Difference between revisions of "Mock AIME 2 2006-2007 Problems/Problem 7"
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==Solution== | ==Solution== | ||
− | {{ | + | "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>: |
+ | <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> | ||
+ | Hence <math>m=51</math>, <math>n=3</math>, <math>m+n=\boxed{054}</math>. | ||
==See Also== | ==See Also== |
Latest revision as of 18:10, 10 July 2014
Problem
A right circular cone of base radius cm and slant height cm is given. is a point on the circumference of the base and the shortest path from around the cone and back is drawn (see diagram). If the length of this path is where is squarefree, find
Solution
"Unfolding" this cone results in a circular sector with radius and arc length . Let the vertex of this sector be . The problem is then reduced to finding the shortest distance between the two points and on the arc that are the farthest away from each other. Since is of the circumference of a circle with radius , we must have that . We know that , so we can use the Law of Cosines to find the length of : Hence , , .
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 |