Difference between revisions of "2014 AIME I Problems/Problem 10"
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− | + | <asy> | |
+ | size(150); | ||
+ | pair a=(5,0),b=(2,3*sqrt(3)),c=(6,0),d=(3,3*sqrt(3)),e=(0,0); | ||
+ | draw(circle(e,5)); | ||
+ | draw(circle(c,1)); | ||
+ | draw(circle(d,1)); | ||
+ | dot(a^^b^^c^^d^^e^^(5/2,5*sqrt(3)/2)); | ||
+ | label("$A$",a,W,fontsize(9)); | ||
+ | label("$B$",b,NW,fontsize(9)); | ||
+ | label("$C$",c,E,fontsize(9)); | ||
+ | label("$D$",d,E,fontsize(9)); | ||
+ | label("$E$",e,SW,fontsize(9)); | ||
+ | label("$F$",(5/2,5*sqrt(3)/2),SSW,fontsize(9)); | ||
+ | </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>\alpha = \angle AEF</math>, in radians. The smaller disk must then have rolled along an arc of length <math>5\alpha</math>, since it 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, the length of major arc <math>\overarc{BF}</math> equals the length of minor arc <math>\overarc{AF}</math>, or <math>5\alpha</math>. Since <math>\overline{AC} || \overline{BD}</math>, <math>\angle BDF \cong \angle FEA</math>, 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>. | 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>\alpha = \angle AEF</math>, in radians. The smaller disk must then have rolled along an arc of length <math>5\alpha</math>, since it 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, the length of major arc <math>\overarc{BF}</math> equals the length of minor arc <math>\overarc{AF}</math>, or <math>5\alpha</math>. Since <math>\overline{AC} || \overline{BD}</math>, <math>\angle BDF \cong \angle FEA</math>, 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>. |
Revision as of 11:45, 22 August 2014
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
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 it 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 .
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.