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Difference between revisions of "2004 AIME II Problems/Problem 11"
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== Solution ==
 
== Solution ==
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{{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>}}
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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]]
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Revision as of 08:50, 18 July 2007

Problem

A right circular cone has a base with radius 600 and height $200\sqrt{7}.$ 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 $375\sqrt{2}.$ Find the least distance that the fly could have crawled.

Solution

Label the starting point of the fly as $A$ and the ending as $B$ and the vertex of the cone as $O$.With the give info $OA=125$ and $OB=375\sqrt{2}$ a By Pythagoras the slant height can be calculated by: $200\sqrt{7}^{2} + 600^2=640000$ so the slant height of the cone is 800. The base of the cone has a circumference of $1200\pi$So if we cut the cone along its slant height and through $A$ we get a sector of a circle $O$ with radius 800. Now the sector is $\frac{1200\pi}{1600\pi}=\frac{3}{4}$. So the sector is 270 degrees. Now we know that $A$ and $B$ are on opposite sides therefore since $A$ 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 $BA$ to create $\triangle{ABO}$. Now by Law of Cosines $AB^{2}=(125)^{2}+(375\sqrt{2})^{2}-2(125\cdot375\sqrt{2})(cos 135)$ from there $AB=\sqrt{ (125)^{2}+(375\sqrt{2})^{2}-2(125\cdot375\sqrt{2})(cos 135)}=625$

See also

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