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 "2005 AIME I Problems/Problem 1"
m (solution, box)
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
−
Six circles 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>
+
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]]).
  
 
== Solution ==
 
== Solution ==
 +
{{image}}
 +
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 ==
−
* [[2005 AIME I Problems]]
+
{{AIME box|year=2005|n=I|before=First Question|num-a=2}}

Revision as of 14:16, 4 March 2007

Problem

Six circles form a ring with 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).

Solution

An image is supposed to go here. You can help us out by creating one and editing it in. Thanks.

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 $\lfloor 300\pi \rfloor = 942$.

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
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2005_AIME_I_Problems/Problem_1&oldid=13623"