# 1968 AHSME Problems/Problem 30

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

## 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.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 