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2001 AMC 10 Problem 4

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2001 AMC 10 Problem 4AMC 10 Step by Step

Name: Date: September 13, 2026

Circles · difficulty 1 of 5 · about 1 min · position P1-10

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

A)
$2$
B)
$3$
C)
$4$
D)
$5$
E)
$6$

Answer

Key insight

A line meets a circle at most twice, so three sides give at most six points, and a triangle slightly larger than the circle achieves all six.

Solution

A straight line can cross a circle in at most two points, and each side of a triangle is part of a line. With three sides, the total number of intersection points cannot exceed $3 \cdot 2 = 6$ .

To see that $6$ is actually possible, start with a triangle whose incircle touches all three sides, then enlarge the circle very slightly (keeping its center). Each former tangency point splits into two crossing points, one on each side of it, and all six lie on the sides of the triangle.

The answer is $\boxed{\textbf{(E)}\ 6}$ .

Why this works

A "maximum number of intersections" question has two halves: an upper bound obtained by counting per piece (each segment contributes at most two points) and an explicit configuration that reaches the bound. Without the construction, six is only a ceiling; without the bound, six is only an example.

The trap

Picturing only the incircle (three tangent points) and answering 3.

Common mistakes

Picturing only the incircle (three tangent points) and answering 3.
Believing a segment can only cross a circle once, or forgetting that a circle can cut through all three sides at the same time.

Techniques

Bound the quantity above/below or estimate to pin it down