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Difference between revisions of "1976 AHSME Problems/Problem 28"
m (→‎Solution)
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   <li>We construct all lines in set <math>Z.</math> <p>
 
   <li>We construct all lines in set <math>Z.</math> <p>
 
The lines in set <math>Z</math> can have <math>\binom{50}{2}=1225</math> points of intersection. <p>
 
The lines in set <math>Z</math> can have <math>\binom{50}{2}=1225</math> points of intersection. <p>
−
Moreover, each line in set <math>Z</math> can have <math>1</math> point of intersection with each line in sets <math>Y</math> and <math>Z.</math> So, there are <math>50\cdot50=2500</math> points of intersection. <p>
+
Moreover, each line in set <math>Z</math> can have <math>1</math> point of intersection with each line in sets <math>X</math> and <math>Y.</math> So, there are <math>50\cdot50=2500</math> points of intersection. <p>
 
<b>At this point, we have <math>\boldsymbol{1225+2500=3725}</math> additional points of intersection.</b> <p>
 
<b>At this point, we have <math>\boldsymbol{1225+2500=3725}</math> additional points of intersection.</b> <p>
 
</li>
 
</li>

Revision as of 20:36, 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*} 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 intersect at most once. 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 sets one by one:

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

    Since all 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, 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 $\boldsymbol{1+625=626}$ 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.

    Moreover, each line in set $Z$ can have $1$ point of intersection with each line in sets $X$ and $Y.$ So, there are $50\cdot50=2500$ points of intersection.

    At this point, 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 (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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