Difference between revisions of "2004 AIME II Problems/Problem 11"
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== Problem == | == Problem == | ||
| − | A right circular cone has a base with radius 600 and height <math> 200\sqrt{7}. </math> 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 <math> 375\sqrt{2}. </math> Find the least distance that the fly could have crawled. | + | A right circular cone has a base with radius <math>600</math> and height <math> 200\sqrt{7}. </math> A fly starts at a point on the surface of the cone whose distance from the vertex of the cone is <math>125</math>, 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 <math> 375\sqrt{2}. </math> Find the least distance that the fly could have crawled. |
== 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 | + | Label the starting point of the fly as <math>\displaystyle A</math> and the ending as <math>\displaystyle B </math> and the vertex of the cone as <math> \displaystyle O</math>. With the given information, <math>\displaystyle OA=125</math> and <math>OB=375\sqrt{2}</math>. By the [[Pythagorean Theorem]], the slant height can be calculated by: <math>200\sqrt{7}^{2} + 600^2=640000 </math>, so the slant height of the cone is <math>\displaystyle 800</math>. The base of the cone has a circumference of <math>\displaystyle 1200\pi</math>, so if we cut the cone along its slant height and through <math>\displaystyle A</math>, we get a sector of a circle <math>\displaystyle O</math> with radius <math>\displaystyle 800</math>. Now the sector is <math>\frac{1200\pi}{1600\pi}=\frac{3}{4}</math> of the entire circle. So the degree measure of the sector is <math>\displaystyle 270^\circ</math>. Now we know that <math>\displaystyle A</math> and <math>\displaystyle B</math> are on opposite sides. Therefore, since <math>\displaystyle A</math> lies on a radius of the circle that is the "side" of a 270 degree sector, <math>\displaystyle B</math> will lie exactly halfway between. Thus, the radius through <math>\displaystyle B</math> will divide the circle into two sectors, each with measure <math>\displaystyle 135^\circ</math>. Draw in <math>\displaystyle BA</math> to create <math> \triangle{ABO}</math>. Now by the [[Law of Cosines]], <math>\displaystyle AB^{2}=(125)^{2}+(375\sqrt{2})^{2}-2(125\cdot375\sqrt{2})(cos 135)</math>. From there we have <math>\displaystyle 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]] | ||
Revision as of 11:20, 18 July 2007
Problem
A right circular cone has a base with radius 
and height 
A fly starts at a point on the surface of the cone whose distance from the vertex of the cone is 
, 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 given information, 
and 
. By the Pythagorean Theorem, the slant height can be calculated by: 
, so the slant height of the cone is 
. 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 . Now the sector is 
of the entire circle. So the degree measure of the sector is 
. 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, will lie exactly halfway between. Thus, the radius through will divide the circle into two sectors, each with measure 
. Draw in 
to create 
. Now by the Law of Cosines, 
. From there we have 
