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

(Didn't check for correctness, just made pretty)
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== Solution ==
 
== Solution ==
−
{{image}}
+
<center><asy>defaultpen(fontsize(8));
−
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>.   
+
pair A=(4*sqrt(5),8), B=(0,8), O=(0,0);
 +
draw(circle((0,0),8));
 +
draw(O--A--B--O);
 +
label("A",A,(1,1));label("B",B,(-1,1));label("O",O,(-1,-1));
 +
label("<math>8</math>",A/3,(0.5,-1));label("<math>4</math>",5*A/6,(0.5,-1));
 +
label("<math>8</math>",B/2,(-1,0));label("<math>4\sqrt{5}</math>",B/2+A/2,(0,1));</asy></center>
 +
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>4\sqrt{5}</math>.   
  
−
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>. 
+
<center><asy>
 +
defaultpen(fontsize(8));
 +
pair A=(-4*sqrt(5),4), B=(0,4*(8*sqrt(6)-4*sqrt(5))/(8*sqrt(6))), C=(8*sqrt(6)-4*sqrt(5),0), D=(-4*sqrt(5),0), E=(0,0);
 +
draw(A--C--D--A);draw(B--E);
 +
label("A",A,(-1,1));label("B",B,(1,1));label("C",C,(1,0));label("D",D,(-1,-1));label("E",E,(0,-1));
 +
label("<math>4\sqrt{5}</math>",D/2+E/2,(0,-1));label("<math>8\sqrt{6}-4\sqrt{5}</math>",C/2+E/2,(0,-1));
 +
label("<math>4</math>",D/2+A/2,(-1,0));label("<math>x</math>",C/2+B/2,(1,0.5));label("<math>20-x</math>",0.7*A+0.3*B,(1,0.5));
 +
dot(A^^B^^C^^D^^E);
 +
</asy></center>
  
−
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>  
+
Now look at a side view and "unroll" the cylinder to be a flat surface. Let <math>C</math> be the bottom tether of the rope, let <math>D</math> be the point on the ground below <math>A</math>, and let <math>E</math> be the point directly below <math>B</math>. [[Triangle]]s <math>\triangle CDA</math> and <math>\triangle CEB</math> are [[similar]] [[right triangle]]s. By the Pythagorean Theorem <math>CD=8\cdot\sqrt{6}</math>. 
  
−
Therefore <math>a=60, b=750, c=3, a+b+c=813</math>.
+
Let <math>x</math> be the length of <math>CB</math>. <math>\frac{CA}{CD}=\frac{CB}{CE}\implies \frac{20}{8\sqrt{6}}=\frac{x}{8\sqrt{6}-4\sqrt{5}}\implies x=\frac{60-\sqrt{750}}{3}</math>
 +
 
 +
Therefore <math>a=60, b=750, c=3, a+b+c=\boxed{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 17:35, 19 November 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

defaultpen(fontsize(8));
pair A=(4*sqrt(5),8), B=(0,8), O=(0,0);
draw(circle((0,0),8));
draw(O--A--B--O);
label("A",A,(1,1));label("B",B,(-1,1));label("O",O,(-1,-1));
label("<math>8</math>",A/3,(0.5,-1));label("<math>4</math>",5*A/6,(0.5,-1));
label("<math>8</math>",B/2,(-1,0));label("<math>4\sqrt{5}</math>",B/2+A/2,(0,1)); (Error making remote request. Unknown error_msg)

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 $4\sqrt{5}$. 

defaultpen(fontsize(8));
pair A=(-4*sqrt(5),4), B=(0,4*(8*sqrt(6)-4*sqrt(5))/(8*sqrt(6))), C=(8*sqrt(6)-4*sqrt(5),0), D=(-4*sqrt(5),0), E=(0,0);
draw(A--C--D--A);draw(B--E);
label("A",A,(-1,1));label("B",B,(1,1));label("C",C,(1,0));label("D",D,(-1,-1));label("E",E,(0,-1));
label("<math>4\sqrt{5}</math>",D/2+E/2,(0,-1));label("<math>8\sqrt{6}-4\sqrt{5}</math>",C/2+E/2,(0,-1));
label("<math>4</math>",D/2+A/2,(-1,0));label("<math>x</math>",C/2+B/2,(1,0.5));label("<math>20-x</math>",0.7*A+0.3*B,(1,0.5));
dot(A^^B^^C^^D^^E);
 (Error making remote request. Unknown error_msg)

Now look at a side view and "unroll" the cylinder to be a flat surface. Let $C$ be the bottom tether of the rope, let $D$ be the point on the ground below $A$, and let $E$ be the point directly below $B$. Triangles $\triangle CDA$ and $\triangle CEB$ are similar right triangles. By the Pythagorean Theorem $CD=8\cdot\sqrt{6}$.

Let $x$ be the length of $CB$. $\frac{CA}{CD}=\frac{CB}{CE}\implies \frac{20}{8\sqrt{6}}=\frac{x}{8\sqrt{6}-4\sqrt{5}}\implies x=\frac{60-\sqrt{750}}{3}$

Therefore $a=60, b=750, c=3, a+b+c=\boxed{813}$.

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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