Difference between revisions of "1986 AHSME Problems/Problem 18"

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A plane intersects a right circular cylinder of radius <math>1</math> forming an ellipse.  
 
A plane intersects a right circular cylinder of radius <math>1</math> forming an ellipse.  
If the major axis of the ellipse of <math>50\%</math> longer than the minor axis, the length of the major axis is
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If the major axis of the ellipse is <math>50\%</math> longer than the minor axis, the length of the major axis is
  
 
<math>\textbf{(A)}\ 1\qquad
 
<math>\textbf{(A)}\ 1\qquad
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\textbf{(C)}\ 2\qquad
 
\textbf{(C)}\ 2\qquad
 
\textbf{(D)}\ \frac{9}{4}\qquad
 
\textbf{(D)}\ \frac{9}{4}\qquad
\textbf{(E)}\ 3  </math>
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\textbf{(E)}\ 3  </math>
  
 
==Solution==
 
==Solution==
  
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===Solution 1===
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The length of the minor axis is the distance from the opposite points of the cylinder, which is simply <math>2 \cdot 1 =2</math>, and according to the question, the length of the major axis is simply <math>2 \cdot 150\% = \boxed{(E) 3}</math>
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~ <math>shalomkeshet</math>
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===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 <math>2(1) = 2</math>. Therefore, our answer is <math>2(1.5) = 3</math>, and so our answer is <math>\boxed{E}</math>.
 
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 <math>2(1) = 2</math>. Therefore, our answer is <math>2(1.5) = 3</math>, and so our answer is <math>\boxed{E}</math>.
  

Latest revision as of 03:26, 31 October 2024

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 $\boxed{E}$.

See also

1986 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 17
Followed by
Problem 19
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