The greatest number of points of intersection of 8 straight lines and 4 circles are? My attempt:Assuming every line cuts all the four circles at two points each, the points of intersection of lines with circles=64. Now, each circle cuts the other 3 circles at 6 points.Then points of intersection of circles is 6*4=24 making total number of points of intersection 88. The text book answer is 104. How is that?
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While each circle could cut another circle at 6 points, the number of circle-circle intersections is not $6 \times 4 = 24$. That would over count by a factor of 2. Line-line intersections have also been neglected here. – zahbaz Apr 11 '15 at 05:44
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The number of intersections of lines with lines is $7+6+5+4+3+2+1={8 \choose 2}$ and the number of intersection of the circles with circles is $6+4+2=2 {4\choose 2}$. This is because every line intersects with every other line and every circle intersects with every other circle. Your calculations on the number of intersections of lines with circles is correct at 64.
Archaick
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if we have $k$ circles and $n$ lines, the answer is $2{k\choose2}+{n\choose2}+2nk$, since every two circles intersect at most at two points, every two lines at most one point and every line and circle at most two.
$2{8\choose2}+{4\choose2}+2(8)(4) = 104 $
user373141
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