FIRST PROBLEM: Set 102 straight lines on a plane, so they intersect in exactly 2002 points.
Draw 24 equally-spaced parallel lines, and cross them with 76 other
equally-spaced lines perpendicular to the first group, for 24*76 =
1824 crossings. Draw the other two lines at a 45-degree angle to all
the others (and parallel to each other). Place one so that it crosses 11 of the intersections of the first two groups of lines. It makes new intersections with the other 78 lines. Place the other line so that it crosses all 100 of the original lines. 1824 + 178 = 2002.
As it is, it's quite difficult to solve -- I guess the professor made a typo when he wrote the problem, so here is a... SECOND PROBLEM: What number would have made that a reasonable High School level problem?
102 points crossing in 2001 intersections -- a problem he may have given the year before. This works with 26 parallel lines crossed by 75 parallel lines (1900 intersections) plus one additional line which makes new intersections with each of the first 101 lines.
Joseph DeVincentis
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Hi Ed,
"Set 102 straight lines on a plane, so they intersect in exactly 2002
points. As it is, it's quite difficult to solve -- I guess the professor
made a typo when he wrote the problem"
The problem is extremely simple to solve using a program.
One has to draw 5 groups of lines, the lines is each group being parallel to
each other but not to the lines in the other groups. The number of lines in
each group is 2,5,7,9 and 79.
Here's the C++ program:
#include
using namespace std;
#define MAX_LEVEL 5
int n[MAX_LEVEL];
void search(int level, int prev, int remainder) {
if (level==(MAX_LEVEL-1)) {
n[MAX_LEVEL-1] = remainder;
int sum = 0;
for (int i=0;i