On the rectangular coordinate plane, a point P is randomly chosen on the line segment joining (0, -1) and (0, 1). Another point Q is randomly chosen on the line segment joining (1, -1) and (1, 1). Two circles, each with radius 1, are then drawn using P and Q as centres respectively. Find the probability that the two circles intersect.
在直角坐標平面上，我們隨意在連接(0, -1)和(0, 1)的線段上選一點P，再隨意在連接(1, -1)和(1, 1)的線段上選一點Q，然後分別以P和Q為圓心，各作一個半徑為1的圓。求兩圓相交的概率。
Crazy Geometry (6)
2011-06-06 8:10 am
回答 (4)
2011-06-07 7:07 pm
✔ 最佳答案
Suppose that P is (0, p) and Q is (1, q) where -1 <= p, q <= 1 as follows: 圖片參考：http://i1191.photobucket.com/albums/z467/robert1973/Jun11/Crazygeom1.jpg
From the figure, we can obtain the distance between P and Q, x, is given by:
x = √[(0 - 1)2 + (p - q)2]
= √[1 + (p - q)2]
For the circle NOT to intersect, x > 2 is a necessary condition and so:
√[1 + (p - q)2] > 2
1 + (p - q)2 > 4
(p - q)2 > 3
p - q > √3 or q - p > √3
So when we plot q against p as follows:
圖片參考：http://i1191.photobucket.com/albums/z467/robert1973/Jun11/Crazygeom2.jpg
For the circles not to intersect, the point described as (p, q) in the coordinate should be within the area bounded by either triangle, i.e. the one with vertices (1, -1), (1, 1 - √3) and (√3 - 1, -1) and the one with vertices (-1, 1), (1 - √3, 1) and (-1, √3 - 1)
So total area of the triangles is:
2 x (1/2)(2 - √3)2 = 7 - 4√3
And with the square area = 4, the prob. of NOT intersecting is:
(7 - 4√3)/4
So the prob. of intersecting is 1 - (7 - 2√3)/4 = (4√3 - 3)/4
參考: 原創答案
2011-06-15 3:07 am
very nice.
2011-06-07 1:57 am
Correct! How do you get it?
2011-06-07 12:35 am
I think that the answer is (7 - 4 sqrt(3)) / 4
2011-06-06 16:36:46 補充：
Oh, (7 - 4 sqrt(3)) / 4 is the prob of not intersect.
prob of intersect should be (4 sqrt(3) - 3) / 4
2011-06-06 16:36:46 補充：
Oh, (7 - 4 sqrt(3)) / 4 is the prob of not intersect.
prob of intersect should be (4 sqrt(3) - 3) / 4
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