Difference between revisions of "2009 UNCO Math Contest II Problems/Problem 6"
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== Solution == | == 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 | ||
| + | |||
| + | <math>\binom{m}{2} \binom{n}{2}</math> | ||
== See also == | == See also == | ||
Revision as of 12:46, 12 February 2017
Problem
Let each of 
distinct points on the positive
-axis be joined to each of 
distinct points on
the positive 
-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 
when 
and 
![[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]](http://latex.artofproblemsolving.com/9/a/4/9a4f1e83e4540dd6d84dee9c1a84b446076fed15.png)
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
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 | ||
