Difference between revisions of "2001 AMC 10 Problems/Problem 4"

m (Dnhjan)
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<math> \textbf{(A) }2\qquad\textbf{(B) }3\qquad\textbf{(C) }4\qquad\textbf{(D) }5\qquad\textbf{(E) }6 </math>
 
<math> \textbf{(A) }2\qquad\textbf{(B) }3\qquad\textbf{(C) }4\qquad\textbf{(D) }5\qquad\textbf{(E) }6 </math>
==Solutions==
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=== Solution 1 ===
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== Solution 1 ==
  
 
[[File:circle-triangle problem.PNG]]
 
[[File:circle-triangle problem.PNG]]
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Therefore, <math> 2 \times 3 = \boxed{\textbf{(E) }6} </math>.
 
Therefore, <math> 2 \times 3 = \boxed{\textbf{(E) }6} </math>.
  
=== Solution 2 ===
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== Solution 2 ==
  
 
We know that the maximum amount of points that a circle and a line segment can intersect is <math>2</math>. Therefore, because there are <math>3</math> line segments in a triangle, the maximum amount of points of intersection is <math>\boxed{\textbf{(E) }6}</math>.
 
We know that the maximum amount of points that a circle and a line segment can intersect is <math>2</math>. Therefore, because there are <math>3</math> line segments in a triangle, the maximum amount of points of intersection is <math>\boxed{\textbf{(E) }6}</math>.

Revision as of 15:53, 30 April 2021

Problem

What is the maximum number of possible points of intersection of a circle and a triangle?

$\textbf{(A) }2\qquad\textbf{(B) }3\qquad\textbf{(C) }4\qquad\textbf{(D) }5\qquad\textbf{(E) }6$

Solution 1

Circle-triangle problem.PNG

We can draw a circle and a triangle, such that each side is tangent to the circle. This means that each side would intersect the circle at one point.

You would then have $3$ points, but what if the circle was bigger? Then, each side would intersect the circle at 2 points.

Therefore, $2 \times 3 = \boxed{\textbf{(E) }6}$.

Solution 2

We know that the maximum amount of points that a circle and a line segment can intersect is $2$. Therefore, because there are $3$ line segments in a triangle, the maximum amount of points of intersection is $\boxed{\textbf{(E) }6}$.

See Also

2001 AMC 10 (ProblemsAnswer KeyResources)
Preceded by
Problem 3
Followed by
Problem 5
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
All AMC 10 Problems and Solutions

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