5. Xander draws five points and a number of infinitely long lines on an infinite sheet of paper, in such a way that at least two of those points lie on each line and that the lines intersect only at points that Xander has drawn.

What is the maximum number of lines that Xander can have drawn?
A) 3
B) 4
C) 5
D) 6
E) 7
A5. D) 6 We can get 6 lines as follows. We take the extended sides of a square together with the two diagonals. The 5 points are now the four corners of the square and the center, see also the figure below.

We will show that 7 (or more) lines are not possible. We do this by contradiction: assume that a situation with 7 lines does exist. In

![](https://cdn.mathpix.com/cropped/2024_04_17_a4374d618d1fd957f1eeg-2.jpg?height=317&width=319&top_left_y=1378&top_left_x=1528)
general, 7 lines intersect in $\frac{7 \cdot 6}{2}=21$ points, but there are two reasons why there could be fewer.

First, lines can be parallel, meaning they do not intersect. Since each line must contain at least 2 of the 5 points, it is not possible for 3 lines to be parallel (as that would require $3 \cdot 2=6$ points). Therefore, of the 7 lines, there can be at most 3 pairs that are parallel, and thus at most 3 fewer intersection points due to parallelism. This leaves at least $21-3=18$ intersection points.

Second, intersection points can coincide. This happens when more than two lines pass through the same point. If 3 lines pass through one point, 3 intersection points coincide. If 4 lines pass through one point, 6 intersection points coincide, and so on. The at least 18 remaining intersection points must coincide in the 5 points we started with. This means that there must be a point, say $P$, where at least 4 intersection points coincide (since $5 \cdot 3=15<18$). Therefore, at least 4 different lines, say $\ell_{1}, \ell_{2}, \ell_{3}$, and $\ell_{4}$, pass through this point.

On each of these 4 lines, there must be another one of the 5 points, say points $P_{1}, P_{2}, P_{3}$, and $P_{4}$. These points cannot coincide. Since we have now found all 5 points, there can be no fifth line passing through the point $P$. Each of the remaining 3 lines intersects at least 3 of the lines $\ell_{1}, \ell_{2}, \ell_{3}$, and $\ell_{4}$ (since it can be parallel to at most one of those lines). This means that the remaining lines must contain at least 3 of the points $P_{1}, P_{2}, P_{3}$, and $P_{4}$. However, this implies that the 3 remaining lines are all the same, which is a contradiction.

We conclude that 7 (or more) lines are not possible and that 6 is the maximum number of lines.