1976 AHSME Problems/Problem 28

Revision as of 11:29, 9 September 2021 by Aopssheriff (talk | contribs) (Undo revision 161945 by Polpot123 (talk))
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

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 intersect at most once. To maximize the number of points of intersection, note that each point must be 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 sets one by one:

  1. We construct all lines in set $X.$

    Since the lines in set $X$ are parallel to each other, they have $0$ points of intersection.

  2. We construct all lines in set $Y.$

    The lines in set $Y$ have $1$ point of intersection with each other, namely $A.$

    Moreover, each line in set $Y$ can intersect each line in set $X$ once. So, there are $25^2=625$ points of intersection.

    Now, we have $\boldsymbol{1+625=626}$ additional points of intersection.

  3. We construct all lines in set $Z.$

    The lines in set $Z$ can have $\binom{50}{2}=1225$ points of intersection with each other.

    Moreover, each line in set $Z$ can intersect each line in sets $X$ and $Y$ once. So, there are $50^2=2500$ points of intersection.

    Now, we have $\boldsymbol{1225+2500=3725}$ additional points of intersection.

Together, the answer is $626+3725=\boxed{\textbf{(B) }4351}.$

~MRENTHUSIASM

See also

1976 AHSME (ProblemsAnswer KeyResources)
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. AMC logo.png

Invalid username
Login to AoPS