1986 AHSME Problems/Problem 18

Problem

A plane intersects a right circular cylinder of radius $1$ forming an ellipse. If the major axis of the ellipse is $50\%$ longer than the minor axis, the length of the major axis is

$\textbf{(A)}\ 1\qquad \textbf{(B)}\ \frac{3}{2}\qquad \textbf{(C)}\ 2\qquad \textbf{(D)}\ \frac{9}{4}\qquad \textbf{(E)}\ 3$

Solution

Solution 1

The length of the minor axis is the distance from the opposite points of the cylinder, which is simply $2 \cdot 1 =2$ , and according to the question, the length of the major axis is simply $2 \cdot 150\% = \boxed{(E) 3}$

~ $shalomkeshet$

Solution 2

We note that we can draw the minor axis to see that because the minor axis is the minimum distance between two opposite points on the ellipse, we can draw a line through two opposite points of the cylinder, and so the minor axis is $2(1) = 2$ . Therefore, our answer is $2(1.5) = 3$ , and so our answer is $\boxed{E}$ .

1986 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 17	Followed by Problem 19
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1986 AHSME Problems/Problem 18

Contents

Problem

Solution

Solution 1

Solution 2

See also