Difference between revisions of "2004 AIME II Problems/Problem 11"
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== Solution == | == Solution == | ||
| − | {{ | + | {{Label the starting point of the fly as <math>A</math> and the ending as <math>B </math> and the vertex of the cone as <math>O</math>.With the give info <math>OA=125</math> and <math>OB=375\sqrt{2}</math> a By Pythagoras the slant height can be calculated by: <math>200\sqrt{7}^{2} + 600^2=640000 </math> so the slant height of the cone is 800. The base of the cone has a circumference of <math>1200\pi</math>So if we cut the cone along its slant height and through <math>A</math> we get a sector of a circle <math>O</math> with radius 800. Now the sector is <math>\frac{1200\pi}{1600\pi}=\frac{3}{4}</math>. So the sector is 270 degrees. Now we know that <math>A</math> and <math>B</math> are on opposite sides therefore since <math>A</math> lies on a radius of the circle that is the "side" of a 270 degree sector B will lie exactly halfway between so the radius through B will divide the circle into two sectors each with measure 135. Draw in <math>BA</math> to create <math>\triangle{ABO}</math>. Now by Law of Cosines <math>AB^{2}=(125)^{2}+(375\sqrt{2})^{2}-2(125\cdot375\sqrt{2})(cos 135)</math> from there <math>AB=\sqrt{ (125)^{2}+(375\sqrt{2})^{2}-2(125\cdot375\sqrt{2})(cos 135)}=625</math>}} |
== See also == | == See also == | ||
* [[2004 AIME II Problems/Problem 10| Previous problem]] | * [[2004 AIME II Problems/Problem 10| Previous problem]] | ||
* [[2004 AIME II Problems/Problem 12| Next problem]] | * [[2004 AIME II Problems/Problem 12| Next problem]] | ||
* [[2004 AIME II Problems]] | * [[2004 AIME II Problems]] | ||
Revision as of 23:25, 17 July 2007
Problem
A right circular cone has a base with radius 600 and height 
A fly starts at a point on the surface of the cone whose distance from the vertex of the cone is 125, and crawls along the surface of the cone to a point on the exact opposite side of the cone whose distance from the vertex is 
Find the least distance that the fly could have crawled.
Solution
{{Label the starting point of the fly as 
and the ending as 
and the vertex of the cone as 
.With the give info 
and 
a By Pythagoras the slant height can be calculated by: 
so the slant height of the cone is 800. The base of the cone has a circumference of 
So if we cut the cone along its slant height and through we get a sector of a circle with radius 800. Now the sector is 
. So the sector is 270 degrees. Now we know that and are on opposite sides therefore since lies on a radius of the circle that is the "side" of a 270 degree sector B will lie exactly halfway between so the radius through B will divide the circle into two sectors each with measure 135. Draw in 
to create 
. Now by Law of Cosines 
from there 
}}
