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
.
Therefore .
NOTE: Just because this problem is #14 doesn't mean it's too hard. Attempt it, and you will get an elegant solution!
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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