Circles

Geometry Level 3

Two concentric circles have radii 1 1 and 4 4 . Six congruent circles form a ring where each of the six circles is tangent to the two circles adjacent to it as shown. The three lightly shaded circles are internally tangent to the circle with radius 4 4 while the three darkly shaded circles are externally tangent to the circle with radius 1 1 . The radius of the six congruent circles can be written k + m n \large\frac{k+\sqrt{m}}{ n} , where k , m , and n k, m, \text { and } n are integers with k k and n n relatively prime. Find k + m + n = ? k+m+n=?

206 304 126 200 128

Hana Wehbi by Hana Wehbi , 51, USA

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2 solutions

Hosam Hajjir
Oct 12, 2017

We draw a triangle (shown with different side colors) with its vertices at the centers of the two concentric circles and of a lightly shaded circle and of an adjacent darkly shaded circle. Let the radius of the six congruent circles be r r .

The length of the red side is 1 + r 1 + r . The length of the blue side is 4 − r 4 - r , and the length of the orange side is 2 r 2 r . In addition, we know that the angle between the red side and the blue side is 6 0 ∘ 60^{\circ} . Using the Law of Cosines, we can write,

( 2 r ) 2 = ( 4 − r ) 2 + ( 1 + r ) 2 − 2 ( 4 − r ) ( 1 + r ) cos ⁡ 6 0 ∘ (2 r)^2 = (4 - r)^2 + (1 + r)^2 - 2 (4-r)(1+r) \cos 60^{\circ}

so that,

4 r 2 = 16 − 8 r + r 2 + 1 + 2 r + r 2 − 4 − 3 r + r 2 4 r^2 = 16 - 8 r + r^2 + 1 + 2 r + r^2 - 4 - 3 r + r^2

simplifying,

r 2 + 9 r − 13 = 0 r^2 + 9 r - 13 = 0

Using the quadratic formula,

r = 1 2 ( − 9 + 133 ) r = \dfrac{1}{2} (-9 + \sqrt{133} )

Therefore, k = − 9 , m = 133 k = -9, m = 133 and n = 2 n = 2 , making the answer − 9 + 133 + 2 = 126 -9 + 133 + 2 = 126

4 Helpful 0 Interesting 1 Brilliant 0 Confused

Thank you.

Hana Wehbi - 3 years, 8 months ago
Munem Shahriar
Oct 13, 2017

Connect the centers of the white circle and two adjacent shaded circles to form a triangle. The radius of the congruent circles is r r .

Now,

( 2 r ) 2 = ( 1 + r ) 2 + ( 4 − r ) 2 − 2 ( 1 + r ) ( 4 − r ) cos ⁡ 6 0 ∘ (2r)^2=(1+r)^2+(4-r)^2-2(1+r)(4-r)\cos{60^\circ}

which solves to r = − 9 + 133 2 r=\dfrac{-9+\sqrt{133}}{2} .

Hence k + m + n = ( − 9 + 133 + 2 ) = 126 k+m+n= (-9+133+2) = \boxed{126}

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Thank you.

Hana Wehbi - 3 years, 8 months ago

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