Question

20£®ÔÚƽÃæÖ±½Ç×ø±êϵÖУ¬ÉèA£¨x₁£¬y₁£©£¬B£¨x₂£¬y₂£©£®¶¨Ò壺${d_¦Á}£¨A£¬B£©={£¨{|{{x_1}-{x_2}}|^¦Á}+{|{{y_1}-{y_2}}|^¦Á}£©^{\frac{1}{¦Á}}}$£¬ÆäÖЦÁ¡ÊR⁺£¨R⁺±íʾÕýʵÊý£©£®
£¨¢ñ£©ÉèA£¨1£¬1£©£¬B£¨2£¬3£©£¬Çód₁£¨A£¬B£©ºÍd₂£¨A£¬B£©µÄÖµ£»
£¨¢ò£© ÇóÖ¤£º¶ÔƽÃæÖÐÈÎÒâÁ½µãAºÍB¶¼ÓÐ${d_2}£¨A£¬B£©¡Ü{d_1}£¨A£¬B£©¡Ü\sqrt{2}{d_2}£¨A£¬B£©$£»
£¨¢ó£©ÉèM£¨x£¬y£©£¬OΪԭµã£¬¼Ç${D_¦Á}=\{M£¨x£¬y£©|{d_¦Á}£¨M£¬O£©¡Ü1£¬¦Á¡Ê{R^+}\}$£®Èô0£¼¦Á£¼¦Â£¬ÊÔд³öD_¦ÁÓëD_¦ÂµÄ¹Øϵ£¨Ö»Ðèд³ö½áÂÛ£¬²»±ØÖ¤Ã÷£©£®

Answer 1

·ÖÎö £¨¢ñ£©d_¦Á£¨A£¬B£©µÄ¶¨Òå´úÈë¼´¿ÉµÃ³ö£®
£¨¢ò£©ÉèA£¨x₁£¬y₁£©£¬B£¨x₂£¬y₂£©£¬Ôòd₁£¨A£¬B£©=|x₁-x₂|+|y₁-y₂|£¬${d_2}£¨A£¬B£©={£¨{|{{x_1}-{x_2}}|^2}+{|{{y_1}-{y_2}}|^2}£©^{\frac{1}{2}}}$£®Í¨¹ý¼ÆËãÕ¹¿ª¼´¿ÉÖ¤Ã÷£®
£¨¢ó£©D_¦Á?D_¦ÂÕæ×Ó¼¯£®ÈÎÈ¡£¨x₀£¬y₀£©¡ÊD_¦Á£¬${d_¦Á}£¨M£¬O£©={£¨{|{x_0}|^¦Á}+{|{y_0}|^¦Á}£©^{\frac{1}{¦Á}}}¡Ü1$£®¶Ôx₀£¬y₀·ÖÀàÌÖÂÛ£¬¼´¿ÉÖ¤Ã÷£®

½â´ð £¨¢ñ£©½â£ºd₁£¨A£¬B£©=|1-2|+|1-3|=3£¬
d₂£¨A£¬B£©=$[|1-2{|}^{2}+|1-3{|}^{2}]^{\frac{1}{2}}$=${5}^{\frac{1}{2}}$=$\sqrt{5}$£®
£¨¢ò£©Ö¤Ã÷£ºÉèA£¨x₁£¬y₁£©£¬B£¨x₂£¬y₂£©£¬Ôòd₁£¨A£¬B£©=|x₁-x₂|+|y₁-y₂|£¬${d_2}£¨A£¬B£©={£¨{|{{x_1}-{x_2}}|^2}+{|{{y_1}-{y_2}}|^2}£©^{\frac{1}{2}}}$£®
${d_1}^2£¨A£¬B£©={£¨|{{x_1}-{x_2}}|+|{{y_1}-{y_2}}|£©^2}$=${x_1}^2+{x_2}^2+{y_1}^2+{y_2}^2-2{x_1}{x_2}-2{y_1}{y_2}+2|{{x_1}-{x_2}}||{{y_1}-{y_2}}|$£®
${d_2}^2£¨A£¬B£©=£¨{|{{x_1}-{x_2}}|^2}+{|{{y_1}-{y_2}}|^2}£©$=${x_1}^2+{x_2}^2+{y_1}^2+{y_2}^2-2{x_1}{x_2}-2{y_1}{y_2}$£®
ËùÒÔd₂£¨A£¬B£©¡Üd₁£¨A£¬B£©³ÉÁ¢£®
ÒòΪ${£¨\sqrt{2}{d_2}£¨A£¬B£©£©^2}=2{x_1}^2+2{x_2}^2+2{y_1}^2+2{y_2}^2-4{x_1}{x_2}-4{y_1}{y_2}$£¬
ËùÒÔ${£¨\sqrt{2}•{d_2}£¨A£¬B£©£©^2}-{d_1}{£¨A£¬B£©^2}$=${x_1}^2+{x_2}^2+{y_1}^2+{y_2}^2-2{x_1}{x_2}-2{y_1}{y_2}-2|{{x_1}-{x_2}}||{{y_1}-{y_2}}|$=${£¨{x_1}-{x_2}£©^2}+{£¨{y_1}-{y_2}£©^2}-2|{{x_1}-{x_2}}||{{y_1}-{y_2}}|$=${£¨|{x_1}-{x_2}|-|{y_1}-{y_2}|£©^2}¡Ý0$£®
ËùÒÔ${d_1}£¨A£¬B£©¡Ü\sqrt{2}{d_2}£¨A£¬B£©$³ÉÁ¢£®
£¨¢ó£©D_¦Á?D_¦ÂÕæ×Ó¼¯£®
Ö¤Ã÷ÈçÏ£º
ÈÎÈ¡£¨x₀£¬y₀£©¡ÊD_¦Á£¬${d_¦Á}£¨M£¬O£©={£¨{|{x_0}|^¦Á}+{|{y_0}|^¦Á}£©^{\frac{1}{¦Á}}}¡Ü1$£®
µ±x₀=1£¬y₀=0ʱ£¬d_¦Á£¨M£¬O£©=0£¬d_¦Â£¨M£¬O£©=0£¬´ËʱD_¦Á⊆D_¦Â£®
µ±|x₀|=1£¬|y₀|=0ʱ£¬${d_¦Á}£¨M£¬O£©={£¨{|{x_0}|^¦Á}+{|{y_0}|^¦Á}£©^{\frac{1}{¦Á}}}=1$£¬d_¦Â£¨M£¬O£©=1£®
´ËʱD_¦Á⊆D_¦Â£®
ͬÀí¿ÉµÃ£¬µ±|x₀|=0£¬|y₀|=1ʱ£¬D_¦Á⊆D_¦Â£®
µ±|x₀|¡Ù1£¬|y₀|¡Ù1ʱ£¬ÒòΪ${d_¦Á}£¨M£¬O£©={£¨{|{x_0}|^¦Á}+{|{y_0}|^¦Á}£©^{\frac{1}{¦Á}}}¡Ü1$£¬ËùÒÔ ${|{x_0}|^¦Á}+{|{y_0}|^¦Á}¡Ü1$£®
ÓÖÒòΪ0£¼¦Á£¼¦Â£¬ËùÒÔ${|{x_0}|^¦Â}+{|{y_0}|^¦Â}£¼{|{x_0}|^¦Á}+{|{y_0}|^¦Á}¡Ü1$£®´ËʱD_¦Á⊆D_¦Â£®
·´Ö®²»³ÉÁ¢£®
ËùÒÔD_¦Á?D_¦Â£®

µãÆÀ ±¾Ì⿼²éÁËж¨Òå¡¢¼¯ºÏÖ®¼äµÄ¹Øϵ¡¢Á½µãÖ®¼äµÄ¾àÀ빫ʽ¡¢·ÖÀàÌÖÂÛ·½·¨¡¢²»µÈʽµÄ½â·¨£¬¿¼²éÁËÍÆÀíÄÜÁ¦Óë¼ÆËãÄÜÁ¦£¬ÊôÓÚÄÑÌ⣮