1968 AHSME Problems/Problem 30

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 $\fbox{A}$

1968 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 29	Followed by Problem 31
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1968 AHSME Problems/Problem 30

Contents

Problem

Solution 1

Solution 2

See also