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Difference between revisions of "1976 AHSME Problems/Problem 28"
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== Solution ==
 
== Solution ==
−
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 partition <math>\{L_1,L_2,\dots,L_{100}\}</math> into three sets. Let
−
 
 
−
We partition <math>L_1,L_2,\dots,L_{100}</math> into three sets. Let
 
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
 
A &= \{L_n\mid n\equiv0\pmod{4}\}, \\
 
A &= \{L_n\mid n\equiv0\pmod{4}\}, \\
 
B &= \{L_n\mid n\equiv1\pmod{4}\}, \\
 
B &= \{L_n\mid n\equiv1\pmod{4}\}, \\
−
C &= \{L_n\mid n\equiv2\text{ or }3\pmod{4}\},
+
C &= \{L_n\mid n\equiv2,3\pmod{4}\},
 
\end{align*}</cmath>
 
\end{align*}</cmath>
 
from which <math>|A|=|B|=25</math> and <math>|C|=50.</math>
 
from which <math>|A|=|B|=25</math> and <math>|C|=50.</math>
 +
 +
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:
 +
<cmath>\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}</cmath>
 +
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
  
 
== See also ==
 
== See also ==

Revision as of 15:55, 8 September 2021

Problem

Lines $L_1,L_2,\dots,L_{100}$ are distinct. All lines $L_{4n}, n$ a positive integer, are parallel to each other. All lines $L_{4n-3}, n$ a positive integer, pass through a given point $A.$ The maximum number of points of intersection of pairs of lines from the complete set $\{L_1,L_2,\dots,L_{100}\}$ is

$\textbf{(A) }4350\qquad \textbf{(B) }4351\qquad \textbf{(C) }4900\qquad \textbf{(D) }4901\qquad \textbf{(E) }9851$

Solution

We partition $\{L_1,L_2,\dots,L_{100}\}$ into three sets. Let \begin{align*} A &= \{L_n\mid n\equiv0\pmod{4}\}, \\ B &= \{L_n\mid n\equiv1\pmod{4}\}, \\ C &= \{L_n\mid n\equiv2,3\pmod{4}\}, \end{align*} from which $|A|=|B|=25$ and $|C|=50.$

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: \[\begin{array}{c|c} & \\ [-2ex] \textbf{Set} & \textbf{\# of Points of Intersection} \\ [0.5ex] \hline & \\ [-2ex] A & 0 \\ [1ex] B & 1 \\ [1ex] C & \binom{50}{2} \\ [1ex] A\cap B & 25\cdot25 \\ [1ex] A\cap C & 25\cdot50 \\ [1ex] B\cap C & 25\cdot50 \\ \end{array}\] Together, the answer is \[1+\binom{50}{2}+25\cdot25+25\cdot50+\25\cdot50=1+1225+625+1250+1250=\boxed{\textbf{(B) }4351}.\] ~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

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