如下 As follows:
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求出一條線段的最小值. (難題)
2013-05-28 8:59 pm
回答 (3)
2013-05-28 9:47 pm
✔ 最佳答案
設 C 為原點, B 座標為 (4, 0) , A 座標為 (0, 3)設 N 座標為 (4-n, 0), BN=n, BM=m (0<n<4)
tri BMN=(1/2) tri ABC ---> n/4*m/5=1/2
nm=10 --> m=10/n, M 座標為 (4-4m/5, 0+3m/5)=(4-8/n, 6/n)
MN=sqrt((n-8/n)^2+(6/n)^2)
n^2-16+64/n^2+36/n^2
=n^2+100/n^2-16
=(n-10/n)^2+4
minimum value of MN=sqrt(4)=2 (when n=sqrt 10)
參考: Paul
2013-05-28 9:45 pm
Sol
△ABC=3*4/2=6
SinB=3/5
△BMN=6/2=3
△BMN=(1/2)*BM*BN*SinB
3=(1/2)*BM*BN*(3/5)
BM*BN=10
CosB=4/5=(BM^2+BN^2－MN^2)/(2*BM*BN)
(BM^2+BN^2－MN^2)/20=4/5
BM^2+BN^2－MN^2=16
MN^2=BM^2+BN^2－16
>=2√(BM^2*BN^2)－16
=2*BM*BN－16
=2*10－16
=4
MN>=2
(A)
△ABC=3*4/2=6
SinB=3/5
△BMN=6/2=3
△BMN=(1/2)*BM*BN*SinB
3=(1/2)*BM*BN*(3/5)
BM*BN=10
CosB=4/5=(BM^2+BN^2－MN^2)/(2*BM*BN)
(BM^2+BN^2－MN^2)/20=4/5
BM^2+BN^2－MN^2=16
MN^2=BM^2+BN^2－16
>=2√(BM^2*BN^2)－16
=2*BM*BN－16
=2*10－16
=4
MN>=2
(A)
2013-05-28 9:44 pm
令BM=x , BN=y
面積一半：x*y / 5*4 = 1/2 ==> xy=10
餘弦定理：MN^2= x^2+y^2-2xy * cosB
= x^2+y^2-2*10*4/5= x^2+y^2-16
算幾不等式：
(x^2+y^2)/2 >= √(x^2 y^2) = xy = 10
x^2+y^2>=20
MN^2=x^2+y^2-16>=24
MN>=2
[>= ：大於等於]
面積一半：x*y / 5*4 = 1/2 ==> xy=10
餘弦定理：MN^2= x^2+y^2-2xy * cosB
= x^2+y^2-2*10*4/5= x^2+y^2-16
算幾不等式：
(x^2+y^2)/2 >= √(x^2 y^2) = xy = 10
x^2+y^2>=20
MN^2=x^2+y^2-16>=24
MN>=2
[>= ：大於等於]
收錄日期: 2021-04-11 19:43:47
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20130528000015KK01642