2014 AIME I Problems/Problem 10
Problem 10
A disk with radius is externally tangent to a disk with radius
. Let
be the point where the disks are tangent,
be the center of the smaller disk, and
be the center of the larger disk. While the larger disk remains fixed, the smaller disk is allowed to roll along the outside of the larger disk until the smaller disk has turned through an angle of
. That is, if the center of the smaller disk has moved to the point
, and the point on the smaller disk that began at
has now moved to point
, then
is parallel to
. Then
, where
and
are relatively prime positive integers. Find
.
Solution 1
Let be the new tangency point of the two disks. The smaller disk rolled along minor arc
on the larger disk. Let
, in radians. The smaller disk must then have rolled along an arc of length
, since the larger disk has a radius of
. Since all of the points on major arc
on the smaller disk have come into contact with the larger disk at some point during the rolling, and none of the other points on the smaller disk did, the length of major arc
equals the length of minor arc
, or
. Since
,
, so the angles of minor arc
and minor arc
are equal, so minor arc
has an angle of
. Since the smaller disk has a radius of
, the length of minor arc
is
. This means that
equals the circumference of the smaller disk, so
, or
.
Now, to find , we construct
. Also, drop a perpendicular from
to
, and call this point
. Since
,
is right, and
,
and
. Now drop a perpendicular from
to
, and call this point
. Since
,
, and
. Thus, we know that
, and by using the Pythagorean Theorem on
, we get that
. Thus,
, so
, and our answer is
.
Solution 2 (Trigonometry)
First, we determine how far the small circle goes. For the small circle to rotate completely around the circumference, it must rotate times (the circumference of the small circle is
while the larger one has a circumference of
) plus the extra rotation the circle gets for rotating around the circle, for a total of
times. Therefore, one rotation will bring point
from
.
Now, draw , and call
, in degrees. We know that
is 6, and
is 1. Since
,
. By the Law of Cosines,
, and since lengths are positive,
.
By the Law of Sines, we know that , so
. As
is clearly between
and
,
is positive. As
,
.
Now we use the angle sum formula to find the sine of :
.
Finally, we square this to get , so our answer is
.
See also
2014 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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