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Difference between revisions of "2005 AIME I Problems/Problem 1"

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== Problem ==
 
== Problem ==
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Six [[circle]]s form a ring with with each circle [[externally tangent]] to two circles adjacent to it. All circles are [[internally tangent]] to a circle <math> C </math> with [[radius]] 30. Let <math> K </math> be the area of the region inside circle <math> C </math> and outside of the six circles in the ring. Find <math> \lfloor K \rfloor</math> (the [[floor function]]).
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Six [[congruent]] [[circle]]s form a ring with each circle [[externally tangent]] to two circles adjacent to it. All circles are [[internally tangent]] to a circle <math> C </math> with [[radius]] 30. Let <math> K </math> be the area of the region inside circle <math> C </math> and outside of the six circles in the ring. Find <math> \lfloor K \rfloor</math> (the [[floor function]]).
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[[Image:2005 AIME I Problem 1.png]]
  
 
== Solution ==
 
== Solution ==
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{{image}}
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Define the radii of the six congruent circles as <math>r</math>. If we draw all of the radii to the points of external tangency, we get a [[regular polygon|regular]] [[hexagon]]. If we connect the [[vertex|vertices]] of the hexagon to the [[center]] of the circle <math>C</math>, we form several [[equilateral triangle]]s. The length of each side of the triangle is <math>2r</math>. Notice that the radius of circle <math>C</math> is equal to the length of the side of the triangle plus <math>r</math>. Thus, the radius of <math>C</math> has a length of <math>3r = 30</math>, and so <math>r = 10</math>. <math>K = 30^2\pi - 6(10^2\pi) = 300\pi</math>, so <math>\lfloor 300\pi \rfloor = \boxed{942}</math>.
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Define the radii of the six congruent circles as <math>r</math>. If we draw all of the radii to the points of external tangency, we get a [[regular polygon|regular]] [[hexagon]]. If we connect the [[vertex|vertices]] of the hexagon to the [[center]] of the circle <math>C</math>, we form several [[equilateral triangle]]s. The length of each side of the triangle is <math>2r</math>. Notice that the radius of circle <math>C</math> is equal to the length of the side of the triangle plus <math>r</math>. Thus, the radius of <math>C</math> has a length of <math>3r = 30</math>, and so <math>r = 10</math>. <math>K = 30^2\pi - 6(10^2\pi) = 300\pi</math>, so <math>\lfloor 300\pi \rfloor = 942</math>.
 
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2005|n=I|before=First Question|num-a=2}}
 
{{AIME box|year=2005|n=I|before=First Question|num-a=2}}
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[[Category:Intermediate Geometry Problems]]
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{{MAA Notice}}

Latest revision as of 12:17, 22 July 2017

Problem

Six congruent circles form a ring with each circle externally tangent to two circles adjacent to it. All circles are internally tangent to a circle $C$ with radius 30. Let $K$ be the area of the region inside circle $C$ and outside of the six circles in the ring. Find $\lfloor K \rfloor$ (the floor function).

2005 AIME I Problem 1.png

Solution

Define the radii of the six congruent circles as $r$. If we draw all of the radii to the points of external tangency, we get a regular hexagon. If we connect the vertices of the hexagon to the center of the circle $C$, we form several equilateral triangles. The length of each side of the triangle is $2r$. Notice that the radius of circle $C$ is equal to the length of the side of the triangle plus $r$. Thus, the radius of $C$ has a length of $3r = 30$, and so $r = 10$. $K = 30^2\pi - 6(10^2\pi) = 300\pi$, so $\lfloor 300\pi \rfloor = \boxed{942}$.

See also

2005 AIME I (Problems • Answer Key • Resources)
Preceded by
First Question 		Followed by
Problem 2
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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