Three cones and a sphere - Unknown radius

Geometry Level pending

Three right circular cones are placed on the floor with their bases touching. The first cone with a base radius 6 6 , and height 14 14 is placed with its base centered at the origin ( 0 , 0 , 0 ) (0, 0, 0 ) . The second cone with a base radius of 8 8 and height 14 14 is placed with its base centered at ( 14 , 0 , 0 ) (14, 0, 0) . The third cone with a base radius of 7 7 and height 14 14 is placed with its base centered at ( 5 , 12 , 0 ) (5, 12, 0) . A hollow spherical ball of radius R R is placed on the floor in between the three cones such that it is tangent to each of them. Find the radius R R .


The answer is 1.7297.

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1 solution

Hosam Hajjir
Mar 21, 2021

Taking each cone separately, and viewing a cut through the cone and the sphere in a cutting plane that includes the axis of the cone and the center of the sphere, we obtain the above figure. The radius of cone base is r r and the radius of the sphere is R R . The height of the cone is H H . The semi-vertical angle of the cone is θ = tan 1 ( r H ) \theta = \tan^{-1} \left( \dfrac{r}{H} \right) . Let A A be the apex of the cone, and P P be the point of tangency between the cone and the sphere. Further, let s = A P s = | AP | . Finally, point B = ( x 1 , y 1 , z 1 ) B = (x_1, y_1, z_1) is the center of the sphere, and z 1 = R z_1 = R . Writing two expressions for horizontal separation and vertical separation between point C C (the cone base center) and point B B , we obtain the following:

( x 1 C x ) 2 + ( y 1 C y ) 2 = s sin θ + R cos θ \sqrt{ (x_1 - C_x)^2 + (y_1 - C_y)^2 } = s \sin \theta + R \cos \theta

H s cos θ + R sin θ = z 1 H - s \cos \theta + R \sin \theta = z_1

Eliminating s s , we obtain a single equation in x 1 , y 1 , z 1 x_1, y_1, z_1 ,

( x 1 C x ) 2 + ( y 1 C y ) 2 = tan θ ( H + R sin θ z 1 ) + R cos θ \sqrt{ (x_1 - C_x)^2 + (y_1 - C_y)^2 } = \tan \theta (H + R \sin \theta - z_1) + R \cos \theta

Since z 1 = R z_1 = R , this last equation becomes,

( x 1 C x ) 2 + ( y 1 C y ) 2 = tan θ H + R ( tan θ ( sin θ 1 ) ) + cos θ ) \sqrt{ (x_1 - C_x)^2 + (y_1 - C_y)^2 } = \tan \theta H + R ( \tan \theta (\sin \theta - 1) ) + \cos \theta )

Applying this equation for each of the three cones results in three simultaneous equations in x 1 , y 1 , R x_1, y_1, R . These solve numerically to,

x 1 = 5.925692639 , y 1 = 3.984234786 , R = 1.729742283 1.7297 x_1 = 5.925692639, y_1= 3.984234786, R= 1.729742283 \approx \boxed{1.7297}

Thanks! Fine problem.

geogebra.org_3d

Yuriy Kazakov - 2 months, 2 weeks ago

Nice visualization. Thanks for taking the time to share it.

Hosam Hajjir - 2 months, 2 weeks ago

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