试解

如下图所示，注意到圆弧部分的面积可相互抵消(S1=S4, S2=S3)，因此此题比初看要好处理些。

  S_blue - S_yellow
= ( S_AFC + S_GDEH + S1 + S3) - (S_FCDG + S_HEBK + S2 + S4)
= S_AFC + S_GDEH - S_FCDG - S_HEBK + (S1-S4) + (S3 - S2)
= S_AFC + S_GDEH - S_FCDG - S_HEBK
= S_AFC + (S_AHE - S_AGD) - (S_AGD - S_AFC) - (S_ABK - S_AHE)
= S_AFC + S_AHE - S_AGD - S_AGD + S_AFC - S_ABK + S_AHE
= 2*S_AFC + 2*S_AHE - 2*S_AGD  - S_ABK
   ##### let t := S_ABK
= 2t*1/16 + 2t*9/16 - 2t*1/4 - t
= t * 1/16 * (2 + 18 - 8 - 16)
= t/16 * (-4)
= -1/4 * t
= -1/4 * 1/2 * AB*DK
= -1/4 * AD^2
= -1/4 * 2^2
= -1

所有跟帖： 

• 对！ -万斤油- ♂ (0 bytes) () 12/19/2023 postreply 21:24:00

• 何必那么复杂呢 -monseigneur- ♂ (92 bytes) () 12/20/2023 postreply 20:44:07

• 太棒了 -yma16- ♂ (0 bytes) () 12/25/2023 postreply 08:28:38