**Question: **There is a perfect sphere of diameter 40 cms. resting up against a perfectly straight wall and a perfectly straight floor i.e. the wall and the floor make a perfect right angle. Can a perfect sphere of diameter 7 cms. pass through the space between the big sphere, the wall and the floor?

**Answer:**

For the sake of simplicity, consider two-dimension i.e view sphere as a two dimensional circle with diameter 40 cms.

From Figure I, (40 cms diameter sphere)

OC2 = OD2 + CD2

OC2 = 202 + 202

OC = 28.28427 cms

Also, X is the closest point to origin O on the sphere.

CX = 20 cms (radius)

OX = OC - CX

OX = 28.28427 - 20

OX = 8.28427 cms

From Figure II, (7 cms diameter sphere)

OP2 = OQ2 + PQ2

OP2 = (3.5)2 + (3.5)2

OP = 4.94974 cms

Also, Y is the farthest point to origin O on the sphere.

PY = 3.5 cms (radius)

OY = OP + PY

OY = 4.94974 + 3.5

OY = 8.44974 cms

Now, as OY > OX i.e. smaller sphere requires more space than the space available. Hence, smaller sphere of 7 cms diameter can not pass through the space between the big sphere, the wall and the floor. **Another method. **

Draw a line tangent to the big sphere at the point X such that X is the closest point to the origin O on sphere. The tanget will cut X and Y axes at A and B respectively such that OA=OB. [See Fig III] From above, OX=8.28427 cms.

From the right angle triangle OAB, we can deduct that

OA = OB = 11.71572 cms

AB = 16.56854 cms

Now, the diameter of the inscribed circle of right angle triangle is given by d = a + b - c where a <= b < c

The maximum possible diameter of the circle which can pass through the space between the big sphere, the wall and the floor is

= OA + OB - AB

= 11.71572 + 11.71572 - 16.56854

= 6.86291 cms

Hence, the sphere with 7 cms diameter can not pass through the space between the big sphere, the wall and the floor.