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Difference between revisions of "2004 AIME I Problems/Problem 14"

m (→‎Solution: change ^(1/2) to \sqrt{}, though I might've misplaced some of them)
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== Solution ==
 
== Solution ==
 
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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=\sqrt{80} </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*\sqrt{6}</math>.  Let the length of <math>x=AB</math>. <math>\frac{CA}{CD}=\frac{AB}{AE}</math>: <math>\frac{5}{2\sqrt{6}}=\frac{x}{\sqrt{80}}</math>. <math>20-x=(60-\sqrt{750})/3</math>, Therefore <math>a=60, b=750, c=3, a+b+c=813</math>.
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Looking from an overhead view, call the [[center]] of the [[circle]] <math>O</math>, the tether point to the unicorn <math>A</math> and the last point where the rope touches the tower <math>B</math>.  <math>\triangle OAB</math> is a [[right triangle]] because <math>OB</math> is a radius and <math>BA</math> is a [[tangent line]] at point <math>B</math>.  We use the [[Pythagorean Theorem]] to find the horizontal component of <math>AB</math> has length <math>\sqrt{80}</math>.   
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Now, looking at a side view and "unrolling" the cylinder to be a flat surface, call the bottom tether of the rope <math>C</math>, the point on the ground below <math>A</math>, <math>D</math>, and the point directly above <math>B</math> and 4 feet off the ground <math>E</math>.  [[Triangle]]s <math>\triangle CDA</math> and <math>\triangle AEB</math> are [[similar]] [[right triangle]]s.  By the Pythagorean Theorem <math>CD=8\cdot\sqrt{6}</math>.   
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Let <math>x</math> be the length of <math>AB</math>. <math>\frac{CA}{CD}=\frac{AB}{AE}</math>: <math>\frac{5}{2\sqrt{6}}=\frac{x}{\sqrt{80}}</math>. <math>20-x=(60-\sqrt{750})/3</math>  
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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}}
 
{{AIME box|year=2004|n=I|num-b=13|num-a=15}}

Revision as of 13:21, 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 $\frac{a-\sqrt{b}}c$ feet, where $a, b,$ and $c$ are positive integers, and $c$ is prime. Find $a+b+c.$

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 $B$. $\triangle OAB$ is a right triangle because $OB$ is a radius and $BA$ is a tangent line at point $B$. We use the Pythagorean Theorem to find the horizontal component of $AB$ has length $\sqrt{80}$.

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 $\triangle CDA$ and $\triangle AEB$ are similar right triangles. By the Pythagorean Theorem $CD=8\cdot\sqrt{6}$.

Let $x$ be the length of $AB$. $\frac{CA}{CD}=\frac{AB}{AE}$: $\frac{5}{2\sqrt{6}}=\frac{x}{\sqrt{80}}$. $20-x=(60-\sqrt{750})/3$

Therefore $a=60, b=750, c=3, a+b+c=813$.

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
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