Question: Similar to a 96 National Mathcounts question: Circles of the same colors are congruent and tangent to each other. What is the ratio of the area of the largest circle to the area of the smallest circle?
Solution :
The most difficult part might be to find the leg that is "2R - r".
R and r being the radius of the median and the smallest circle.
Using Pythagorean theorem, you have \(\left( R+r\right) ^{2}= R^{2}+\left( 2R-r\right) ^{2}\)
\(R^{2}+2Rr+r^{2}=R^{2}+4R^{2}-4Rr+r^{2}\)
Consolidate and you have 6Rr = \(4R^{2}\)\(\rightarrow\) 3r = 2R \(\rightarrow\) r =\(\dfrac {2R} {3}\)
The ratio of the area of the largest circle to the area of the smallest circle is thus
\(\rightarrow\)\(\dfrac {\left( 2R\right) ^{2}} {\left( \dfrac {2R} {3}\right) ^{2}}\) = 9 .
Monday, December 31, 2012
Harder Geometry Question from National Mathcounts
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