Sphere.....Sphere....

Geometry Level 4

Eight spheres of radius 100 are placed on a flat surface so that each sphere is tangent to two others and their centers are the vertices of a regular octagon. A ninth sphere is placed on the flat surface so that it is tangent to each of the other eight spheres. The radius of this last sphere is a+b\ sqrt{c} , where _a,b and c are positive integers, and c is not divisible by the square of any prime. Find a+b+c .


The answer is 152.

Abhishek Sharma by Abhishek Sharma , 20, India

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1 solution

Horizontal distance between the centers of and any other sphere and the 9th, is the radius of the circum circle of the octagonal with sides 200 = 100 ∗ 2 2 − 2 . R as the radius of 9th , the actual distance would be R + 100. Center of 9th is R - 100 above the plane of the octagon. These three sides form a right angle triangle with R +100 as hypotenuse. ⟹ ( R + 100 ) 2 = ( R − 100 ) 2 + ( 100 ∗ 2 2 − 2 . ) 2 . ⟹ 2 ∗ 100 ∗ R = 100 ∗ 2 2 − 2 . S o l v i n g f o r R , w e g e t R = 100 + 50 ∗ 2 = 152 \text{Horizontal distance between the centers of and any other sphere and the}\\ \text{9th, is the radius of the circum circle of the octagonal with sides 200 }\\=100*\dfrac{2}{\sqrt{2-\sqrt2}}. \text{ R as the radius of 9th , the actual distance would} \\ \text{ be R + 100. Center of 9th is R - 100 above the plane of the octagon.}\\\text{These three sides form a right angle triangle with R +100 as hypotenuse.}\\\implies~(R+100)^2= (R-100)^2 + (100*\dfrac{2}{\sqrt{2-\sqrt2}}. )^2.\\\implies~2*\sqrt{100*R}=100*\dfrac{2}{\sqrt{2-\sqrt2}}.\\Solving~for~~R, ~we~get~~R=100+50*\sqrt2=\large \boxed{\color{#D61F06}{~~~~152~~~~} }

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