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

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
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

Page

Toolbox

Search

2001 AMC 10 Problems/Problem 4

Contents

Problem

Solution

Solution 1

Solution 2

See Also