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

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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.
 
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.
  
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== Solution ==
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== Solution 1==
 
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>.
 
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>.
  
 
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>.
 
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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== Solution 2==
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Image here
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To find the shortest length from the red to blue points, the net of the side of the cone could be drawn.
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Image here2
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The angle <math>YVX</math> is equal to <math>360^\circ \cdot \frac{1200\pi}{1600\pi} \cdot \frac{1}{2}</math>, or <math>135^\circ</math>. Therefore, the law of cosines could be utilized.
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<cmath>
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AB = \sqrt{(375\sqrt{2})^2 + 125^2 - 2 \cdot (375\sqrt{2})(125)(\cos 135^\circ)} = \boxed{625}
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</cmath>
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~Image and Solution by MaPhyCom
  
 
== See also ==
 
== See also ==

Revision as of 04:45, 25 June 2025

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 1

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 $\sqrt{(-375-125)^2+(-375-0)^2}=125\sqrt{4^2+3^2}=\boxed{625}$.

Solution 2

Image here To find the shortest length from the red to blue points, the net of the side of the cone could be drawn. Image here2 The angle $YVX$ is equal to $360^\circ \cdot \frac{1200\pi}{1600\pi} \cdot \frac{1}{2}$, or $135^\circ$. Therefore, the law of cosines could be utilized. \[AB = \sqrt{(375\sqrt{2})^2 + 125^2 - 2 \cdot (375\sqrt{2})(125)(\cos 135^\circ)} = \boxed{625}\]

~Image and Solution by MaPhyCom

See also

2004 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 10 		Followed by
Problem 12
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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