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

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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);
 
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);
  
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</asy>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<asy>
 
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pointpen = black; pathpen = black + linewidth(0.7);
 
  
 
pair O=(0,0), A=(-125,0), B=375*2^.5*expi(pi/4);  
 
pair O=(0,0), A=(-125,0), B=375*2^.5*expi(pi/4);  

Revision as of 11:08, 13 September 2015

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 $A$ and the ending as $B$ and the vertex of the cone as $O$. With the given information, $OA=125$ 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 $800$. The base of the cone has a circumference of $1200\pi$, so if we cut the cone along its slant height and through $A$, we get a sector of a circle $O$ with radius $800$.

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; 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);


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>

Now the sector is $\frac{1200\pi}{1600\pi}=\frac{3}{4}$ of the entire circle. So the degree measure of the sector is $270^\circ$. Now we know that $A$ and $B$ are on opposite sides. Therefore, since $A$ lies on a radius of the circle that is the "side" of a 270 degree sector, $B$ will lie exactly halfway between. Thus, the radius through $B$ will divide the circle into two sectors, each with measure $135^\circ$. Draw in $BA$ to create $\triangle{ABO}$. Now, by the Law of Cosines, $AB^{2}=(125)^{2}+(375\sqrt{2})^{2}-2(125\cdot375\sqrt{2})(\cos {135})$. From there we have $AB=\sqrt{ (125)^{2}+(375\sqrt{2})^{2}-2(125\cdot375\sqrt{2})(\cos {135})}=\boxed{625}$.

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

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