Let a-BCD be a regular triangular pyramid with side length a and bottom length b,

Then the center of the receiving ball must be at the height of this triangular pyramid. Set the height AM, connect DM and BC at E, and connect AE, and then make the height AM of the triangular pyramid intersecting with the middle vertical line of the side AD in the plane ADE, then 0 is the center of the circumscribed sphere, and AO and DO are the radii of the circumscribed sphere.

(When the included angle between the side of the triangular pyramid and its opposite surface is less than 90 degrees, that is, when the included angle DAE is less than 90 degrees, the center of the sphere is inside the pyramid; When the line-plane angle is equal to 90 degrees, the center of the sphere is just on the center m of the regular triangle on the bottom; When the line-plane angle is greater than 90 degrees, the center of the sphere is outside the pyrAMid and on the extension line of the pyramid height am. The solution I give below is the first case, and the center of the sphere is inside the pyramid. The other two cases can be deduced by themselves. )

Let ao = do = r

Then, DM = 2/3de = 2/3 * b=b/ root number 3 times root number 2.

Am = root number (a 2-b 2/3),

OM=AM-A0= radical sign (A 2-B 2/3)-R.

From do 2 = om 2+DM 2,

R = root number 3 times a2÷2(3a 2-B2)

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