1964 AMC 12 Problem 36

Question

In this figure the radius of the circle is equal to the altitude of the equilateral triangle

ABC.

The circle is made to roll along the side

AB,

remaining tangent to it at a variable point

T

and intersecting lines

AC

and

BC

in variable points

M

and

N,

respectively. Let

n

be the number of degrees in arc

MTN.

Then

n,

for all permissible positions of the circle:

varies from

30^\circ

to

90^\circ

varies from

30^\circ

to

60^\circ

varies from

60^\circ

to

90^\circ

remains constant at

30^\circ

remains constant at

60^\circ

Answer: E

Concepts:circle coordinate geometry equilateral triangle inscribed angle arc

Difficulty rating: 2130

Small hint:

The circle’s center

O

and vertex

C

are the same distance above

AB

, so

CO\parallel AB

Big hint:

Extend

NC

through

C

to meet the circle again at

D

, then use reflection across line

CO

Solution:

Let

O

be the circle’s center. Both

O

and

C

are one triangle altitude above

AB,

so

CO\parallel AB.

Extend

NC

through

C

to meet the circle again at

D.

Since

CD

is opposite to

CN,

\begin{aligned} \angle MCD &=180^\circ-\angle MCN\\ &=120^\circ. \end{aligned}

The line

CO,

parallel to

AB,

bisects this angle. Reflection across

CO

fixes the circle and interchanges rays

CM

and

CD,

so it interchanges

M

and

D.

Hence

CM=CD,

and isosceles triangle

MCD

has base angles

30^\circ.

Because

D,C,N

are collinear,

\angle MDN=30^\circ.

This inscribed angle subtends arc

MTN,

whose measure is therefore

60^\circ

for every permissible circle position.

Thus, the correct answer is E.

1964 AMC 12 Problem 36

Answer: E

Difficulty rating: 2130

Small hint:

Big hint:

Solution:

Problem 36 in other years