Pe laturile patrulaterului convex \( ABCD \) se considera punctele \( M\in (AB) \) , \( N\in (BC) \) , \( P\in (CD) \) , \( Q\in (DA) \) astfel
incat \( \frac{MA}{MB}=\frac{NB}{NC}=\frac{PC}{PD}=\frac{QD}{QA}=k. \) Sa se determine \( k \) astfel incat aria patrulaterului \( MNPQ \) sa fie minima.
Patrulater cu arie minima
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Mateescu Constantin
- Newton
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Patrulater cu arie minima
Last edited by Mateescu Constantin on Tue Nov 24, 2009 2:03 pm, edited 1 time in total.
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Mateescu Constantin
- Newton
- Posts: 307
- Joined: Tue Apr 21, 2009 8:17 am
- Location: Pitesti
Conform conditei din enunt avem:
\( \left\|\ \begin{array}{cc}
\frac{MA}{AB} & = & \frac{NB}{BC} & = & \frac{PC}{CD} & = & \frac{QD}{DA} & = & \frac{k}{k+1} \\\\\\\\
\frac{MB}{AB} & = & \frac{NC}{BC} & = & \frac{PD}{CD} & = & \frac{QA}{DA} & = & \frac{1}{k+1}\ \end{array}\right|\ \Longrightarrow\ \frac{A_{AMQ}}{A_{ABD}}=\frac{AM\cdot AQ}{AB\cdot AD}=\frac{k}{{(k+1)}^{2}}\ \Longleftrightarrow\ A_{AMQ}=\frac{k}{{(k+1)}^{2}}\cdot A_{ABD} \)
Scriind si relatiile analoage rezulta:
\( \left\|\ \begin{array}{cccc}
A_{AMQ} & = & \frac{k}{{(k+1)}^{2}}\cdot A_{ABD} \\\\\\\
A_{BMN} & = & \frac{k}{{(k+1)}^{2}}\cdot A_{ABC} \\\\\\\\
A_{CNP} & = & \frac{k}{{(k+1)}^{2}}\cdot A_{BCD} \\\\\\\\
A_{DPQ} & = & \frac{k}{{(k+1)}^{2}}\cdot A_{ADC}\ \end{array}\right|\ \bigoplus\ \Longrightarrow\ A_{AMQ}+A_{BMN}+A_{CNP}+A_{DPQ}=\frac{2k}{{(k+1)}^{2}}\cdot A_{ABCD} \)
\( \Longrightarrow\ A_{MNPQ}=A_{ABCD}-\frac{2k}{{(k+1)}^{2}}\cdot A_{ABCD}=\frac{k^{2}+1}{{(k+1)}^{2}}\cdot A_{ABCD}. \)
Tinand seama de faptul ca \( {(1+k)}^{2}\le 2(k^{2}+1),\ \forall k\in \mathbb{R} \) deducem ca \( A_{MNPQ} \) admite un minim egal cu \( \frac{1}{2}\cdot A_{ABCD} \)
care se realizeaza pentru \( k=1. \)
\( \left\|\ \begin{array}{cc}
\frac{MA}{AB} & = & \frac{NB}{BC} & = & \frac{PC}{CD} & = & \frac{QD}{DA} & = & \frac{k}{k+1} \\\\\\\\
\frac{MB}{AB} & = & \frac{NC}{BC} & = & \frac{PD}{CD} & = & \frac{QA}{DA} & = & \frac{1}{k+1}\ \end{array}\right|\ \Longrightarrow\ \frac{A_{AMQ}}{A_{ABD}}=\frac{AM\cdot AQ}{AB\cdot AD}=\frac{k}{{(k+1)}^{2}}\ \Longleftrightarrow\ A_{AMQ}=\frac{k}{{(k+1)}^{2}}\cdot A_{ABD} \)
Scriind si relatiile analoage rezulta:
\( \left\|\ \begin{array}{cccc}
A_{AMQ} & = & \frac{k}{{(k+1)}^{2}}\cdot A_{ABD} \\\\\\\
A_{BMN} & = & \frac{k}{{(k+1)}^{2}}\cdot A_{ABC} \\\\\\\\
A_{CNP} & = & \frac{k}{{(k+1)}^{2}}\cdot A_{BCD} \\\\\\\\
A_{DPQ} & = & \frac{k}{{(k+1)}^{2}}\cdot A_{ADC}\ \end{array}\right|\ \bigoplus\ \Longrightarrow\ A_{AMQ}+A_{BMN}+A_{CNP}+A_{DPQ}=\frac{2k}{{(k+1)}^{2}}\cdot A_{ABCD} \)
\( \Longrightarrow\ A_{MNPQ}=A_{ABCD}-\frac{2k}{{(k+1)}^{2}}\cdot A_{ABCD}=\frac{k^{2}+1}{{(k+1)}^{2}}\cdot A_{ABCD}. \)
Tinand seama de faptul ca \( {(1+k)}^{2}\le 2(k^{2}+1),\ \forall k\in \mathbb{R} \) deducem ca \( A_{MNPQ} \) admite un minim egal cu \( \frac{1}{2}\cdot A_{ABCD} \)
care se realizeaza pentru \( k=1. \)