Question

14£®Èçͼ£¬Ä³»úе³§Òª½«³¤6m£¬¿í2mµÄ³¤·½ÐÎÌúƤABCD½øÐвüô£®ÒÑÖªµãFΪADµÄÖе㣬µãEÔÚ±ßBCÉÏ£¬²Ã¼ôʱÏȽ«ËıßÐÎCDFEÑØÖ±ÏßEF·­ÕÛµ½MNFE´¦£¨µãC£¬D·Ö±ðÂäÔÚÖ±ÏßBCÏ·½µãM£¬N´¦£¬FN½»±ßBCÓÚµãP£©£¬ÔÙÑØÖ±ÏßPE²Ã¼ô£®
£¨1£©µ±¡ÏEFP=$\frac{¦Ð}{4}$ʱ£¬ÊÔÅжÏËıßÐÎMNPEµÄÐÎ×´£¬²¢ÇóÆäÃæ»ý£»
£¨2£©Èôʹ²Ã¼ôµÃµ½µÄËıßÐÎMNPEÃæ»ý×î´ó£¬Çë¸ø³ö²Ã¼ô·½°¸£¬²¢ËµÃ÷ÀíÓÉ£®

Answer 1

·ÖÎö £¨1£©µ±¡ÏEFP=$\frac{¦Ð}{4}$ʱ£¬ÓÉÌõ¼þµÃ¡ÏEFP=¡ÏEFD=¡ÏFEP=$\frac{¦Ð}{4}$£®¿ÉµÃFN¡ÍBC£¬ËıßÐÎMNPEΪ¾ØÐΣ®¼´¿ÉµÃ³ö£®
£¨2£©½â·¨Ò»£ºÉè$¡ÏEFD=¦È\;\;£¨0£¼¦È£¼\frac{¦Ð}{2}£©$£¬ÓÉÌõ¼þ£¬Öª¡ÏEFP=¡ÏEFD=¡ÏFEP=¦È£®¿ÉµÃ$PF=\frac{2}{sin£¨¦Ð-2¦È£©}=\frac{2}{sin2¦È}$£¬$NP=NF-PF=3-\frac{2}{sin2¦È}$£¬$ME=3-\frac{2}{tan¦È}$£®ËıßÐÎMNPEÃæ»ýΪ$S=\frac{1}{2}£¨NP+ME£©MN$=$\frac{1}{2}[{£¨3-\frac{2}{sin2¦È}£©+£¨3-\frac{2}{tan¦È}£©}]¡Á2$=$6-\frac{2}{tan¦È}-\frac{2}{sin2¦È}$£¬»¯¼òÀûÓûù±¾²»µÈʽµÄÐÔÖʼ´¿ÉµÃ³ö£®
½â·¨¶þ£ºÉèBE=tm£¬3£¼t£¼6£¬ÔòME=6-t£®¿ÉµÃPE=PF£¬¼´$\sqrt{£¨3-BP{£©^2}+{2^2}}=t-BP$£®$BP=\frac{{13-{t^2}}}{2£¨3-t£©}$£¬NP=3-T+$\frac{13-{t}^{2}}{2£¨3-t£©}$£¬ËıßÐÎMNPEÃæ»ýΪ$S=\frac{1}{2}£¨NP+ME£©MN$=$\frac{1}{2}[{£¨3-t+\frac{{13-{t^2}}}{2£¨3-t£©}£©+£¨6-t£©}]¡Á2$=$6-[{\frac{3}{2}£¨t-3£©+\frac{2}{t-3}}]$£¬ÀûÓûù±¾²»µÈʽµÄÐÔÖʼ´¿ÉµÃ³ö£®

½â´ð ½â£º£¨1£©µ±¡ÏEFP=$\frac{¦Ð}{4}$ʱ£¬ÓÉÌõ¼þµÃ¡ÏEFP=¡ÏEFD=¡ÏFEP=$\frac{¦Ð}{4}$£®
ËùÒÔ¡ÏFPE=$\frac{¦Ð}{2}$£®ËùÒÔFN¡ÍBC£¬
ËıßÐÎMNPEΪ¾ØÐΣ®¡­3·Ö
ËùÒÔËıßÐÎMNPEµÄÃæ»ýS=PN•MN=2m²£®¡­5·Ö
£¨2£©½â·¨Ò»£º
Éè$¡ÏEFD=¦È\;\;£¨0£¼¦È£¼\frac{¦Ð}{2}£©$£¬ÓÉÌõ¼þ£¬Öª¡ÏEFP=¡ÏEFD=¡ÏFEP=¦È£®
ËùÒÔ$PF=\frac{2}{sin£¨¦Ð-2¦È£©}=\frac{2}{sin2¦È}$£¬$NP=NF-PF=3-\frac{2}{sin2¦È}$£¬$ME=3-\frac{2}{tan¦È}$£® ¡­8·Ö
ÓÉ$\left\{\begin{array}{l}3-\frac{2}{sin2¦È}£¾0\\ 3-\frac{2}{tan¦È}£¾0\\ 0£¼¦È£¼\frac{¦Ð}{2}\end{array}\right.$µÃ$\left\{\begin{array}{l}sin2¦È£¾\frac{2}{3}\\ tan¦È£¾\frac{2}{3}£¬\;\;\;\;\;\;\;\;\;£¨*£©\\ 0£¼¦È£¼\frac{¦Ð}{2}.\end{array}\right.$
ËùÒÔËıßÐÎMNPEÃæ»ýΪ$S=\frac{1}{2}£¨NP+ME£©MN$=$\frac{1}{2}[{£¨3-\frac{2}{sin2¦È}£©+£¨3-\frac{2}{tan¦È}£©}]¡Á2$=$6-\frac{2}{tan¦È}-\frac{2}{sin2¦È}$=$6-\frac{2}{tan¦È}-\frac{{2£¨{{sin}^2}¦È+{{cos}^2}¦È£©}}{2sin¦Ècos¦È}$=$6-£¨tan¦È+\frac{3}{tan¦È}£©$¡­12·Ö
$¡Ü6-2\sqrt{tan¦È\frac{3}{tan¦È}}=6-2\sqrt{3}$£®
µ±ÇÒ½öµ±$tan¦È=\frac{3}{tan¦È}$£¬¼´$tan¦È=\sqrt{3}\;£¬¦È=\frac{¦Ð}{3}$ʱȡ¡°=¡±£®¡­14·Ö
´Ëʱ£¬£¨*£©³ÉÁ¢£®
´ð£ºµ±$¡ÏEFD=\frac{¦Ð}{3}$ʱ£¬ÑØÖ±ÏßPE²Ã¼ô£¬ËıßÐÎMNPEÃæ»ý×î´ó£¬
×î´óֵΪ$6-2\sqrt{3}$m²£®  ¡­16·Ö
½â·¨¶þ£º
ÉèBE=tm£¬3£¼t£¼6£¬ÔòME=6-t£®
ÒòΪ¡ÏEFP=¡ÏEFD=¡ÏFEP£¬ËùÒÔPE=PF£¬¼´$\sqrt{£¨3-BP{£©^2}+{2^2}}=t-BP$£®
ËùÒÔ$BP=\frac{{13-{t^2}}}{2£¨3-t£©}$£¬$NP=3-PF=3-PE=3-£¨t-BP£©=3-t+\frac{{13-{t^2}}}{2£¨3-t£©}$£®  ¡­8·Ö
ÓÉ$\left\{\begin{array}{l}3£¼t£¼6\\ \frac{{13-{t^2}}}{2£¨3-t£©}£¾0\\ 3-t+\frac{{13-{t^2}}}{2£¨3-t£©}£¾0\end{array}\right.$µÃ$\left\{\begin{array}{l}3£¼t£¼6\\ t£¾\sqrt{13}£¬\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;£¨*£©\\{t^2}-12t+31£¼0.\end{array}\right.$
ËùÒÔËıßÐÎMNPEÃæ»ýΪ$S=\frac{1}{2}£¨NP+ME£©MN$=$\frac{1}{2}[{£¨3-t+\frac{{13-{t^2}}}{2£¨3-t£©}£©+£¨6-t£©}]¡Á2$=$\frac{{3{t^2}-30t+67}}{2£¨3-t£©}$¡­12·Ö
=$6-[{\frac{3}{2}£¨t-3£©+\frac{2}{t-3}}]$$¡Ü6-2\sqrt{3}$£®
µ±ÇÒ½öµ±$\frac{3}{2}£¨t-3£©=\frac{2}{t-3}$£¬¼´$t=3+\sqrt{\frac{4}{3}}\;=3+\frac{{2\sqrt{3}}}{3}$ʱȡ¡°=¡±£® ¡­14·Ö
´Ëʱ£¬£¨*£©³ÉÁ¢£®
´ð£ºµ±µãE¾àBµã$3+\frac{{2\sqrt{3}}}{3}$mʱ£¬ÑØÖ±ÏßPE²Ã¼ô£¬ËıßÐÎMNPEÃæ»ý×î´ó£¬
×î´óֵΪ$6-2\sqrt{3}$m²£®  ¡­16·Ö£®

µãÆÀ ±¾Ì⿼²éÁ˺¯ÊýµÄÐÔÖÊ¡¢¾ØÐεÄÃæ»ý¼ÆË㹫ʽ¡¢»ù±¾²»µÈʽµÄÐÔÖÊ¡¢Èý½Çº¯ÊýµÄµ¥µ÷ÐÔÓ¦ÓëÇóÖµ£¬¿¼²éÁËÍÆÀíÄÜÁ¦Óë¼ÆËãÄÜÁ¦£¬ÊôÓÚÖеµÌ⣮