Difference between revisions of "1976 AHSME Problems/Problem 28"
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+ | == Problem == | ||
+ | Lines <math>L_1,L_2,\dots,L_{100}</math> are distinct. All lines <math>L_{4n}, n</math> a positive integer, are parallel to each other. | ||
+ | All lines <math>L_{4n-3}, n</math> a positive integer, pass through a given point <math>A.</math> The maximum number of points of intersection of pairs of lines from the complete set <math>\{L_1,L_2,\dots,L_{100}\}</math> is | ||
+ | <math>\textbf{(A) }4350\qquad | ||
+ | \textbf{(B) }4351\qquad | ||
+ | \textbf{(C) }4900\qquad | ||
+ | \textbf{(D) }4901\qquad | ||
+ | \textbf{(E) }9851</math> | ||
+ | |||
+ | == Solution == | ||
+ | We partition <math>\{L_1,L_2,\dots,L_{100}\}</math> into three sets. Let | ||
+ | <cmath>\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*}</cmath> | ||
+ | from which <math>|X|=|Y|=25</math> and <math>|Z|=50.</math> | ||
+ | |||
+ | 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: | ||
+ | <ol style="margin-left: 1.5em;"> | ||
+ | <li>We construct all lines in set <math>X.</math> <p> | ||
+ | Since the lines in set <math>X</math> are parallel to each other, they have <math>0</math> points of intersection. | ||
+ | </li> | ||
+ | <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> | ||
+ | Moreover, each line in set <math>Y</math> can intersect each line in set <math>X</math> once. So, there are <math>25^2=625</math> points of intersection. <p> | ||
+ | <b>Now, we have <math>\boldsymbol{1+625=626}</math> additional points of intersection.</b> <p> | ||
+ | </li> | ||
+ | <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> | ||
+ | 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>50^2=2500</math> points of intersection. <p> | ||
+ | <b>Now, we have <math>\boldsymbol{1225+2500=3725}</math> additional points of intersection.</b> <p> | ||
+ | </li> | ||
+ | </ol> | ||
+ | Together, the answer is <math>626+3725=\boxed{\textbf{(B) }4351}.</math> | ||
+ | |||
+ | ~MRENTHUSIASM | ||
+ | |||
+ | == See also == | ||
+ | |||
+ | {{AHSME box|year=1976|num-b=27|num-a=29}} | ||
+ | {{MAA Notice}} | ||
+ | [[Category:AHSME]] | ||
+ | [[Category:AHSME Problems]] |
Latest revision as of 11:29, 9 September 2021
Problem
Lines are distinct. All lines a positive integer, are parallel to each other. All lines a positive integer, pass through a given point The maximum number of points of intersection of pairs of lines from the complete set is
Solution
We partition into three sets. Let from which and
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 all lines in set
Since the lines in set are parallel to each other, they have points of intersection.
- We construct all lines in set
The lines in set have point of intersection with each other, namely
Moreover, each line in set can intersect each line in set once. So, there are points of intersection.
Now, we have additional points of intersection.
- We construct all lines in set
The lines in set can have points of intersection with each other.
Moreover, each line in set can intersect each line in sets and once. So, there are points of intersection.
Now, we have additional points of intersection.
Together, the answer is
~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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