2004 AIME I Problems/Problem 14
Problem
A unicorn is tethered by a 
-foot silver rope to the base of a magician's cylindrical tower whose radius is 
feet. The rope is attached to the tower at ground level and to the unicorn at a height of 
feet. The unicorn has pulled the rope taut, the end of the rope is 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
![[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]](http://latex.artofproblemsolving.com/e/7/a/e7a34c3d4ddfe19837ca395ee9dcd4db7baccec9.png)
![[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]](http://latex.artofproblemsolving.com/7/8/f/78f5d0489f526f664af2aad1a7e6378370937e7d.png)
Looking from an overhead view, call the center of the circle 
, the tether point to the unicorn 
and the last point where the rope touches the tower 
. 
is a right triangle because 
is a radius and 
is a tangent line at point . We use the Pythagorean Theorem to find the horizontal component of 
has length 
.
![[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]](http://latex.artofproblemsolving.com/9/3/0/930d93ee9d48b6fc6a0869685c1358eccd0d16ac.png)
Now look at a side view and "unroll" the cylinder to be a flat surface. Let 
be the bottom tether of the rope, let 
be the point on the ground below , and let 
be the point directly below . Triangles 
and 
are similar right triangles. By the Pythagorean Theorem 
.
Let 
be the length of 
. ![\[\frac{CA}{CD}=\frac{CB}{CE}\implies \frac{20}{8\sqrt{6}}=\frac{x}{8\sqrt{6}-4\sqrt{5}}\implies x=\frac{60-\sqrt{750}}{3}\]](http://latex.artofproblemsolving.com/b/7/b/b7bcbd740a90d19e6b4fafdf46b8f734b3864365.png)
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 | ||
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