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