§Ú¶R¥ª¥» Maths mock paper¾Þ¡A ¦³ÉN¤HÃÑ¥§ÃD¡H¡H

This is one way to do it: https://imgur.com/a/EvgSx30

There might be other ways, I'll try to look for them.

Interesting tidbit: Length of CM (half of what we're required to calculate), (1+sqrt5)/2, is the golden ratio.

[

¦n¼ô±¤f¡A¼Ó¥D¥§¥»«Y«}®Ô¤åÓ¥»¡H§Ú³£ø«Y¦nÃÑ><

ì©«¥Ñ·¤§¨k©ó 2021-1-29 07:56 PM µoªí

This is one way to do it: https://imgur.com/a/EvgSx30

There might be other ways, I'll try to look for them.

Interesting tidbit: Length of CM (half of what we're required to calculate), (1+sqrt5) ...

ì©«¥Ñ·¤§¨k©ó 2021-1-29 07:56 PM µoªí

This is one way to do it: https://imgur.com/a/EvgSx30

There might be other ways, I'll try to look for them.

Interesting tidbit: Length of CM (half of what we're required to calculate), (1+sqrt5) ...

¦³¸ÑÃD¼v¤ù°Ú¡I¦³ÉN¿³½ì¤@»ô½Òª÷¶R¾¤Ú»¡H¡H

So there's another way, although it's kind of cheating.

You can redraw the figure (or even, use the figure given, because it can't not be in scale in this particular question), and you'd find out AB is roughly 3.5 times of MN. Hence B is the correct option.

Now this isn't the best practice when it comes to understanding math, but this technique is sufficient in reaching the correct answer for geometry MC questions in general. Also, math is a lot about intuition and inspiration, so this isn't all bad

EDIT: Oops, 1+sqrt5 is definitely not over 4 lmao.

[

ì©«¥Ñ·¤§¨k©ó 2021-2-3 04:22 AM µoªí

So there's another way, although it's kind of cheating.

You can redraw the figure (or even, use the figure given, because it can't not be in scale in this particular question), and you'd find out A ...

HI¡I§Ú喺«×¤À¨Éúï§ÚûH°µªk¡G

º¥ý¡A§Anª¾¹D¥H¤U¨âÓ«e´£¡]¦pªG§Aaim 5-5**¡A¤@©wn¤£°²«ä¯ÁËÝ¼g¥X嚟¡^

1¡B¹Ï1Åã¥ÜÃäªø¬°rûHµ¥Ãä¤T¨¤§Î¤TÂI®y¼Ð

2¡B¹Ï2Åã¥Ü¥~±µµ¥Ãä¤T¨¤§ÎûH¶êûH¶ê¤ß®y¼Ð

¤§«á¡A§Ú哋´N¤Þ¤Jª½¨¤®y¼Ð¡A¥HA¬°ìÂI¡AB¦ì©óx-¶b¤§¤W¡A³]O¬°¶ê¤ß¡AABªø¬°r

ËÝ¼Ë¡A§Ú哋§Q¥Î¥H¤W¨âÓ«e´£¥i¥H»´©ö±o¨ì¹Ï3¡AO®y¼Ð¬° (r/2,sqrt(3)r/6)¡AN®y¼Ð¬°¡]3r/4+1,sqrt(3)r/4) (MN¥¦æ©óAB¦n©öÃÒ¡A¦]¬°«YMC´NøÃÒ°Õ)

¦]¬°ON¡×OA¡A§Ú哋¥i¥H±o¨ì

(r/2)^2+(sqrt(3)r/6)^2=(r/4+1)^2+(sqrt(3)r/4-sqrt(3)r/6)^2

§Ú´Nø·|ª½±µ¸Ñ¨Ì±ø¤èµ{¡A·|±N¿ï¶µ³vÓ¥N¤J¥h¡AÚ»úïÃäÓ啱¡Aµ²ªG«Yr=1+sqrt(5)

Åwªï¥úÁ{ ¤p¨ò¸ê°T½×¾Â (http://lsforum.net/board/) | Powered by Discuz! 6.0.0 |