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

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label("\(E\)",E,S);  dot(E);
 
label("\(E\)",E,S);  dot(E);
 
label("\(O\)",O,NW);  dot(O);
 
label("\(O\)",O,NW);  dot(O);
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</asy>&nbsp;&nbsp;&nbsp;<asy>defaultpen(fontsize(10)+linewidth(0.62));
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pair A=(4*sqrt(5),-8), B=(0,-8), O=(0,0);
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pair O1 = (25,4), A1=O1+(4*sqrt(5),-8), B1=O1+(0,-8);
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draw(circle((0,0),8));
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draw(circle(O1,8));
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draw(O--A--B--O);
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draw(O1--A1--B1--O1);
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label("\(A\)",A,(1,1));label("\(B\)",B,(-1,1));label("\(O\)",O,(-1,-1));
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label("\(A\)",A1,SE);label("\(B\)",B1,SW);label("\(O\)",O1,N);
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label("$8$",A/3,(1,0.5));label("$4$",5*A/6,(1,0.5));
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label("$8$",O1--B1,W);label("$4$",(5*A1+O1)/6,0.8*(1,1));
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label("$8$",B/2,(-1,0));label("$4\sqrt{5}$",B/2+A/2,(0,-1));
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label("$8$",O1--A1,0.8*(-0.5,2));label("$4\sqrt{5}$",B1/2+A1/2,(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>.   
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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>.
  
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NOTE: Just because this problem is #14 doesn't mean it's too hard. Attempt it, and you will get an elegant solution!
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== Solution 2 ==
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Note that by Power of a Point, the point the unicorn is at has power <math>4 \cdot 20 = 80</math> which implies that the tangent from that point to the tower is of length <math>\sqrt{80}=4\sqrt{5},</math> however this is length of the rope projected into 2-D. If we let <math>\theta</math> be the angle between the horizontal and the rope, we have that <math>\cos\theta=\frac{1}{5}</math> which implies that <math>\sin\theta=\frac{2\sqrt{6}}{5}.</math> Note that the portion of rope not on the tower is <math>4\sqrt{5} \cdot \frac{5}{2\sqrt{6}}= \frac{5\sqrt{30}}{3},</math> the requested length of rope is <math>20-\frac{5\sqrt{30}}{3}=\frac{60-\sqrt{750}}{3}</math> thus the requested sum is <math>\boxed{813}.</math>
  
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~ Dhillonr25
 
== 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}}

Latest revision as of 00:00, 27 November 2022

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); pair O1 = (25,4), A1=O1+(4*sqrt(5),-8), B1=O1+(0,-8); draw(circle(O1,8)); draw(O1--A1--B1--O1); label("\(A\)",A1,SE);label("\(B\)",B1,SW);label("\(O\)",O1,N); label("$8$",O1--B1,W);label("$4$",(5*A1+O1)/6,0.8*(1,1)); label("$8$",O1--A1,0.8*(-0.5,2));label("$4\sqrt{5}$",B1/2+A1/2,(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 $a=60, b=750, c=3, a+b+c=\boxed{813}$.

Solution 2

Note that by Power of a Point, the point the unicorn is at has power $4 \cdot 20 = 80$ which implies that the tangent from that point to the tower is of length $\sqrt{80}=4\sqrt{5},$ however this is length of the rope projected into 2-D. If we let $\theta$ be the angle between the horizontal and the rope, we have that $\cos\theta=\frac{1}{5}$ which implies that $\sin\theta=\frac{2\sqrt{6}}{5}.$ Note that the portion of rope not on the tower is $4\sqrt{5} \cdot \frac{5}{2\sqrt{6}}= \frac{5\sqrt{30}}{3},$ the requested length of rope is $20-\frac{5\sqrt{30}}{3}=\frac{60-\sqrt{750}}{3}$ thus the requested sum is $\boxed{813}.$

~ Dhillonr25

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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