Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(asymptote diagrams)
(→‎Solution)
 
(7 intermediate revisions by 3 users not shown)
Line 3: Line 3:
  
 
== 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 given information, <math>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>800</math>. 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 <math>800</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>.
−
<center>
 
−
<asy> size(200);
 
−
import three; pointpen = black; pathpen = black + linewidth(0.7); currentprojection = perspective(0,-20,5);
 
  
−
triple O=(0,0,0), P=(0,0,2*7^.5), Q=(-6,0,0), R=(6,0,0), A=(5*Q + 27*P)/32, B=(15*2^.5 * R + (32-15*2^.5)*P)/32;
+
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>.
−
D(circle(O,6)); D(Q--P--R--O); D(arc((A+B)/2,B,A,(0.52,0,1),CW),linewidth(0.7)+linetype("4 4")); dot(A); dot(B); dot(O); MP("600",(3,0,0),S); MP("125",(A+P)/2,NW); MP("375\sqrt{2}",(B+P)/2,NE); MP("A",A,W); MP("B",B,E);
 
−
 
 
−
</asy>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<asy>
 
−
pointpen = black; pathpen = black + linewidth(0.7);
 
−
 
 
−
pair O=(0,0), A=(-125,0), B=375*2^.5*expi(pi/4);
 
−
D(arc(O,800,180,-90)); D(D(MP("A",A))--D(MP("B",B))--O); D((-800,0)--D(MP("O",O,SE))--(0,-800)); MP("800",(0,-400),W);
 
−
 
 
−
</asy></center>
 
−
 
 
−
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>270^\circ</math>. 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, <math>B</math> will lie exactly halfway between. Thus, the radius through <math>B</math> will divide the circle into two sectors, each with measure <math>135^\circ</math>. Draw in <math>BA</math> to create <math> \triangle{ABO}</math>. Now, by the [[Law of Cosines]], <math>AB^{2}=(125)^{2}+(375\sqrt{2})^{2}-2(125\cdot375\sqrt{2})(\cos {135})</math>. From there we have <math>AB=\sqrt{ (125)^{2}+(375\sqrt{2})^{2}-2(125\cdot375\sqrt{2})(\cos {135})}=\boxed{625}</math>.
 
  
 
== See also ==
 
== See also ==
Line 25: Line 11:
  
 
[[Category:Intermediate Geometry Problems]]
 
[[Category:Intermediate Geometry Problems]]
 +
{{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
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2004_AIME_II_Problems/Problem_11&oldid=191235"