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Difference between revisions of "2004 AIME II Problems/Problem 11"

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== Problem ==
 
== Problem ==
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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.
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A [[right cone|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 ==
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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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The easiest way is to unwrap the cone into a circular sector. Center the sector at the origin with one radius on the positive <math>x</math>-axis and the angle <math>\theta</math> going counterclockwise. The circumference of the base is <math>C=1200\pi</math>. The sector's radius (cone's sweep) is <math>R=\sqrt{r^2+h^2}=\sqrt{600^2+(200\sqrt{7})^2}=\sqrt{360000+280000}=\sqrt{640000}=800</math>. Setting <math>\theta R=C\implies 800\theta=1200\pi\implies\theta=\frac{3\pi}{2}</math>.
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If the starting point <math>A</math> is on the positive <math>x</math>-axis at <math>(125,0)</math> then we can take the end point <math>B</math> on <math>\theta</math>'s bisector at <math>\frac{3\pi}{4}</math> radians along the <math>y=-x</math> line in the second quadrant. Using the distance from the vertex puts <math>B</math> at <math>(-375,-375)</math>. Thus the shortest distance for the fly to travel is along segment <math>AB</math> in the sector, which gives a distance <math>\sqrt{(-375-125)^2+(-375-0)^2}=125\sqrt{4^2+3^2}=\boxed{625}</math>.
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== See also ==
 
== See also ==
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* [[2004 AIME II Problems/Problem 10| Previous problem]]
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{{AIME box|year=2004|n=II|num-b=10|num-a=12}}
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* [[2004 AIME II Problems/Problem 12| Next problem]]
 
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* [[2004 AIME II Problems]]
 
  
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{{wikify}}
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[[Category:Intermediate Geometry Problems]]
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{{MAA Notice}}

Latest revision as of 21:20, 23 March 2023

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

The easiest way is to unwrap the cone into a circular sector. Center the sector at the origin with one radius on the positive $x$-axis and the angle $\theta$ going counterclockwise. The circumference of the base is $C=1200\pi$. The sector's radius (cone's sweep) is $R=\sqrt{r^2+h^2}=\sqrt{600^2+(200\sqrt{7})^2}=\sqrt{360000+280000}=\sqrt{640000}=800$. Setting $\theta R=C\implies 800\theta=1200\pi\implies\theta=\frac{3\pi}{2}$.

If the starting point $A$ is on the positive $x$-axis at $(125,0)$ then we can take the end point $B$ on $\theta$'s bisector at $\frac{3\pi}{4}$ radians along the $y=-x$ line in the second quadrant. Using the distance from the vertex puts $B$ at $(-375,-375)$. Thus the shortest distance for the fly to travel is along segment $AB$ in the sector, which gives a distance $\sqrt{(-375-125)^2+(-375-0)^2}=125\sqrt{4^2+3^2}=\boxed{625}$.

See also

2004 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 10 		Followed by
Problem 12
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