Take the maximal angle made by a red and blue line and inscribe a sufficiently large a circle between these two lines.

if we draw the circle which tangent to the greatest angle between tow lines red and blue. and the further intersection point between tow lines must lie inside of it ...so we obtain the circle ..

Here are two solutions. It is obvious that the desired circle will intersect exactly $n-1$ lines of each color and touch exactly one of each color. Two lines have to be equidistant from the center of that circle, therefore the center $C$ lies on the angle bisector of a red line and a blue line. We need to take $C$ so that it doesn't lie on any other angle bisector of the pairs. Now, the problem is to make sure that it intersects the rest of $n-1$ pairs of lines. To do that, we can just take the maximum angle between all possible pairs of lines. Then the circle touching that maximum angle related lines will definitely intersect the $n-1$ red and $n-1$ blue lines. Alternatively, we could consider the convex hull of all the intersection points of red and blue lines. Then just take any angle of the hull and it will work. Prove or disprove: This problem can be generalized for any $k$ where $k\leq2n$. Replace $2n-1$ by $k$.

I suggest this solution:we can easily prove that it exists a circle which touch 2 lines AB,AC and cut the line BC(because the center of that circle moves on the bisector of the angle between AB,AC).Firstly,we choose any 2 different coloured lines AB,AC and draw a circle touch them.If it touches any lines of the remain 2n-2 lines,enlarge it until it doesn't touch any lines except AB,AC.Each line except AB,AC cuts the circle at 2 points.So that we have the solution for the problem.

Take an intersection A of red lines and an intersection B of blue lines. Then a circle thru A,B with a large enough radius works. Or?

We need to construct a circle where n-1 red lines have a point inside the circle, and one is tangent, and the same goes for blue lines. This is obviously true for 1 red and 1 blue line. We simply prove we can add a line of any color and make a new circle. If the new line is in the old circle were done. If its outside the old circle, dilate the old circle about the tangent point of the opposite color. If it is tangent, dilate about the intersection of the tangents by an arbitrarily small amount. das it we proved it lmoa

randomusername wrote: Take an intersection A of red lines and an intersection B of blue lines. Then a circle thru A,B with a large enough radius works. Or? Exacty one line must be tangent to the circle to get 2n-1 points ^^'

Since there are finitely many intersections of lines, all of these intersections can be contained within some circle $\mathcal{C}$. Going around the boundary of $\mathcal{C}$, we read off the colors of the lines. There must exist two ``adjacent'' lines with opposite colors. Call these two lines $m$ and $n$. Now draw a circle (lying outside $\mathcal{C}$) tangent to $m$ and $n$ such that it intersects all the lines; this is our desired circle. $\blacksquare$

There exists a red line such that on rotating it clockwise about a point, the first other line that it becomes parallel to is blue colored. Call these lines the good red and blue line respectively. Take a Cartesian coordinate system with the good red line as the x-axis. For any line (not necessarily one of the red and blue lines), define its angle to be the angle $a$ (in degrees) such that $0 \le a < 180$, and the slope of the line is $tan(a)$. Call a line right if its angle is positive and less than $90$, and call it left its angle is greater than $90$. Fact 1: For any right line, there exists $h > 0$ such that any point $(0, y)$ with $y > h$ lies to the left of that line. Fact 2: For any left line, there exists $h > 0$ such that any point $(0, y)$ with $y > h$ lies to the right of that line. Call this $h$ corresponding to a right/left line the height of that line. Fact 3: For any two right lines $L_1$ and $L_2$, where angle of $L_1$ is greater than that of $L_2$, there exists $d > 0$ such that for any $c > d$, the point of intersection of the line $y = c$ with $L_1$ lies to the left of the point of intersection of $y = c$ with $L_2$. Call this $d$ the difference of $L_1$ and $L_2$. Consider the angle bisectors of all the red/blue lines (other than the x-axis) with x-axis. Note that for any red/blue line (other than the x-axis), one its angle bisectors with the x-axis is left, and the other is right. Also, the good blue line's right angle bisector with the x-axis is the one with the largest angle among all the right angle bisectors. Let $H$ be the maximum of the heights of all these angle bisectors, and let $D$ be the maximum of the differences of all pairs of right angle bisectors. Choose a positive real $R$ with $R > max(H, D)$. Consider the circle with center $(0, R)$ and radius $R$. It is tangent to the good red line, and intersects all other red/blue lines in two points. Move the circle to the right by keeping its radius same, and moving its center to the right on the line $y = R$. The first angle angle bisector the center will come to lie on is the right angle bisector of the good blue line. At this moment, the circle satisfies the required conditions, and we're done.

Easy solution: Among all pairs of red-blue lines, choose with the most obtuse angle(with the biggest measure).Assume they intersect at $A$ and $B,R$ are blue,red points such that $\angle RAB$ is the biggest.If there is line $\alpha$ intersecting with rays $AX,AY$ let say at $M,N$ and wlog with color blue, then $AR\cap \alpha$ makes bigger angle which is contradiction. Now, draw a circle $\omega$ tangent to rays $AB,AR$. By increasing its size, we can afford that all other lines will intersect $\omega$ at two distinct points.And we will get our claim>>

flyingpurplepeopleeater wrote: randomusername wrote: Take an intersection A of red lines and an intersection B of blue lines. Then a circle thru A,B with a large enough radius works. Or? Exacty one line must be tangent to the circle to get 2n-1 points ^^' I feel like this works, circle passing through the intersection also results in $2n-1$ points

We say $\mathcal{B}$ is trivial if all the blue lines passes through one point, simiarly define triviality on $\mathcal{R}$. We divide into three cases: Case I: Both $\mathcal{B}$ and $\mathcal{R}$ are nontrivial. By the dual of Sylvester-Gallai Theorem there exists a point $P_1$ in $\mathcal{B}$ which lies on exactly two blue lines, and a point $P_2$ in $\mathcal{R}$ which lies on exactly two red lines. We show that there exists a circle through $P_1$ and $P_2$ which intersect all other lines in two points, which clearly satisfies the requirement of the problem, by showing the following lemma: Lemma 1. Suppose $A,B$ are two points such that the $y$-axis is its perpendicular bisector. For each real number $x$ define the $x$-circle to the the circle centered at $(0,x)$ and passes through $A,B$. Suppose $l$ is a line. Then there exists a constant $k$ such that if $|x|> k$ then the $x$-circle intersects $l$ at two points. Proof. If that line $l$ passes through the interior of the segment $AB$ then every circle through $A,B$ will intersect $l$ at two points. Otherwise, by Appollonius problem there exists two circles, one with center having positive y-coordinate and another with center having negative y-coordinate, that are tangent to $l$ and passes through $A,B$. Suppose they centered at $(x_1,0)$ and $(x_2,0)$. then its suffices to take $$k=\max{|x_1|,|x_2|}$$[asy][asy] size(6cm); real labelscalefactor = 0.3; /* changes label-to-point distance */pen dps = linewidth(0.7) + fontsize(0); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -22.499793631259898, xmax = 22.3439885834916, ymin = -12.581374867507975, ymax = 25.511730454700285; /* image dimensions */ /* draw figures */draw(circle((0,4.380618324183532), 6.116770182167897), linewidth(0.8)); draw((xmin, 1.1871032214460897*xmin + 13.874850719287913)--(xmax, 1.1871032214460897*xmax + 13.874850719287913), linewidth(0.8)); /* line */draw(circle((0,8.514490421689638), 9.52478911579308), linewidth(0.8)); /* dots and labels */dot((-4.269081934009437,0),dotstyle); label("$A$", (-4.096162578969407,0.3976051104174153), NE * labelscalefactor); dot((4.269081934009437,0),dotstyle); label("$B$", (4.422548737811344,0.3976051104174153), NE * labelscalefactor); dot((0,4.380618324183532),dotstyle); label("$C$", (0.16319307942096858,4.7775085704603475), NE * labelscalefactor); dot((-4.678137180148465,8.321419002366945),dotstyle); label("$D$", (-6.1052926065120365,8.916316427198165), NE * labelscalefactor); dot((0,8.514490421689638),dotstyle); label("$E$", (0.16319307942096858,8.916316427198165), NE * labelscalefactor); dot((-8.360128128122685,3.9505156866914057),linewidth(4pt) + dotstyle); label("$F$", (-9.802091857190476,4.014039159994149), NE * labelscalefactor); dot((3.0776577263834852,17.528348120786198),linewidth(4pt) + dotstyle); label("$G$", (2.3732361097178614,18.43959275775023), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy] Therefore, by taking $|x|$ sufficiently large we can find such a circle Case II: $\mathcal{B}$ is trivial while $\mathcal{R}$ is nontrivial. This is very similar to Case I. Lemma 2. Suppose $A$ is a point and for every real number $x$ define the $x$-circle to be the circle through $A$ and tangent to the $y$-axis at $(0,x)$. Suppose $l$ is a line. Then there exists a constant $k$ such that if $|x|>k$ then the $x$-circle intersects $l$ at two points. Proof. Similar to Case I. Now by dual of Sylvester-Gallai Theorem we can find a point $A$ in $\mathcal{R}$ lying in exactly one red line. Now by selecting a circle with sufficiently large radius it will be tangent to one blue line and intersect all others. Case III: Both $\mathcal{B}$ and $\mathcal{R}$ are trivial, then all blue lines intersect at a point $B$ and all red lines intersect at a point $R$. Let $\mathcal{I}$ be the set of intersection points between blue lines and red lines. Let $P$ be a point on the convex hull of $\mathcal I\cup B\cup R$ which is distinct from $B$ and $R$. [asy][asy] size(6cm); real labelscalefactor = 0.5; /* changes label-to-point distance */pen dps = linewidth(0.7) + fontsize(0); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -100.03212876887468, xmax = 44.864230594854526, ymin = -6.462390463522458, ymax = 116.62161372717223; /* image dimensions */ /* draw figures */draw((xmin, 1.245614035087721*xmin + 121.41160781482247)--(xmax, 1.245614035087721*xmax + 121.41160781482247), linewidth(0.8) + blue); /* line */draw((xmin, -0.5485232067510549*xmin + 22.271574817691974)--(xmax, -0.5485232067510549*xmax + 22.271574817691974), linewidth(0.8) + red); /* line */draw((xmin, -3.447684424557947*xmin-63.17449255841388)--(xmax, -3.447684424557947*xmax-63.17449255841388), linewidth(0.8)); /* line */ /* dots and labels */dot((-26.972962089360898,87.81370766842552),dotstyle); label("$A$", (-26.41543005988324,89.09649886596203), NE * labelscalefactor); dot((-55.25777553980421,52.581747054715386),dotstyle); label("$P$", (-55.888076632971526,47.0298138893955), NE * labelscalefactor); dot((3.544862949275375,20.32713522526244),dotstyle); label("$C$", (4.095900092873355,21.582066187521917), NE * labelscalefactor); dot((-39.32971703384321,72.42196028143876),dotstyle); label("$E$", (-38.749797568444414,73.77591606585447), NE * labelscalefactor); dot((-29.47268570346792,38.43802689132417),dotstyle); label("$F$", (-29.012139009054014,39.75902883171734), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy] Suppose $AP$ and $PC$ are sides of the convex hull, then for any line $l$ they intersect $PA$ in the ray $PA$ and intersect $PC$ in the ray $PC$. Therefore we can pick some point $K$ on $PC$ such that every line $l$ intersects the interior of segment $KP$, the circle tangent to both $PA$ and $PC$ will satisfy the condition. Charmander3333 wrote: flyingpurplepeopleeater wrote: randomusername wrote: Take an intersection A of red lines and an intersection B of blue lines. Then a circle thru A,B with a large enough radius works. Or? Exacty one line must be tangent to the circle to get 2n-1 points ^^' I feel like this works, circle passing through the intersection also results in $2n-1$ points The lines may all intersect at one point

Extremely reminiscent of the "future" IMO 2016 P6. Draw a sufficiently large circle to contain all intersections. Pick a red and blue line whose intersections with this circle are adjacent through a minor arc, and draw a tangent circle to both whose radius approaches infinity and whose center goes in the direction that is not the one that subtends the minor arc.

Inconsistent wrote: Extremely reminiscent of the "future" IMO 2016 P6. Draw a sufficiently large circle to contain all intersections. Pick a red and blue line whose intersections with this circle are adjacent through a minor arc, and draw a tangent circle to both whose radius approaches infinity and whose center goes in the direction that is not the one that subtends the minor arc. Then it is the IMO problem which is reminiscent of this