2004 AIME II Problems/Problem 11: Difference between revisions
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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. | ||
== Solution == | == 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}=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>. | ||
== Solution 2== | |||
[[File:2004_AIME_II_Problem_11_Diagram_1.png|200px|thumb|center]] | |||
To find the shortest length from the red to blue points, the net of the side of the cone could be drawn. | |||
[[File:2004_AIME_II_Problem_11_Diagram_2.png|200px|thumb|center]] | |||
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. | |||
<cmath> | |||
AB = \sqrt{(375\sqrt{2})^2 + 125^2 - 2 \cdot (375\sqrt{2})(125)(\cos 135^\circ)} = \boxed{625} | |||
</cmath> | |||
~Diagram and Solution by MaPhyCom | |||
== See also == | == See also == | ||
Latest revision as of 03:53, 25 June 2025
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 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 
-axis and the angle 
going counterclockwise. The circumference of the base is 
. The sector's radius (cone's sweep) is 
. Setting 
.
If the starting point 
is on the positive -axis at 
then we can take the end point 
on 's bisector at 
radians along the 
line in the second quadrant. Using the distance from the vertex puts at 
. Thus the shortest distance for the fly to travel is along segment 
in the sector, which gives a distance 
.
Solution 2
To find the shortest length from the red to blue points, the net of the side of the cone could be drawn.
The angle 
is equal to 
, or 
. Therefore, the law of cosines could be utilized.
![\[AB = \sqrt{(375\sqrt{2})^2 + 125^2 - 2 \cdot (375\sqrt{2})(125)(\cos 135^\circ)} = \boxed{625}\]](http://latex.artofproblemsolving.com/0/7/8/0781c1bc8c553c441595491e9f8901c8fdd15a15.png)
~Diagram 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 | ||
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