Green Light, Red Light
In the large circle, three green circles, each sharing one point of tangency with the outer circumference, and one red circle are arranged symmetrically, such that the red circle is tangent to each of the three circles. Each of the two green circles shares one point of tangency with the red circle at the midpoint of its respective chord. These two chords shares an intersection point at the circumference.
If the radius ratio of the green circle to the red circle is
b a ( c + d e )
where a , b , c , d , e are positive integers, g cd ( a , b ) = g cd ( c , d ) = 1 and e is square free, input a + b + c + d + e as your answer.
The answer is 10.
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1 solution
How did you arrive directly at the ratio, I got confused at the second last step, after proving the similarity
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cross-multiply 2 R + 3 G G = R + 2 G R to get G ( R + 2 G ) = R ( 2 R + 3 G ) .
distribute to get G R + 2 G 2 = 2 R 2 + 3 G R
add/subtract to get 2 G 2 − 2 G R − 2 R 2 = 0
divide everything by 2 to get G 2 − G R − R 2 = 0
use the quadratic formula to get G = 2 1 + 5 ⋅ R
oh yes :p, i got confused since in the solution in the rhs, the comma makes the 2G seem like 2G'
I just set red radius as 1 and solved
Let R be the radius of the red circle and G be the radius of the green circles, and label the diagram as follows:
Then O E = O B = O A + A B = R + 2 G , and C E = C O + O E = G + R + R + 2 G = 2 R + 3 G .
Since △ C D E ∼ △ O A E by AA similarity, C E C D = O E O A , or 2 R + 3 G G = R + 2 G R , which rearranges to R G = 2 1 ( 1 + 1 5 ) = ϕ , the golden ratio.
Therefore, a = 1 , b = 2 , c = 1 , d = 1 , e = 5 , and a + b + c + d + e = 1 0 .