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); | ||
− | + | ||
− | + | pair O1 = (25,4), A1=O1+(4*sqrt(5),-8), B1=O1+(0,-8); | |
− | draw(circle( | + | draw(circle(O1,8)); |
− | draw( | + | draw(O1--A1--B1--O1); |
− | label("\(A\)", | + | label("\(A\)",A1,SE);label("\(B\)",B1,SW);label("\(O\)",O1,N); |
− | label("$8$", | + | label("$8$",O1--B1,W);label("$4$",(5*A1+O1)/6,0.8*(1,1)); |
− | label("$8$", | + | 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>. | ||
− | + | == Solution 2 == | |
+ | |||
+ | 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> | ||
+ | ~ 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 01:00, 27 November 2022
Contents
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); 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]](http://latex.artofproblemsolving.com/3/6/0/360facfdf852b7ec7101d59f66672bd121c68849.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 .
Solution 2
Note that by Power of a Point, the point the unicorn is at has power which implies that the tangent from that point to the tower is of length
however this is length of the rope projected into 2-D. If we let
be the angle between the horizontal and the rope, we have that
which implies that
Note that the portion of rope not on the tower is
the requested length of rope is
thus the requested sum is
~ 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.