2001 AMC 10 Problem 21

Question

A right circular cylinder with its diameter equal to its height is inscribed in a right circular cone. The cone has diameter

10

and altitude

12,

and the axes of the cylinder and cone coincide. Find the radius of the cylinder.

\dfrac83

\dfrac{30}{11}

3

\dfrac{25}{8}

\dfrac72

In an axial cross-section, the cylinder has radius

r

and height

2r

Similar triangles give

\dfrac{12-2r}{r}=\dfrac{12}{5}

Take an axial cross-section. The cone has base radius

5

and height

12;

the cylinder appears as a rectangle of width

2r

and height

2r.

By similar triangles,

\dfrac{12-2r}{r}=\dfrac{12}{5},

so

5(12-2r)=12r,

giving

60=22r

and

r=\dfrac{30}{11}.

Thus, the correct answer is B.

Problem 21 in other years