Art of Problem Solving
Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus
During AMC 10A/12A testing, the AoPS Wiki is in read-only mode and no edits can be made.

2009 UNCO Math Contest II Problems/Problem 6

Problem

Let each of $m$ distinct points on the positive $x$-axis be joined to each of $n$ distinct points on the positive $y$-axis. Assume no three segments are concurrent (except at the axes). Obtain with proof a formula for the number of interior intersection points. The diagram shows that the answer is $3$ when $m=3$ and $n=2.$

[asy] draw((0,0)--(0,3),arrow=Arrow()); draw((0,0)--(4,0),arrow=Arrow()); for(int x=0;x<4;++x){ for(int y=0;y<3;++y){ D((x,0)--(0,y),black); }} dot(IP((2,0)--(0,1),(1,0)--(0,2))); dot(IP((3,0)--(0,1),(1,0)--(0,2))); dot(IP((3,0)--(0,1),(2,0)--(0,2))); [/asy]


Solution

Solution 1

Notice that choosing two points on the x axis and two points on the y axis, then, after constructing all possible lines, there will be only one point of intersection. So the answer is

$\binom{m}{2} \binom{n}{2}$

Solution 2

Let the points be $x_1 < x_2 \dots < x_m$ and $y_1<y_2 \dots < y_n$. Let $(x_i,y_j)$ represent the segment formed by joining $(x_i,0), (0,y_j)$. We first take $(x_1,y_1)$. No segments intersect this. For $(x_1,y_2)$, all the segments $(x_i,y_1)$, $i \in \{2,3 \dots m\}$ will intersect. Thus, we get $(m-1)(1)$ intersections. For $(x_1,y_3)$, all the segments with $(x_i,y_1),(x_i,y_2), x_i \neq x_1$ will intersect this. Thus we get $(m-1)(2)$ intersections. We keep going like this and finally for $(x_1,y_n)$, we get $(m-1)(n-1)$ intersections. When we move to $(x_2,y_i)$, we note that only points to the right intersect, thus we will be doing the same steps with $(m-2)$ instead. From this, we get the total intersections to be \[(1 + 2 + \dots + n-1)(m-1) + (1 + 2 + \dots + n-1)(m-2) \dots (1 + 2 + \dots + n-1)(1) = \boxed{\frac{(n-1)(n)}{2} \cdot \frac{(m-1)(m)}{2}}\]

See also

2009 UNCO Math Contest II (Problems • Answer Key • Resources)
Preceded by
Problem 5 		Followed by
Problem 7
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10
All UNCO Math Contest Problems and Solutions
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2009_UNCO_Math_Contest_II_Problems/Problem_6&oldid=259113"