2001 AMC 10 Problems/Problem 4

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

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 $\boxed{\textbf{(E) }6}$ .

2001 AMC 10 (Problems • Answer Key • Resources)
Preceded by Problem 3	Followed by Problem 5
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All AMC 10 Problems and Solutions

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2001 AMC 10 Problems/Problem 4

Contents

Problem

Solution 1

Solution 2

See Also