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Difference between revisions of "1976 AHSME Problems/Problem 28"

m (→‎Solution)
m (→‎Solution: Made the solution more concise.)
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from which <math>|X|=|Y|=25</math> and <math>|Z|=50.</math>
 
from which <math>|X|=|Y|=25</math> and <math>|Z|=50.</math>
  
−
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.
+
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:
 
We construct the sets one by one:
Line 27: Line 27:
 
   <li>We construct all lines in set <math>Y.</math> <p>
 
   <li>We construct all lines in set <math>Y.</math> <p>
 
The lines in set <math>Y</math> have <math>1</math> point of intersection with each other, namely <math>A.</math> <p>
 
The lines in set <math>Y</math> have <math>1</math> point of intersection with each other, namely <math>A.</math> <p>
−
Moreover, each line in set <math>Y</math> can have <math>1</math> point of intersection with each line in set <math>X.</math> So, there are <math>25\cdot25=625</math> points of intersection. <p>
+
Moreover, each line in set <math>Y</math> can intersect each line in set <math>X</math> once. So, there are <math>25\cdot25=625</math> points of intersection. <p>
 
<b>Now, we have <math>\boldsymbol{1+625=626}</math> additional points of intersection.</b> <p>
 
<b>Now, we have <math>\boldsymbol{1+625=626}</math> additional points of intersection.</b> <p>
 
</li>
 
</li>
 
   <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 with each other. <p>
 
The lines in set <math>Z</math> can have <math>\binom{50}{2}=1225</math> points of intersection with each other. <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>
+
Moreover, each line in set <math>Z</math> can intersect each line in sets <math>X</math> and <math>Y</math> once.  So, there are <math>50\cdot50=2500</math> points of intersection. <p>
 
<b>Now, we have <math>\boldsymbol{1225+2500=3725}</math> additional points of intersection.</b> <p>
 
<b>Now, we have <math>\boldsymbol{1225+2500=3725}</math> additional points of intersection.</b> <p>
 
</li>
 
</li>

Revision as of 02:54, 9 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 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 $25\cdot25=625$ 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 $50\cdot50=2500$ 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 (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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