道路設計 (光線版) 002 - 拼圖
By Charlotte
at 2008-01-03T16:10
at 2008-01-03T16:10
Table of Contents
目前自己想到的(a)(b)...有更短的請提出
(a)
A_______B CF交AE,BD在XY, (牆FX=50km, 牆CY=50km)
∕ ﹨ 作B到AD垂線交AD在Z (牆BZ=50√3km)
∕ ﹨ 接著在ADE裡找一點O 使AO+DO+EO最短
F∕ X Y ﹨C 設E(0,0) D(100,0) A(0,100√3)
﹨ ∕ 另外O在BE連線上(參照前一題的正方形)
﹨ ∕ 所以O的座標為(a,a√3)
﹨_______∕
E D
L(AO)^2= (100√3-a√3)^2 + a^2
L(EO)^2= (a√3)^2 + a^2 = 4 a^2
L(DO)^2= (a√3)^2 + (100-a)^2
-----------------------------------------
Total = 40000 - 800a + 12a^2
a=800/24=100/3時 Total最少 => AO = 120.185 km ; EO = 66.66km ; DO = 74.535km
AO+EO+DO+FX+CY+BZ= 447.98km
(b)
A_______B CF交AE,BD在XY, (牆FX=50km, 牆CY=50km)
∕ ﹨ O為正六邊形的中心點
∕ ﹨ P為正三角形ABO的重心
F∕ X Y ﹨C Q為正三角形DEO的重心
﹨ ∕
﹨ ∕
﹨_______∕
E D
AP+BP+PQ+EQ+DQ+FX+CY = (100/√3)*6 + 100 = 446.4km
結論: (b)比較短...(a)在算好玩的嗎...
--
(a)
A_______B CF交AE,BD在XY, (牆FX=50km, 牆CY=50km)
∕ ﹨ 作B到AD垂線交AD在Z (牆BZ=50√3km)
∕ ﹨ 接著在ADE裡找一點O 使AO+DO+EO最短
F∕ X Y ﹨C 設E(0,0) D(100,0) A(0,100√3)
﹨ ∕ 另外O在BE連線上(參照前一題的正方形)
﹨ ∕ 所以O的座標為(a,a√3)
﹨_______∕
E D
L(AO)^2= (100√3-a√3)^2 + a^2
L(EO)^2= (a√3)^2 + a^2 = 4 a^2
L(DO)^2= (a√3)^2 + (100-a)^2
-----------------------------------------
Total = 40000 - 800a + 12a^2
a=800/24=100/3時 Total最少 => AO = 120.185 km ; EO = 66.66km ; DO = 74.535km
AO+EO+DO+FX+CY+BZ= 447.98km
(b)
A_______B CF交AE,BD在XY, (牆FX=50km, 牆CY=50km)
∕ ﹨ O為正六邊形的中心點
∕ ﹨ P為正三角形ABO的重心
F∕ X Y ﹨C Q為正三角形DEO的重心
﹨ ∕
﹨ ∕
﹨_______∕
E D
AP+BP+PQ+EQ+DQ+FX+CY = (100/√3)*6 + 100 = 446.4km
結論: (b)比較短...(a)在算好玩的嗎...
--
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