Difference between revisions of "1968 AHSME Problems/Problem 30"
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== Solution 1 == | == Solution 1 == | ||
| − | Notice how <math>P_2</math> can pass through each line segment of <math>P_1</math> at most twice | + | Notice how <math>P_2</math> can pass through each line segment of <math>P_1</math> at most twice. To have more than two intersections, the line passing through <math>P_1</math> would have a zigzag shape which is impossible for convex polygons. Therefore, the intersections does not depend on <math>P_2</math> and the answer is <math>\fbox{A}</math> |
== Solution 2 == | == Solution 2 == | ||
Revision as of 20:09, 1 November 2018
Contents
Problem
Convex polygons 
and 
are drawn in the same plane with 
and 
sides, respectively, 
. If and do not have any line segment in common, then the maximum number of intersections of and is:
Solution 1
Notice how can pass through each line segment of at most twice. To have more than two intersections, the line passing through would have a zigzag shape which is impossible for convex polygons. Therefore, the intersections does not depend on and the answer is 
Solution 2
Try to get the answer experimentally. Draw two of the simplest shapes: a square and a triangle and maximize the number of intersections. You will discover it is 6, and the only expression provided that will give 6 is
See also
| 1968 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 29 |
Followed by Problem 31 | |
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
