Difference between revisions of "2009 UNCO Math Contest II Problems/Problem 6"

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(Solution)
 
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== Solution ==
 
== Solution ==
{{solution}}
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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
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<math>\binom{m}{2} \binom{n}{2}</math>
  
 
== See also ==
 
== See also ==

Latest revision as of 13:46, 12 February 2017

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

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

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

See also

2009 UNCO Math Contest II (ProblemsAnswer KeyResources)
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