Difference between revisions of "2005 PMWC Problems/Problem I4"

(Created page with "==Problem== The larger circle has radius 12 cm. Each of the six identical smaller circles touches its two neighbours and the larger circle. What is the radius of the smaller circ...")
 
(Problem)
 
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</asy>
 
</asy>
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==Solution==
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<asy>
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unitsize(0.5cm);
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draw((0,3)..(3,0)..(0,-3)..(-3,0)..cycle);
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for (int i=0;i<6;i=i+1){
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draw(dir(60*i)..3*dir(60*i)..cycle);
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}
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draw((0,1)..(0,-1)..cycle,rgb(1,0,0));
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</asy>
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Adding a circle in the middle makes the solution obvious. Since the radius of the big circle is <math>12</math> cm, then the diameter of the big circle is <math>24</math> cm. But this is also <math>3</math> of the little circle's diameter! Therefore, the diameter of the little circle is <math>24\div3=8</math> cm and the radius is therefore <math>\boxed{4 \text{cm}}</math>.
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== See also ==
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{{PMWC box|year=2005|num-b=I3|num-a=I5}}
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[[Category:Introductory Algebra Problems]]

Latest revision as of 18:15, 6 October 2013

Problem

The larger circle has radius 12 cm. Each of the six identical smaller circles touches its two neighbours and the larger circle. What is the radius of the smaller circle?

[asy] unitsize(0.5cm); draw((0,3)..(3,0)..(0,-3)..(-3,0)..cycle); for (int i=0;i<6;i=i+1){ draw(dir(60*i)..3*dir(60*i)..cycle); } [/asy]

Solution

[asy] unitsize(0.5cm); draw((0,3)..(3,0)..(0,-3)..(-3,0)..cycle); for (int i=0;i<6;i=i+1){ draw(dir(60*i)..3*dir(60*i)..cycle); } draw((0,1)..(0,-1)..cycle,rgb(1,0,0)); [/asy] Adding a circle in the middle makes the solution obvious. Since the radius of the big circle is $12$ cm, then the diameter of the big circle is $24$ cm. But this is also $3$ of the little circle's diameter! Therefore, the diameter of the little circle is $24\div3=8$ cm and the radius is therefore $\boxed{4 \text{cm}}$.

See also

2005 PMWC (Problems)
Preceded by
Problem I3
Followed by
Problem I5
I: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
T: 1 2 3 4 5 6 7 8 9 10