## [問題]幾何題目

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### Re: [問題]幾何題目

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AI為角A平分線，AI分別交BC、DE於F及G，DE最短為與AG垂直；

ΔHCB~ΔHIA~ΔEGA (AAA)  =>  AG:GE = AI:IH = BC:CH = 5:1  =>  AG = 5GE
AG*GE/2 = ΔAGE面積=15/2  =>  5*GE2 = 15  =>  GE = √3  => DE = 2√3。
☆子 是也

G@ry

Another method:
We use coordinate Geometry.
Let A be the origin, C be (12,0), B be (5,0)
Also, D will be (12r,5r) E be (a,0) (Where r and a are both reals)
Obviously, ar=30
The shortest distance is √[(12r-a)^2+25r^2]
=169r^2-144+36/r^2
144 is a constant, so the value depends on 169r^2+36/r^2.
By the AM-GM, we can easily find out the minimum.
When r=6/13, 169r^2+36/r^2-144=12
So, DE=√12
追求神乎其技,至高無上的數學境界!~

We use coordinate Geometry.
Let A be the origin, C be (12,0), B be (12,5)
Also, D will be (12r,5r) E be (a,0) (Where r and a are both reals)
Obviously, 5ar=30, ar=6
The shortest distance is √[(12r-a)^2+25r^2]
=√(169r^2-144+36/r^2)
144 is a constant, so the value depends on 169r^2+36/r^2.
By the AM-GM, we can easily find out the minimum.
When r=6/13, 169r^2+36/r^2-144=12
So, DE=√12

some mistakes...
☆子 是也

G@ry

Thanks for correcting my solution...
Though those mistakes don't really matter. XD
追求神乎其技,至高無上的數學境界!~