APMO 2015, problem 4

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Masum
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APMO 2015, problem 4

Unread post by Masum » Thu Aug 20, 2015 1:02 pm

In a plane, there are $2n$ distinct lines where $n$ is a positive integer. Among them, $n$ lines are colored blue and $n$ lines are colored red and no two lines are parallel. Let $\mathcal{B}$($\mathcal{R}$) be the set of all points that lie on at least one blue(red) line. Prove that, there exists a circle that intersects $\mathcal{B}$ and $\mathcal{R}$ in exactly $2n-1$ points.
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Tasnood
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Re: APMO 2015, problem 4

Unread post by Tasnood » Fri Feb 09, 2018 10:50 pm

Basically there are two possible approaches that I considered. Since the circle intersects at most $2n$ points in $A$, we want either (1) the circle to intersect all the lines in $A$ twice and the circle contains exactly one of the points of intersection between lines in $A$ or, (2) the circle intersects all but one line in $A$ twice, and is tangent to exactly one line in $A$. These are the two possible ways we can have for the circle to intersect $A$ in exactly $2n-1$ points.

If we take the first approach it is a bit more difficult. The solution here would be something along the following lines: take the convex hull of the points of intersection between lines in $A$, and draw a circle around the convex hull which contains exactly one vertex on the convex hull. This circle intersects $A$ exactly $2n-1$ times. We can do the same thing with $B$ as long as the convex hulls for $A$ and $B$ are not such that one fits completley inside the other. However I don't see how to get around this problem. :roll:

So we try the second approach. We need the circle to be tangent to one line in $A$ and one line in $B$. So choose a red line and a blue line and inscribe a circle between them. We need to ensure that this circle intersects all the other lines twice and all the points of intersection are distinct. So first the circle has to be sufficiently large so that all intersection points between our $2n$ lines are "below" the circle. Secondly we need all the lines to actually intersect the circle, so choose the angle between the red and blue line to be maximal, and that's it. ;)

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