Question

Dat fiind un triunghi ABC cu I centrul cercului inscris triunghiului. Sa se arate ca punctele I,M,N coliniare (M ∈ [AB] ∧ N ∈ [AC]) daca si numai daca 1/MA + 1/NA = a+b+c/bc . (a,b,c lat triunghiului)

Answer 1

$\displaystyle Este~o~problema~clasica.~Voi~prezenta~o~solutie~vectoriala. \\ \\ Notam~ \frac{AM}{MB}=m~si~\frac{AN}{NC}=n. \\ \\ Avem~ \overrightarrow{IM}= \frac{1}{m+1}\overrightarrow{IA}+\frac{m}{n+1}\overrightarrow{IB}~si~\overrightarrow{IN}= \frac{1}{n+1}\overrightarrow{IA}+\frac{n}{n+1}\overrightarrow{IC}. \\ \\ Se~stie~ca~pentru~orice~punct~X~din~plan~are~loc~relatia \\ \\ \overrightarrow{IX}= \frac{a\overrightarrow{AX}+b\overrightarrow{BX}+c\overrightarrow{CX}}{a+b+c}.$

$\displaystyle Folosim~aceasta~relatie~pentru~X=A,~X=B~si~X=C.~Astfel \\ \\ vom~obtine~o~epresie~a~vectorilor~\overrightarrow{IM}~si~\overrightarrow{IN}~in~functie~de~vectorii \\ \\ \overrightarrow{AB},~\overrightarrow{BC},~\overrightarrow{CA}. \\ \\ Apoi~scriem~\overrightarrow{BC}=\overrightarrow{BA}+\overrightarrow{AC}. \\ \\ (scopul~este~sa~exprimam~vectorii~\overrightarrow{IM}~si~\overrightarrow{IN}~in~functie~de~aceeasi \\ \\ doi~vectori)$

$\displaystyle Voi~scrie~expresiile~finale~direct,~caci~sunt~destule~calcule~pe \\ \\ parcurs~si~sunt~usoare. \\ \\ \overrightarrow{IM}= \frac{-am+b-cm}{(a+b+c)(m+1)}\overrightarrow{BA}+ \frac{c+cm}{(a+b+c)(m+1)}\overrightarrow{CA} \\ \\ \overrightarrow{IN}= \frac{b+bn}{(a+b+c)(n+1)}\overrightarrow{BA}+\frac{-an-bn+c}{(a+b+c)(n+1)}\overrightarrow{CA}.$

$\displaystyle Din~conditia~de~coliniaritate~rezulta~ca~cei~doi~vectori~sunt \\ \\ coliniari~ \bold{daca~si~numai~daca}~ \\ \\ \frac{-am+b-cm}{b+bn} \cdot \frac{n+1}{m+1}= \frac{c+cm}{-an-bn+c} \cdot \frac{n+1}{m+1}. \\ \\ Echivalent~cu~(am-b+cm)(an+bn-c)=bc(m+1)(n+1). \\ \\ Dupa~desfacerea~parantezelor,~reducerea~termenilor~si~factorizare \\ \\ avem~(a+b+c)(mna-bn-cm)=0,~de~unde~mna=bn+cm. \\ \\ Deci~\frac{a+b+c}{bc}= \frac{m+1}{mc}+ \frac{n+1}{nb}.$

$\displaystyle~Ultima~relatie~este~chiar~relatia~pe~care~trebuia~sa~o~demonstram. \\ \\ Din~\frac{AM}{MB}=m~rezulta~prin~proportii~derivate~AM= \frac{mc}{m+1}. \\ \\ Analog~AN= \frac{nb}{n+1}.$