Ratio of Circles

Geometry Level 4

Let a > 0 a > 0 .

In the above diagram the larger circle centered at O O has radius R R and the smaller circle centered at O ′ O' , which is tangent to the larger circle at Q Q , has radius r r and P S = 2 a PS = 2a and T W = 3 a + 3 TW = 3a + 3 .

If R + r = 91 R + r = 91 and there are two values of a a which correspond to two areas between the two circles say A 1 A_{1} and A 2 A_{2} , where A 2 > A 1 A_{2} > A_{1} , and A 2 A 1 = α β \dfrac{A_{2}}{A_{1}} = \dfrac{\alpha}{\beta} , where gcd ⁡ ( α , β ) = 1 \gcd(\alpha,\beta) = 1 , find α + β \alpha + \beta .


The answer is 101.

Rocco Dalto by Rocco Dalto , 67, USA

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1 solution

Rocco Dalto
Jan 2, 2020

Using the diagram above we have:

2 R = 2 r + 3 a + 3 ⟹ R = r + 3 a + 3 2 ⟹ R = r + 3 a + 3 2 = O P + 2 a ⟹ O P = r − ( a − 3 2 ) 2R = 2r + 3a + 3 \implies R = r + \dfrac{3a + 3}{2} \implies R = r + \dfrac{3a + 3}{2} = OP + 2a \implies OP = r - (\dfrac{a - 3}{2}) and O ′ O = 3 ( a + 1 ) 2 O'O = \dfrac{3(a + 1)}{2} .

∴ \therefore In right △ P O O ′ \triangle{POO'} we have r 2 − ( a − 3 ) r + ( a − 3 ) 2 4 + 9 ( a + 1 ) 2 4 = r 2 r^2 - (a - 3)r + \dfrac{(a - 3)^2}{4} + \dfrac{9(a + 1)^2}{4} = r^2

⟹ r = 1 4 ( 9 ( a + 1 ) 2 a − 3 + a − 3 ) = \implies r = \dfrac{1}{4}(\dfrac{9(a + 1)^2}{a - 3} + a - 3) = 5 a 2 + 6 a + 9 2 ( a − 3 ) \dfrac{5a^2 + 6a + 9}{2(a - 3)}

⟹ R = 5 a 2 + 6 a + 9 2 ( a − 3 ) + 3 a + 3 2 = 8 a 2 2 ( a − 3 ) \implies R = \dfrac{5a^2 + 6a + 9}{2(a - 3)} + \dfrac{3a + 3}{2} = \dfrac{8a^2}{2(a - 3)}

⟹ R + r = 8 a 2 2 ( a − 3 ) + 5 a 2 + 6 a + 9 2 ( a − 3 ) = 91 \implies R + r = \dfrac{8a^2}{2(a - 3)} + \dfrac{5a^2 + 6a + 9}{2(a - 3)} = 91

⟹ 13 a 2 + 6 a + 9 = 182 a − 546 ⟹ 13 a 2 − 176 a + 155 = 0 ⟹ \implies 13a^2 + 6a + 9 = 182a - 546 \implies 13a^2 - 176a + 155 = 0 \implies

( a − 5 ) ( 13 a − 111 ) = 0 ⟹ (a - 5)(13a - 111) = 0 \implies

a 1 = 5 , a 2 = 111 13 \boxed{a_{1} = 5, a_{2} = \dfrac{111}{13}}

a 1 = 5 ⟹ R = 200 4 = 50 a_{1} = 5 \implies R = \dfrac{200}{4} = 50 and r = 164 4 = 41 ⟹ r = \dfrac{164}{4} = 41 \implies A 1 = π ( 5 0 2 − 4 1 2 ) = 819 π \boxed{A_{1} = \pi(50^2 - 41^2) = 819\pi}

and

a 2 = 111 13 ⟹ R = 98568 1872 = 12321 234 a_{2} = \dfrac{111}{13} \implies R = \dfrac{98568}{1872} = \dfrac{12321}{234} and r = 71784 1872 = 8973 234 r = \dfrac{71784}{1872} = \dfrac{8973}{234}

⟹ A 2 = π ( 1232 1 2 − 897 3 2 23 4 2 ) = 71292312 54756 = 1302 π \implies \boxed{A_{2} = \pi(\dfrac{12321^2 - 8973^2}{234^2}) = \dfrac{71292312}{54756} = 1302\pi}

⟹ A 2 A 1 = 62 ∗ 21 π 39 ∗ 21 π = 62 39 = α β ⟹ \implies \dfrac{A_{2}}{A_{1}} = \dfrac{62 * 21\pi}{39 * 21\pi} = \dfrac{62}{39} = \dfrac{\alpha}{\beta} \implies α + β = 101 \alpha + \beta = \boxed{101} .

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Just a minor simplification, due to the form of the answer you're looking for: A i = π ( R i 2 − r i 2 ) = π ( R i + r i ) ( R i − r i ) = 3 × 91 2 π ( a i + 1 ) A_i=\pi \left(R_i^2-r_i^2 \right) = \pi (R_i+r_i)(R_i-r_i) = \frac{3\times 91}{2} \pi (a_i+1) , so that the ratio we need is just A 1 A 2 = a 1 + 1 a 2 + 1 \frac{A_1}{A_2}=\frac{a_1+1}{a_2+1} .

Chris Lewis - 1 year, 5 months ago

That's Very Nice. Thanks!

You meant A 1 A 2 = a 1 + 1 a 2 + 1 \dfrac{A_{1}}{A_{2}} = \dfrac{a_{1} + 1}{a_{2} + \boxed{1}}

Rocco Dalto - 1 year, 5 months ago

Log in to reply

I did! Thanks, corrected now.

Chris Lewis - 1 year, 5 months ago

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