Difference between revisions of "2014 AIME I Problems/Problem 10"
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A disk with radius <math>1</math> is externally tangent to a disk with radius <math>5</math>. Let <math>A</math> be the point where the disks are tangent, <math>C</math> be the center of the smaller disk, and <math>E</math> 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 <math>360^\circ</math>. That is, if the center of the smaller disk has moved to the point <math>D</math>, and the point on the smaller disk that began at <math>A</math> has now moved to point <math>B</math>, then <math>\overline{AC}</math> is parallel to <math>\overline{BD}</math>. Then <math>\sin^2(\angle BEA)=\tfrac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>. | A disk with radius <math>1</math> is externally tangent to a disk with radius <math>5</math>. Let <math>A</math> be the point where the disks are tangent, <math>C</math> be the center of the smaller disk, and <math>E</math> 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 <math>360^\circ</math>. That is, if the center of the smaller disk has moved to the point <math>D</math>, and the point on the smaller disk that began at <math>A</math> has now moved to point <math>B</math>, then <math>\overline{AC}</math> is parallel to <math>\overline{BD}</math>. Then <math>\sin^2(\angle BEA)=\tfrac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>. | ||
− | == Solution == | + | == Solution 1 == |
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</asy> | </asy> | ||
− | Let <math>F</math> be the new tangency point of the two disks. The smaller disk rolled along minor arc <math>\overarc{AF}</math> on the larger disk | + | Let <math>F</math> be the new tangency point of the two disks. The smaller disk rolled along minor arc <math>\overarc{AF}</math> on the larger disk. |
− | Now, to find <math>\sin^2{\angle BEA}</math>, we construct <math>\triangle BDE</math>. Also, drop a perpendicular from <math>D</math> to <math>\overline{EA}</math>, and call this point <math>X</math>. Since <math>\alpha = \frac{\pi}{3}</math> | + | Let <math>\alpha = \angle AEF</math>, in radians. The smaller disk must then have rolled along an arc of length <math>5\alpha</math>, since the larger disk has a radius of <math>5</math>. Since all of the points on major arc <math>\overarc{BF}</math> 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,<cmath>\overarc{BF}=\overarc{AF}=5\alpha</cmath>. |
+ | |||
+ | Since <math>\overline{AC} || \overline{BD}</math>, <cmath>\angle BDF \cong \angle FEA</cmath> so the angles of minor arc <math>\overarc{BF}</math> and minor arc <math>\overarc{AF}</math> are equal, so minor arc <math>\overarc{BF}</math> has an angle of <math>\alpha</math>. Since the smaller disk has a radius of <math>1</math>, the length of minor arc <math>\overarc{BF}</math> is <math>\alpha</math>. This means that <math>5\alpha + \alpha</math> equals the circumference of the smaller disk, so <math>6\alpha = 2\pi</math>, or <math>\alpha = \frac{\pi}{3}</math>. | ||
+ | |||
+ | Now, to find <math>\sin^2{\angle BEA}</math>, we construct <math>\triangle BDE</math>. Also, drop a perpendicular from <math>D</math> to <math>\overline{EA}</math>, and call this point <math>X</math>. Since <math>\alpha = \frac{\pi}{3}</math> and <math>\angle DXE</math> is right, <cmath>DE = 6</cmath> <cmath>EX = 3</cmath> and <cmath>DX = 3\sqrt{3}</cmath> | ||
+ | |||
+ | Now drop a perpendicular from <math>B</math> to <math>\overline{EA}</math>, and call this point <math>Y</math>. Since <math>\overline{BD} || \overline{EA}</math>, <cmath>XY = BD = 1</cmath> and <cmath>BY = DX = 3\sqrt{3}</cmath> Thus, we know that <cmath>EY = EX - XY = 3 - 1 = 2</cmath> and by using the Pythagorean Theorem on <math>\triangle BEY</math>, we get that <cmath>BE = \sqrt{31}</cmath> Thus, <cmath>\sin{\angle BEA} = \frac{\sqrt{27}}{\sqrt{31}}</cmath> so <cmath>\sin^2{\angle BEA} = \frac{27}{31}</cmath> and our answer is <math>27 + 31 = \boxed{058}</math>. | ||
+ | |||
+ | == Solution 2 == | ||
+ | |||
+ | First, we determine how far the small circle goes. For the small circle to rotate completely around the circumference, it must rotate <math>5</math> times (the circumference of the small circle is <math>2\pi</math> while the larger one has a circumference of <math>10\pi</math>) plus the extra rotation the circle gets for rotating around the circle, for a total of <math>6</math> times. Therefore, one rotation will bring point <math>D</math> <math>60^\circ</math> from <math>C</math>. | ||
+ | |||
+ | Now, draw <math>\triangle DBE</math>, and call <cmath>\angle BED = x^{\circ}</cmath> We know that <math>\overline{ED}</math> is 6, and <math>\overline{BD}</math> is 1. Since <math>EC || BD</math>, <cmath>\angle BDE = 60^\circ</cmath> | ||
+ | |||
+ | By the Law of Cosines, <cmath>\overline{BE}^2=36+1-2\times 6\times 1\times \cos{60^\circ} = 36+1-6=31</cmath> and since lengths are positive, <cmath>\overline{BE}=\sqrt{31}</cmath> | ||
+ | |||
+ | By the Law of Sines, we know that <cmath>\frac{1}{\sin{x}}=\frac{\sqrt{31}}{\sin{60^\circ}}</cmath> so <cmath>\sin{x} = \frac{\sin{60^\circ}}{\sqrt{31}} = \frac{\sqrt{93}}{62}</cmath> As <math>x</math> is clearly between <math>0</math> and <math>90^\circ</math>, <math>\cos{x}</math> is positive. As <math>\cos{x}=\sqrt{1-\sin^2{x}}</math>, <cmath>\cos{x} = \frac{11\sqrt{31}}{62}</cmath> | ||
+ | |||
+ | Now we use the angle sum formula to find the sine of <math>\angle BEA</math>: <cmath>\sin 60^\circ\cos x + \cos 60^\circ\sin x = \frac{\sqrt{3}}{2}\frac{11\sqrt{31}}{62}+\frac{1}{2}\frac{\sqrt{93}}{62}</cmath> <cmath>= \frac{11\sqrt{93}+\sqrt{93}}{124} = \frac{12\sqrt{93}}{124} = \frac{3\sqrt{93}}{31} = \frac{3\sqrt{31}\sqrt{3}}{31} = \frac{3\sqrt{3}}{\sqrt{31}}</cmath> | ||
+ | |||
+ | Finally, we square this to get <cmath>\frac{9\times 3}{31}=\frac{27}{31}</cmath> so our answer is <math>27+31=\boxed{058}</math>. | ||
== See also == | == See also == | ||
{{AIME box|year=2014|n=I|num-b=9|num-a=11}} | {{AIME box|year=2014|n=I|num-b=9|num-a=11}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 20:02, 4 September 2021
Contents
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,
.
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
and
is right,
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
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
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.