Difference between revisions of "2004 AIME I Problems/Problem 14"
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== Problem == | == Problem == | ||
| − | A unicorn is tethered by a 20-foot silver rope to the base of a magician's cylindrical tower whose radius is 8 feet. The rope is attached to the tower at ground level and to the unicorn at a height of 4 feet. The unicorn has pulled the rope taut, the end of the rope is 4 feet from the nearest point on the tower, and the length of the rope that is touching the tower is <math> \frac{a-\sqrt{b}}c </math> feet, where <math> a, b, </math> and <math> c </math> are positive | + | A unicorn is tethered by a 20-foot silver rope to the base of a magician's [[cylinder|cylindrical]] tower whose [[radius]] is 8 feet. The rope is attached to the tower at ground level and to the unicorn at a [[height]] of 4 feet. The unicorn has pulled the rope taut, the end of the rope is 4 feet from the nearest point on the tower, and the length of the rope that is touching the tower is <math> \frac{a-\sqrt{b}}c </math> feet, where <math> a, b, </math> and <math> c </math> are [[positive]] [[integer]]s, and <math> c </math> is prime. Find <math> a+b+c. </math> |
== Solution == | == Solution == | ||
| − | Looking from an overhead view, call the center of the circle O, the tether point to the unicorn A, and the last point where the rope touches the tower B. | + | {{image}} |
| + | Looking from an overhead view, call the [[center]] of the [[circle]] O, the tether point to the unicorn A, and the last point where the rope touches the tower B. <math>\triangle OAB</math> is a [[right triangle]] because OB is a radius and BA is a [[tangent]] line of point B. We use the [[Pythagorean theorem]] to find the horizontal component of <math>AB=(80)^(1/2) </math>. Now looking at a side view and "unrolling" the cylinder to be a flat surface, call the bottom tether of the rope C, the point on the ground below A D, and the point directly above B and 4 feet off the ground E. Triangles CDA and AEB are similar right triangles. By the Pythagorean theorem <math>CD=8*6^(1/2)</math>. Let the length of <math>x=AB</math>. CA/CD=AB/AE: <math>(5/2)*6^(1/2)=x/80^(1/2)</math>. <math>20-x=(60-750^(1/2))/3</math>, Therefore <math>a=60, b=750, c=3, a+b+c=813</math>. | ||
== See also == | == See also == | ||
| − | + | {{AIME box|year=2004|n=I|num-b=13|num-a=15}} | |
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Revision as of 08:53, 5 March 2007
Problem
A unicorn is tethered by a 20-foot silver rope to the base of a magician's cylindrical tower whose radius is 8 feet. The rope is attached to the tower at ground level and to the unicorn at a height of 4 feet. The unicorn has pulled the rope taut, the end of the rope is 4 feet from the nearest point on the tower, and the length of the rope that is touching the tower is 
feet, where 
and 
are positive integers, and is prime. Find 
Solution
An image is supposed to go here. You can help us out by creating one and editing it in. Thanks.
Looking from an overhead view, call the center of the circle O, the tether point to the unicorn A, and the last point where the rope touches the tower B. 
is a right triangle because OB is a radius and BA is a tangent line of point B. We use the Pythagorean theorem to find the horizontal component of 
. Now looking at a side view and "unrolling" the cylinder to be a flat surface, call the bottom tether of the rope C, the point on the ground below A D, and the point directly above B and 4 feet off the ground E. Triangles CDA and AEB are similar right triangles. By the Pythagorean theorem 
. Let the length of 
. CA/CD=AB/AE: 
. 
, Therefore 
.
See also
| 2004 AIME I (Problems • Answer Key • Resources) | ||
| Preceded by Problem 13 |
Followed by Problem 15 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
