Difference between revisions of "1976 AHSME Problems/Problem 28"
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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 | + | 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 | + | 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> | <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 | + | 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> | <b>Now, we have <math>\boldsymbol{1225+2500=3725}</math> additional points of intersection.</b> <p> | ||
</li> | </li> |
Latest revision as of 02:47, 14 February 2024
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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