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 $y_1<y_2 \dots < y_n$ . 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 $(1 + 2 + \dots + n-1)(m-1) + (1 + 2 + \dots + n-1)(m-2) \dots (1 + 2 + \dots + n-1)(1) = \boxed{\frac{(n-1)(n)}{2} \cdot \frac{(m-1)(m)}{2}}$

2009 UNCO Math Contest II (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
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2009 UNCO Math Contest II Problems/Problem 6

Contents

Problem

Solution

Solution 1

Solution 2

See also