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Difference between revisions of "2004 AIME I Problems/Problem 14"
(→‎Solution: I will try making the 3D rendering later ..)
Line 3: Line 3:
  
 
== Solution ==
 
== Solution ==
−
<center><asy>defaultpen(fontsize(8));
+
<center><!--
 +
<asy>
 +
import three;
 +
defaultpen(fontsize(10)+linewidth(0.62));
 +
currentprojection = perspective(5,-10,30);
 +
 
 +
triple Oxy = (0,0,0), B=(0,-8,HEIGHT), A=(-8,4*5^.5,4), Cxy = 8*expi(RADIAN), C=(Cxy.x,Cxy.y,0), E=(B.x,B.y,0), D=(A.x,A.y,0);
 +
draw(A--Cxy--D--cycle);
 +
</asy>-->
 +
<asy>defaultpen(fontsize(10)+linewidth(0.62));
 
pair A=(4*sqrt(5),8), B=(0,8), O=(0,0);
 
pair A=(4*sqrt(5),8), B=(0,8), O=(0,0);
 
draw(circle((0,0),8));
 
draw(circle((0,0),8));
 
draw(O--A--B--O);
 
draw(O--A--B--O);
−
label("A",A,(1,1));label("B",B,(-1,1));label("O",O,(-1,-1));
+
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("$8$",A/3,(0.5,-1));label("$4$",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>
+
label("$8$",B/2,(-1,0));label("$4\sqrt{5}$",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>.   
 
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>.   
  
 
<center><asy>
 
<center><asy>
−
defaultpen(fontsize(8));
+
defaultpen(fontsize(10)+linewidth(0.6));
 
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);
 
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);
 
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("\(A\)",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("$4\sqrt{5}$",D/2+E/2,(0,-1));label("$8\sqrt{6}-4\sqrt{5}$",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));
+
label("$4$",D/2+A/2,(-1,0));label("$x$",C/2+B/2,(1,0.5));label("$20-x$",0.7*A+0.3*B,(1,0.5));
 
dot(A^^B^^C^^D^^E);
 
dot(A^^B^^C^^D^^E);
 
</asy></center>
 
</asy></center>
Line 24: Line 34:
 
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>.   
 
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>.   
  
−
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>  
+
Let <math>x</math> be the length of <math>CB</math>. <cmath>\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}</cmath>  
  
 
Therefore <math>a=60, b=750, c=3, a+b+c=\boxed{813}</math>.
 
Therefore <math>a=60, b=750, c=3, a+b+c=\boxed{813}</math>.

Revision as of 16:12, 27 April 2008

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

[asy]defaultpen(fontsize(10)+linewidth(0.62)); 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("$8$",A/3,(0.5,-1));label("$4$",5*A/6,(0.5,-1)); label("$8$",B/2,(-1,0));label("$4\sqrt{5}$",B/2+A/2,(0,1)); [/asy]

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 $4\sqrt{5}$.

[asy] defaultpen(fontsize(10)+linewidth(0.6)); 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("$4\sqrt{5}$",D/2+E/2,(0,-1));label("$8\sqrt{6}-4\sqrt{5}$",C/2+E/2,(0,-1)); label("$4$",D/2+A/2,(-1,0));label("$x$",C/2+B/2,(1,0.5));label("$20-x$",0.7*A+0.3*B,(1,0.5)); dot(A^^B^^C^^D^^E); [/asy]

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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