一到英文的几何题A paper rectangle ABCD of area 1 is folded along astraight line so that C and A coincide.Prove that the area of theresulting pentagon is less than 3/4.
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![一到英文的几何题A paper rectangle ABCD of area 1 is folded along astraight line so that C and A coincide.Prove that the area of theresulting pentagon is less than 3/4.](/uploads/image/z/8738063-71-3.jpg?t=%E4%B8%80%E5%88%B0%E8%8B%B1%E6%96%87%E7%9A%84%E5%87%A0%E4%BD%95%E9%A2%98A+paper+rectangle+ABCD+of+area+1+is+folded+along+astraight+line+so+that+C+and+A+coincide.Prove+that+the+area+of+theresulting+pentagon+is+less+than+3%2F4.)
一到英文的几何题A paper rectangle ABCD of area 1 is folded along astraight line so that C and A coincide.Prove that the area of theresulting pentagon is less than 3/4.
一到英文的几何题
A paper rectangle ABCD of area 1 is folded along a
straight line so that C and A coincide.Prove that the area of the
resulting pentagon is less than 3/4.
一到英文的几何题A paper rectangle ABCD of area 1 is folded along astraight line so that C and A coincide.Prove that the area of theresulting pentagon is less than 3/4.
When AB=BC, the rectangle is a square and the pentagon is the triangle BCD whose area is 1/2, as desired. So we can suppose that AB>BC without loss of generality. Let EF be the folding line, where E belongs to AB, and F belongs to CD. Let G be the image of D. Then the desired pentagon is BCGFE, whose area is the sum of areas of the quadrilateral BCFE and the triangle CFG. In notation, let us denote
S_{BCGFE}=S_{BCFE}+S_{CFG}. (1)
Since the quadrilaterals AEFD and BCFE are symmetric along the line EF, we have
S_{BCFE}=S_{ABCD}/2=1/2. (2)
The symmetry along EF also implies that
S_{CFG}=S_{ADF}. (3)
Putting (1), (2) and (3) together, we find it suffices to show that
S_{ADF}DF (which can be easily proved, if necessary, by introducting the natural xy-coordinate system originated at A and show that the slope of EF is negative). Since AB//CD, we have
S_{ADF}