1976 AHSME Problems/Problem 28

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 $\begin{align*} X &= \{L_n\mid n\equiv0\pmod{4}\}, \\ Y &= \{L_n\mid n\equiv1\pmod{4}\}, \\ Z &= \{L_n\mid n\equiv2,3\pmod{4}\}, \\ \end{align*}$ from which $|X|=|Y|=25$ and $|Z|=50.$

Any two distinct lines can have at most one point of intersection. We construct the sets one by one:

We construct all lines in set $X.$
Since all lines in set $X$ are parallel to each other, they have $0$ points of intersection.
We construct all lines in set $Y.$
The lines in set $Y$ have $1$ point of intersection, namely $A.$
Moreover, each line in set $Y$ can have $1$ point of intersection with each line in set $X.$ So, there are $25\cdot25=625$ points of intersection.
At this point, we have $1+625=626$ additional points of intersection.
We construct all lines in set $Z.$
The lines in set $Z$ can have $\binom{50}{2}=1225$ points of intersection.
Moreover, each line in set $Z$ can have $1$ point of intersection with each line in set $Y$ or set $Z.$ So, there are $50\cdot50=2500$ points of intersection.

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
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1976 AHSME Problems/Problem 28

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