Answer / raj kiran

consider 2-d fig for simplicity.. i.e two lines at right
angles and a circle with its circumference touching the
lines(walls).. now the line joining the center of the circle
n the point of contact of the circle wit the walls forms a
square of side 40 cm..

now join the center of the circle n the meeting point of the
wall, this forms the diagonal of the square..
by right angle theorem, we get the diagonal as,
(40*1.414)=56.56..

so the remaining space left ( corner of the wall to the
circumference of th circle) is 16.6cm...

through which a 7cm ball can pass through..

Answer / munna

The answer is yes.
Consider a 2-D diagram,
given that the diameter of the sphere is 40 cms.
so radius = 20 cms.
c0nsider the points that touch the floor and the wall and
the center of the sphere and the intersection of the floor
and wall.This forms a square of length of side 20 cms.
so,
the distance left for any sphere of unknown diameter to
pass through be x.
x=sqrt(800)-20=8.284... cms.
so a sphere of radius 7 cms passes.

Answer / guest

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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.

The puzzle can be solved by 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.

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