求(AP/PD)*(BP/PE)*(CP/PF)=?

rainy

2014-12-23 3:47 pm

D,E,F三點分別在三角形ABC的邊BC,AC,AB上,

若AD,BE,CF三線段交於三角形內部一點P
,
且滿足AP/PD+BP/PE+CP/PF=103
,
求(AP/PD)*(BP/PE)*(CP/PF)=?

回答 (1)

用戶2053

2014-12-24 4:51 am

✔ 最佳答案

設 AP/PD = a , BP/PE = b , CP/PF = c , 則 a + b + c = 103。
圖片參考：https://s.yimg.com/rk/HA04628698/o/280063908.jpg
考慮 PD/AD + PE/BE + PF/CF
= △PBC/△ABC + △PAC/△ABC + △PAB/△ABC
= (△PBC + △PAC + △PAB) / △ABC
= △ABC / △ABC
= 1 而 PD/AD + PE/BE + PF/CF
= PD/(AP+PD) + PE/(BP+PE) + PF/(CP+PF)
= 1/(a+1) + 1/(b+1) + 1/(c+1)故 1/(a+1) + 1/(b+1) + 1/(c+1) = 1
(b+1)(c+1) + (a+1)(c+1) + (a+1)(b+1) = (a+1)(b+1)(c+1)
3 + 2(a + b + c) + ab + bc + ca = abc + ab + bc + ca + a + b + c + 1
2 + a + b + c = abc
2 + 103 = abc
abc = 105
(AP/PD) (BP/PE) (CP/PF) = 105

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