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Difference between revisions of "1968 AHSME Problems/Problem 30"

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== Solution 1 ==
 
== Solution 1 ==
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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 polyons. Therefore, the intersections does not depend on <math>P_2</math> and the answer is <math>\fbox{A}</math>
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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 ==
 
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 <math>\fbox{A}</math>
 
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 <math>\fbox{A}</math>
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Note : this solution isn’t risk free, as the option of “None of These” is given as well.
  
 
== See also ==
 
== See also ==
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{{AHSME box|year=1968|num-b=29|num-a=31}}   
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{{AHSME 35p box|year=1968|num-b=29|num-a=31}}   
  
 
[[Category: Intermediate Combinatorics Problems]]
 
[[Category: Intermediate Combinatorics Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 00:53, 16 August 2023

Problem

Convex polygons $P_1$ and $P_2$ are drawn in the same plane with $n_1$ and $n_2$ sides, respectively, $n_1\le n_2$. If $P_1$ and $P_2$ do not have any line segment in common, then the maximum number of intersections of $P_1$ and $P_2$ is:

$\text{(A) } 2n_1\quad \text{(B) } 2n_2\quad \text{(C) } n_1n_2\quad \text{(D) } n_1+n_2\quad \text{(E) } \text{none of these}$

Solution 1

Notice how $P_2$ can pass through each line segment of $P_1$ at most twice. To have more than two intersections, the line passing through $P_1$ would have a zigzag shape which is impossible for convex polygons. Therefore, the intersections does not depend on $P_2$ and the answer is $\fbox{A}$

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 $\fbox{A}$

Note : this solution isn’t risk free, as the option of “None of These” is given as well.

See also

1968 AHSC (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 • 31 • 32 • 33 • 34 • 35
All AHSME Problems and Solutions

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