Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution: I will try making the 3D rendering later ..)
(Asymptote (3d))
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
−
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>
+
A unicorn is tethered by a <math>20</math>-foot silver rope to the base of a magician's [[cylinder|cylindrical]] tower whose [[radius]] is <math>8</math> feet. The rope is attached to the tower at ground level and to the unicorn at a [[height]] of <math>4</math> feet. The unicorn has pulled the rope taut, the end of the rope is <math>4</math> 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 ==
−
<center><!--
+
<center>
 
<asy>
 
<asy>
−
import three;
+
  /* Settings */
−
defaultpen(fontsize(10)+linewidth(0.62));
+
import three; defaultpen(fontsize(10)+linewidth(0.62));  
−
currentprojection = perspective(5,-10,30);
+
currentprojection = perspective(-2,-50,15); size(200);
−
 
+
  /* Variables */
−
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);
+
real x = 20 - ((750)^.5)/3, CE = 8*(6^.5) - 4*(5^.5), CD = 8*(6^.5), h = 4*CE/CD;
−
draw(A--Cxy--D--cycle);  
+
pair Cxy = 8*expi((3*pi)/2-CE/8);
−
</asy>-->
+
triple Oxy = (0,0,0), A=(4*5^.5,-8,4), B=(0,-8,h), C=(Cxy.x,Cxy.y,0), D=(A.x,A.y,0), E=(B.x,B.y,0), O=(O.x,O.y,h);
−
<asy>defaultpen(fontsize(10)+linewidth(0.62));
+
pair L = 8*expi(pi+0.05), R = 8*expi(-0.22); /* left and right cylinder lines, numbers from trial/error */
−
pair A=(4*sqrt(5),8), B=(0,8), O=(0,0);
+
  /* Drawing */
 +
draw(B--A--D--E--B--C);
 +
draw(circle(Oxy,8));
 +
draw(circle(O,8));
 +
draw((L.x,L.y,0)--(L.x,L.y,h)); draw((R.x,R.y,0)--(R.x,R.y,h));
 +
draw(O--B--(A.x,A.y,h)--cycle,dashed);
 +
  /* Labeling */
 +
label("\(A\)",A,NE);  dot(A);
 +
label("\(B\)",B,NW);  dot(B);
 +
label("\(C\)",C,W);  dot(C);
 +
label("\(D\)",D,E);  dot(D);
 +
label("\(E\)",E,S);  dot(E);
 +
label("\(O\)",O,NW);  dot(O);
 +
</asy>&nbsp;&nbsp;&nbsp;<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(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("$8$",A/3,(0.5,-1));label("$4$",5*A/6,(0.5,-1));
+
label("$8$",A/3,(1,0.5));label("$4$",5*A/6,(1,0.5));
−
label("$8$",B/2,(-1,0));label("$4\sqrt{5}$",B/2+A/2,(0,1));
+
label("$8$",B/2,(-1,0));label("$4\sqrt{5}$",B/2+A/2,(0,-1));
 
</asy></center>
 
</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(10)+linewidth(0.6));
+
defaultpen(fontsize(10)+linewidth(0.62));
 
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);
Line 40: Line 54:
 
== 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}}
 +
 +
[[Category:Intermediate Geometry Problems]]

Revision as of 16:19, 28 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] /* Settings */ import three; defaultpen(fontsize(10)+linewidth(0.62)); currentprojection = perspective(-2,-50,15); size(200); /* Variables */ real x = 20 - ((750)^.5)/3, CE = 8*(6^.5) - 4*(5^.5), CD = 8*(6^.5), h = 4*CE/CD; pair Cxy = 8*expi((3*pi)/2-CE/8); triple Oxy = (0,0,0), A=(4*5^.5,-8,4), B=(0,-8,h), C=(Cxy.x,Cxy.y,0), D=(A.x,A.y,0), E=(B.x,B.y,0), O=(O.x,O.y,h); pair L = 8*expi(pi+0.05), R = 8*expi(-0.22); /* left and right cylinder lines, numbers from trial/error */ /* Drawing */ draw(B--A--D--E--B--C); draw(circle(Oxy,8)); draw(circle(O,8)); draw((L.x,L.y,0)--(L.x,L.y,h)); draw((R.x,R.y,0)--(R.x,R.y,h)); draw(O--B--(A.x,A.y,h)--cycle,dashed); /* Labeling */ label("\(A\)",A,NE); dot(A); label("\(B\)",B,NW); dot(B); label("\(C\)",C,W); dot(C); label("\(D\)",D,E); dot(D); label("\(E\)",E,S); dot(E); label("\(O\)",O,NW); dot(O); [/asy]   [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,(1,0.5));label("$4$",5*A/6,(1,0.5)); 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(0.62)); 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
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2004_AIME_I_Problems/Problem_14&oldid=25715"