Difference between revisions of "1976 AHSME Problems/Problem 28"
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B\cap C & 25\cdot50 \\ | B\cap C & 25\cdot50 \\ | ||
\end{array}</cmath> | \end{array}</cmath> | ||
| − | Together, the answer is <cmath>1+\binom{50}{2}+25\cdot25+25\cdot50+ | + | Together, the answer is <cmath>1+\binom{50}{2}+25\cdot25+25\cdot50+25\cdot50=1+1225+625+1250+1250=\boxed{\textbf{(B) }4351}.</cmath> |
~MRENTHUSIASM | ~MRENTHUSIASM | ||
Revision as of 15:56, 8 September 2021
Problem
Lines 
are distinct. All lines 
a positive integer, are parallel to each other.
All lines 
a positive integer, pass through a given point 
The maximum number of points of intersection of pairs of lines from the complete set 
is
Solution
We partition into three sets. Let
from which 
and 
To maximize the number of points of intersection, note that each point of intersection is passed by exactly two lines. If three or more lines pass through the same point, then we can create more points of intersection by translating the lines.
We construct the following table:
![\[\begin{array}{c|c} & \\ [-2ex] \textbf{Set} & \textbf{\# of Points of Intersection} \\ [0.5ex] \hline & \\ [-2ex] A & 0 \\ [1ex] B & 1 \\ [1ex] C & \binom{50}{2} \\ [1ex] A\cap B & 25\cdot25 \\ [1ex] A\cap C & 25\cdot50 \\ [1ex] B\cap C & 25\cdot50 \\ \end{array}\]](http://latex.artofproblemsolving.com/a/b/8/ab8902492f7643bfbcb489a57e5247f2fd5809ab.png)
Together, the answer is ![\[1+\binom{50}{2}+25\cdot25+25\cdot50+25\cdot50=1+1225+625+1250+1250=\boxed{\textbf{(B) }4351}.\]](http://latex.artofproblemsolving.com/6/f/3/6f3ed869106aaa3253b602a4f8e4f090025d1e5a.png)
~MRENTHUSIASM
See also
| 1976 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 27 |
Followed by Problem 29 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
