42. IMAC 2010 Problema 2.

by Virgil Nicula, Jun 5, 2010, 11:17 PM

Quote:
Fie $M$ si $N$ mijloacele diagonalelor $[AC]$ si $[BD]$ ale patrulaterului inscriptibil $ABCD$ . Sa se arate ca $\frac{|AC-BD|}{2}\le MN\le \frac{AC+BD}{2}$.

Metoda de mai jos este metrica, NU utilizeaza vectorii si NU este la nivelul clasei a VII - a .

Metoda 2. Notam $AB=a$ , $BC=b$ , $CD=c$ , $DA=d$ , $AC=e$ , $BD=f$ si mijloacele $X$ , $Y$ ale laturilor $[AB]$ , $[BC]$

respectiv. Vom folosi relatiile $\boxed {ef=ac+bd}$ (Ptolemeu) si $\boxed {e^2+f^2+4\cdot MN^2=a^2+b^2+c^2+d^2}$ (Euler).

In $\triangle MNX$ avem $XM=\frac b2$ si $N=\frac d2$ . Deci $\boxed {\frac {|b-d|}{2}\le MN\le \frac {b+d}{2}}$ .

In $\triangle MNY$ avem $YM=\frac a2$ si $YN=\frac c2$ . Deci $\boxed {\frac {|a-c|}{2}\le\ MN\le \frac {a+c}{2}}$ .

Relatiile Euler & Ptolemeu $\Longrightarrow$ $(e-f)^2+4\cdot MN^2=$ $(a-c)^2+(b-d)^2 \le 8\cdot MN^2$ $\Longrightarrow$ $\boxed {\frac {|e-f|}{2}\le MN}$ .

Relatiile Euler & Ptolemeu $\Longrightarrow$ $(e+f)^2+4\cdot MN^2=$ $(a+c)^2+(b+d)^2\ge 8\cdot MN^2$ $\Longrightarrow$ $\boxed {\frac {|e+f|}{2}\ \ge\ MN}$ .
This post has been edited 6 times. Last edited by Virgil Nicula, Nov 27, 2015, 7:48 AM

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