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Difference between revisions of "Mock AIME 2 2006-2007 Problems/Problem 7"

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== Problem ==
 
== Problem ==
  
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A right circular cone of base radius <math>\displaystyle 17</math>cm and slant height <math>\displaystyle 34</math>cm is given. <math>\displaystyle P</math> is a point on the circumference of the base and the shortest path from <math>\displaystyle P</math> around the cone and back is drawn (see diagram). If the length of this path is <math>\displaystyle m\sqrt{n},</math> where <math>\displaystyle m</math> and <math>\displaystyle n</math> are relatively prime positive integers, find <math>\displaystyle m+n.</math>
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A right circular cone of base radius <math>17</math>cm and slant height <math>51</math>cm is given. <math>P</math> is a point on the circumference of the base and the shortest path from <math>P</math> around the cone and back is drawn (see diagram). If the length of this path is <math>m\sqrt{n},</math> where <math>n</math> is squarefree, find <math>m+n.</math>
  
 
[[Image:Mock_AIME_2_2007_Problem8.jpg]]
 
[[Image:Mock_AIME_2_2007_Problem8.jpg]]
  
 
==Solution==
 
==Solution==
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{{solution}}
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"Unfolding" this cone results in a circular sector with radius <math>51</math> and arc length <math>17\cdot 2\pi=34\pi</math>. Let the vertex of this sector be <math>O</math>. The problem is then reduced to finding the shortest distance between the two points <math>A</math> and <math>B</math> on the arc that are the farthest away from each other. Since <math>34\pi</math> is <math>1/3</math> of the circumference of a circle with radius <math>51</math>, we must have that <math>\angle AOB=\frac{360^{\circ}}{3}=120^{\circ}</math>. We know that <math>AO=OB=51</math>, so we can use the Law of Cosines to find the length of <math>AB</math>:
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<cmath>AB=\sqrt{AO^2+OB^2-2AO\cdot OB\cdot\cos{120^{\circ}}}=\sqrt{51^2+51^2+51^2}=51\sqrt{3}.</cmath>
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Hence <math>m=51</math>, <math>n=3</math>, <math>m+n=\boxed{054}</math>.
  
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==See Also==
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{{Mock AIME box|year=2006-2007|n=2|num-b=6|num-a=8}}
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*[[Mock AIME 2 2006-2007/Problem 6 | Previous Problem]]
 
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*[[Mock AIME 2 2006-2007/Problem 8 | Next Problem]]
 
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*[[Mock AIME 2 2006-2007]]
 
  
 
[[Category:Intermediate Geometry Problems]]
 
[[Category:Intermediate Geometry Problems]]

Latest revision as of 17:10, 10 July 2014

Problem

A right circular cone of base radius $17$cm and slant height $51$cm is given. $P$ is a point on the circumference of the base and the shortest path from $P$ around the cone and back is drawn (see diagram). If the length of this path is $m\sqrt{n},$ where $n$ is squarefree, find $m+n.$

Mock AIME 2 2007 Problem8.jpg

Solution

"Unfolding" this cone results in a circular sector with radius $51$ and arc length $17\cdot 2\pi=34\pi$. Let the vertex of this sector be $O$. The problem is then reduced to finding the shortest distance between the two points $A$ and $B$ on the arc that are the farthest away from each other. Since $34\pi$ is $1/3$ of the circumference of a circle with radius $51$, we must have that $\angle AOB=\frac{360^{\circ}}{3}=120^{\circ}$. We know that $AO=OB=51$, so we can use the Law of Cosines to find the length of $AB$: \[AB=\sqrt{AO^2+OB^2-2AO\cdot OB\cdot\cos{120^{\circ}}}=\sqrt{51^2+51^2+51^2}=51\sqrt{3}.\] Hence $m=51$, $n=3$, $m+n=\boxed{054}$.

See Also

Mock AIME 2 2006-2007 (Problems, Source)
Preceded by
Problem 6 		Followed by
Problem 8
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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