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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>\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>
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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 17:44, 9 August 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 $\displaystyle A$ and the ending as $\displaystyle B$ and the vertex of the cone as $\displaystyle O$. With the given information, $\displaystyle OA=125$ and $OB=375\sqrt{2}$. By the Pythagorean Theorem, the slant height can be calculated by: $200\sqrt{7}^{2} + 600^2=640000$, so the slant height of the cone is $\displaystyle 800$. The base of the cone has a circumference of $\displaystyle 1200\pi$, so if we cut the cone along its slant height and through $\displaystyle A$, we get a sector of a circle $\displaystyle O$ with radius $\displaystyle 800$. Now the sector is $\frac{1200\pi}{1600\pi}=\frac{3}{4}$ of the entire circle. So the degree measure of the sector is $\displaystyle 270^\circ$. Now we know that $\displaystyle A$ and $\displaystyle B$ are on opposite sides. Therefore, since $\displaystyle A$ lies on a radius of the circle that is the "side" of a 270 degree sector, $\displaystyle B$ will lie exactly halfway between. Thus, the radius through $\displaystyle B$ will divide the circle into two sectors, each with measure $\displaystyle 135^\circ$. Draw in $\displaystyle BA$ to create $\triangle{ABO}$. Now by the Law of Cosines, $\displaystyle AB^{2}=(125)^{2}+(375\sqrt{2})^{2}-2(125\cdot375\sqrt{2})(\cos 135)$. From there we have $\displaystyle AB=\sqrt{ (125)^{2}+(375\sqrt{2})^{2}-2(125\cdot375\sqrt{2})(\cos 135)}=625$

See also

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