport cu triunghiul $ ABC\ \ \Longleftrightarrow$

$ \boxed {\begin{array}{c}
\frac {|x|}{[PBC]}=\frac {|y|}{[PCA]}=\frac {|z|}{[PAB]}\\\\
x+y+z=1\end{array}}$ and $\boxed{\begin{array}{c}
x=0\ \Longleftrightarrow\ P\in BC\\\\
x < 0\ \Longleftrightarrow\ \mathrm {dreapta}\ BC\ \mathrm {separa\ punctele}\ A\ ,\ P\\\\
x > 0\ \Longleftrightarrow\ P\not\in BC\ \mathrm {si\ dreapta}\ BC\ \mathrm {nu\ separa\ punctele}\ A\ ,\ P\end{array}}$

$ \odot$ Daca $ P(x,y,z)$ si $ D\in BC\cap AP$ , atunci $ \left|\begin{array}{c}
y\cdot \overline {DB}+z\cdot\overline {DC}=0\\\\
\overrightarrow {AP}=y\cdot\overrightarrow {AB}+z\cdot\overrightarrow{AC}\end{array}\right|\ \ \wedge\ \ \left|\begin{array}{c}
\frac {\overline {BD}}{z}=\frac {\overline {DC}}{y}=\frac {\overline {BC}}{y+z} \\\\
\frac {\overline{AP}}{y+z}=\frac {\overline {PD}}{x}=\overline {AD}\end{array}\right|\ .$

$ \odot$ Daca $ P(x,y,z)$ si $ D\in BC\cap AP$ , atunci $ \boxed {\ AD^2=\frac {zb^2+yc^2}{y+z}-\frac {yza^2}{(y+z)^2}\ }\ (*)\ .$ Intr-adevar, aplicam relatia Stewart

cevienei $ AD$ in $ \triangle\ ABC\ : \ AD^2\cdot\overline {BC}+c^2\cdot\overline {CD}+b^2\cdot\overline {DB}+\overline {BC}\cdot\overline {CD}\cdot\overline {DB}=0$ $ \implies$

$ AD^2\cdot\overline {BC}-\frac {yc^2}{y+z}\cdot\overline {BC}-\frac {zb^2}{y+z}\cdot\overline {BC}+\frac {yza^2}{(y+z)^2}\cdot\overline {BC}=0\implies$ $ AD^2=\frac {zb^2+yc^2}{y+z}-\frac {yza^2}{(y+z)^2}\ .$

$ \odot$ Coordonatele baricentrice ale unor puncte remarcabile in raport cu triunghiul $ ABC\ : \ \ A(1,0,0)\ \ ;\ \ B(0,1,0)\ \ ;\ \ C(0,0,1)\ ;$

Centrul de greutate $ G\left(\frac 13,\frac 13,\frac 13\right)\ ;$ Incentrul $ I\left(\frac {a}{2p},\frac {b}{2p},\frac {c}{2p}\right)\ ;$ $ A$ - Exincentrul $ I_a\left(\frac {-a}{2(p-a)},\frac {b}{2(p-a)},\frac {c}{2(p-a)}\right)$ etc ;

Punctul lui Nagel $ N\left(\frac {p-a}{p},\frac {p-b}{p},\frac {p-c}{p}\right)\ ;$ Punctul lui Gergonne $ \Gamma\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)\ ;$

Ortocentrul $ H\left(\cot B\cot C,\cot C\cot A,\cot A\cot B\right)\ ;$ Circumcentrul $ O\left(\frac {R^2\sin 2A}{2S},\frac {R^2\sin 2B}{2S},\frac {R^2\sin 2C}{2S}\right)\ ;$

Centrul simedian (Lemoine) $ K\left(\frac {a^2}{S_2},\frac {b^2}{S_2},\frac {c^2}{S_2}\right)\ ,$ unde $ S_p=a^p+b^p+c^p\ ,\ p\in\mathrm N\ .$

$ \bigodot$ Teorema 1. Puterea punctului $ P(x,y,z)$ fata de cercul $ w=C(O,R)$ este $ \boxed {\ p_w(P)=-\left(yza^2+zxb^2+xyc^2\right)\ }\ .$

Quote:
Demonstratie. Fie $ D\in BC\cap AP$ si $ \{D,D'\}=AP\cap w\ .$ Asadar $ p_w(P)=\overline {PA}\cdot\overline {PD'}=$ $ \overline {PA}\cdot\left(\overline {PD}+\overline {DD'}\right)=$

$ \overline {PA}\cdot\overline {PD}+\overline {PA}\cdot\overline {DD'}\ .$ Insa $ \left|\begin{array}{c}
\overline {PA}=(y+z)\cdot\overline {DA}\\\\
\overline {PD}=-x\cdot \overline {DA}\end{array}\right|\ .$ Deci $ p_w(P)=-x(y+z)\cdot DA^2+(y+z)\cdot \overline {DA}\cdot\overline {DD'}=$

$ -x(y+z)\cdot AD^2+p_w(D)\ .$ Insa $ p_w(D)=\overline {DB}\cdot\overline {DC}$ si din relatia $ (*)$ aplicata cevienei $ AD$ se obtine :

$ p_w(P)=-x\left(zb^2+yc^2\right)+\frac {xyza^2}{y+z}-\frac {yza^2}{y+z}=-\left(yza^2+zxb^2+xyc^2\right)$ deoarece $ y+z=1-x\ .$

Observatii. $ OP^2=R^2+p_w(P)=R^2-$ $ \left(yza^2+zxb^2+xyc^2\right)\ge 0\ \implies\ \boxed {\ yza^2+zxb^2+xyc^2\le R^2\ }\ .$

$ \odot$ Puterile unor puncte remarcabile in raport cu circumcercul $ w=C(O,R)$ al triunghiului $ ABC\ :$

$ p_w(G)=-$ $ \frac {a^2+b^2+c^2}{9}\ \implies\ \boxed {\ a^2+b^2+c^2\le R^2\ }\ \ ;\ \ p_w(I)=$ $ -2Rr\ \implies\ \boxed {\ R\ge 2r\ }\ ;$

$ p_w(I_a)=2Rr_a\ \ ;\ \ p_w(N)=-4r(R-2r)\ \ ;\ \ p_w$ $ (\Gamma )=-r(R+r)\left(\frac {2p}{4R+r}\right)^2\ \implies\ \boxed {\ p^2\le\frac {R^2(4R+r)^2}{4r(R+r)}\ }\ .$

$ p_w(H)=-8R^2\cos A\cos B\cos C\ \implies\ \boxed {\ \cos A\cos B\cos C\le\frac 18\ }\ ;$

$ \ p_w(K)=-3\left(\frac {abc}{a^2+b^2+c^2}\right)^2$ $ \implies$ $ \boxed {\ a^2+b^2+c^2\ge 4S\sqrt 3\ }\ .$

$ \bigodot$ Teorema 2. $ \boxed {\ P(x,y,z)\ \implies\ x\cdot PA^2+y\cdot PB^2+z\cdot PC^2=-p_w(P)\ }\ .$

Quote:
Demonstratie. Pentru $ M\in\{A,B,C\}$ in identitatea $ x\cdot \overrightarrow {MA}+y\cdot\overrightarrow {MB}+z\cdot\overrightarrow {MC}=\overrightarrow {MP}$ obtinem $ \left|\begin{array}{c}
y\cdot\overrightarrow {AB}+z\cdot\overrightarrow{AC}=\overrightarrow{AP}\\\\
x\cdot\overrightarrow {BA}+z\cdot\overrightarrow{BC}=\overrightarrow{BP}\\\\
x\cdot\overrightarrow {CA}+y\cdot\overrightarrow{CB}=\overrightarrow{CP}\end{array}\right|\ .$

Realizam produsul scalar $ PA^2=\overrightarrow{PA}\circ\overrightarrow{PA}=\left(y\cdot\overrightarrow {AB}+z\cdot\overrightarrow{AC}\right)\circ\left(y\cdot\overrightarrow {AB}+z\cdot\overrightarrow{AC}\right)=$ $ z^2b^2+y^2c^2+2yz\cdot \overrightarrow {AB}\circ\overrightarrow {AC}=$

$ z^2b^2+y^2c^2+yz\cdot 2bc\cdot\cos A=$ $ z^2b^2+y^2c^2+yz\left(b^2+c^2-a^2\right)\ .$ Asadar,

$ \left|\begin{array}{c}
PA^2=z^2b^2+y^2c^2+yz\left(b^2+c^2-a^2\right)\\\\
PB^2=x^2c^2+z^2a^2+zx\left(c^2+a^2-b^2\right)\\\\
PC^2=y^2a^2+x^2b^2+xy\left(a^2+b^2-c^2\right)\end{array}\right|\ \begin{array}{c}
\cdot\ x\\\\
\cdot\ y\\\\
\cdot\ z\end{array}\ \bigoplus\ \implies$ $ \sum x\cdot PA^2=$ $ \sum \left(-xyz+z^2y+zxy+y^2z+xyz\right)\cdot a^2=$

$ \sum yz\cdot a^2=-p_w(P)\ .$ In concluzie, $ \sum x\cdot PA^2=-p_w(P)\ .$ Acum suntem pregatiti sa dam teorema fundamentala a distantei.

$ \bigodot$ Teorema lui Leibniz. Pentru orice punct $ M$ avem : $ \boxed {\ P(x,y,z)\ \implies\ x\cdot MA^2+y\cdot MB^2+z\cdot MC^2=MP^2-p_w(P)\ }\ .$
Quote:
Metoda I. Aplicam relatia Pitagora generalizata ; $ \left|\begin{array}{c}
MA^2=PM^2+PA^2-2\cdot\overrightarrow {PM}\circ\overrightarrow {PA}\\\\ 
MB^2=PM^2+PB^2-2\cdot\overrightarrow {PM}\circ\overrightarrow {PB}\\\\
MC^2=PM^2+PC^2-2\cdot\overrightarrow {PM}\circ\overrightarrow {PC}\end{array}\right|\begin{array}{c}
\odot\ x\\\\
\odot\ y\\\\
\odot\ z\end{array}\ \bigoplus\ \implies$ $ \sum x\cdot MA^2=$

$ MP^2+\sum x\cdot PA^2-2\cdot \overrightarrow {PM}\circ\sum x\cdot \overrightarrow {PA}=MP^2-p_w(P)$ deoarece $ \sum x\cdot \overrightarrow {PA}=\overrightarrow 0$ si $ \sum x\cdot PA^2=-p_w(P)\ .$


Metoda II. Se stie ca $ \left|\begin{array}{c}
y\cdot\overline {DB}+z\cdot\overline {DC}=0\\\\
(y+z)\cdot \overline {PA}+x\cdot\overline {PD}=0\end{array}\right|\ .$ Aplicam relatia $ (*)$ cevienelor din triunghiurilor mentionate :

$ \left|\begin{array}{cc}
AD/ABC\ : \ & (y+z)\cdot AD^2=zb^2+yc^2-\frac {yza^2}{y+z}\\\\
MD/BMC\ : \ & (y+z)\cdot MD^2=y\cdot MB^2+z\cdot MC^2-\frac {yza^2}{y+z}\\\\
MP/AMD\ : \ & MP^2=x\cdot MA^2+(y+z)MD^20-x(y+z)\cdot AD^2\end{array}\right|$ $ \implies$

$ MP^2=x\cdot MA^2+ y\cdot MB^2+z\cdot MC^2-\frac {yza^2}{y+z}-x\cdot \left[zb^2+yc^2-\frac {yza^2}{y+z}\right]=$ $ \sum x\cdot MA^2-\sum yza^2=$

$ \sum x\cdot MA^2+p_w(P)\ \implies\ \sum x\cdot MA^2=MP^2-p_w(P)\ .$

Observatii.

$ \odot\ \ P(x,y,z)\ \implies\ (\forall )\ M\ ,\ \sum yza^2\ \le\ \sum x\cdot MA^2$ (inegalitatea Klamkin).

$ \odot$ Ilustram teorema lui Leibniz si inegalitatea lui Klamkin prin cateva situatii particulare.

$ M: =A\ \implies\ PA^2=\left(zb^2+yc^2\right)-\sum yza^2\ .$ De exemplu, $ \left|\begin{array}{c}
AI^2=\frac {bc(p-a)}{p}\\\\
AN^2=(b-c)^2+4r^2\end{array}\right|$

$ M: =O\ \implies\ OP^2=R^2-\sum yza^2\ \implies\ \sum yza^2\ \le\ R^2\ .$ De exemplu, $ \left|\begin{array}{c}
OI^2=R^2-2Rr\\\\
ON=R-2r\end{array}\right|\ ,$

$ OH^2=9\cdot OG^2=9R^2+ 2\cdot\left[r(4R+r)-p^2\right]\ ,\ O\Gamma^2=$ $ R^2-\frac {4p^2r(R+r)}{(4R+r)^2}\ \implies\ p^2\ \le\ \frac {R^2(4R+r)^2}{4r(R+r)}\ .$



$ \bigodot$ Teorema 3. Distanta intre punctele $ M_k\left(x_k,y_k,z_k\right)$ , unde $ k\in\overline {1,2}$ este data de relatia

$ \boxed {\ M_1M_2^2=-\left[\left(y_1-y_2\right)\left(z_1-z_2\right)a^2+\left(z_1-z_2\right)\left(x_1-x_2\right)b^2+\left(x_1-x_2\right)\left(y_1-y_2\right)c^2\right]\ }\ .$


Example. If the point $ M$ is the Mittenpunkt's point of $\triangle ABC$ , then

$\left\{\begin{array}{c}
MI^2=\frac{4R\left[(R+r)p^2-16R^2r-r^3-8r^2R\right]}{(4R+r)^2}\\\\
MO^2=\frac{R\left[2(r-2R)p^2r+16R^3+8R^2r+r^2R\right]}{(4R+r)^2}\\\\
MH^2=\frac{(2R+r)^2(16R^2+8Rr+r^2-3p^2)}{(4R+r)^2}\end{array}\right\|$ and $\left\{\begin{array}{c}
MG^2=\frac{(4R^2-8Rr+5r^2)p^2-r^4-48R^2r^2-12r^3R-64R^3r}{9(4R+r)^2}\\\\
MS^2=\frac{4p^4R((2R+r)p^2-r^3-8r^2R-16R^2r)}{(4R+r)^2(4Rr-p^2+r^2)^2}\\\\
MF^2=\frac{4r(R+r)p^2-8R^3r-16R^4-R^2r^2}{4(4R+r)^2}\end{array}\right\|$ .



Problema propusa. Consideram doua puncte $ E$ , $ F$ astfel incat dreapta $ EF$ nu traverseaza triunghiul $ ABC$.

Sa se arate ca $ P(x,y,z)\ \implies\ x\cdot [AEF] + y\cdot [BEF] + z\cdot [CEF] = [PEF]$ (vezi
aici ).

==========================================================================

Bibliografie. Geometrie plana (sintetica, vectoriala, analitica). Culegere de probleme , Editura GIL, Zalau, 2002.
This post has been edited 30 times. Last edited by Virgil Nicula, Nov 22, 2015, 6:25 PM

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