中五數學題part b

2007-10-20 6:49 am

b) C1 is the inscribed circle of the equilateral triangle D1E1F1. C1 and C2 are the circumscribed circle and the inscribed circle of the equilateral triangle D2E2F2 respectively, C2 and C3 are the circumscribed circle and inscribed circle of the equilateral trigangle D3E3F3 respectively, and so on. Using the result in part a) , how many times of the perimeter of triangle D6E6F6 is that of triangle D1E1F1?

回答 (1)

2007-10-20 7:02 pm

✔ 最佳答案

As we do not know what the question in part a is, we can only proceed without it.

The problem can be readily solved with the help of a well drawn diagram.
Let O be the centre of the circles,
H1 the mid-point of D1-E1, so that
O-D1-H1 is a right triangle with angles equal to 60-30-90 degrees respectively.

In the circumscribed triangle D2-E2-F2,
O-D2 is the radius of the circumscribed circle C1,
O-H1 is the radius of the inscribed circle C1, so
O-D2 = O-H1

Solving the right triangle,
O-D2=O-H1=O-D1*sin(30)=O-D1/2
So the perimeter of triangle D2-E2-F2=perimeter of triangle D1-E1-F1 / 2

By repeating the process,
perimeter of D6-E6-F6
=D1-E1-F1/(2^(6-1))
=D1-E1-F1/32

Thus the perimeter of triangle D6-E6-F6 is 1/32 th of that of triangle D1-E1-F1.

參考: me

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