Interesting challenging Problem

[maxhk]

"Two great circles lying in planes that are perpendicular to each other are drawn on a wooden sphere of radius "a".

Part of the sphere is then shaved off in such a way that each cross section of the remaining solid that is perpendicular to the common diameter of the two great circles is a square whose vertices lie on these circles.

Find the volume of this solid."

I don't understand the geometry of the problem.

Can you please explain the problem and if possible draw a diagram for me ?

Thanks

 

[maxhk]

Any genius able to solve this problem ?

 

[daon2]

Think about the circles themselves being perpendicular, centered on the yz and xz axes. The common diameter is the z axis from -r to r. The "squares" lie on the planes z=t, for each t between -r and r. At z=0 the square (or diamond) has side length \(\displaystyle r\sqrt{2}\).

 

[maxhk]

Thanks for your insightful answer.

I tried to visualise the problem for an hour so if I understood you well the volume would be

V = \int_{-a}^{a} 2 * (a^2 - t^2) dt

because at z = t the square would have a side length of sqrt(2 * (a^2 - t^2))

then V = (8*a^3)/3

Is this right ?

Thanks

 

[soroban]

Hello, maxhk!

If I visualize the solid correctly,
. . we have two congruent pyramids with square bases,
. . with their bases together (an octahedron).

One pyramid has a square base with side \(\displaystyle a\sqrt{2}\)
Its height is \(\displaystyle a.\)
Hence, its volume is: .\(\displaystyle V \:=\:\frac{1}{3}Bh \:=\:\frac{1}{3}(2a^2)(a) \:=\:\frac{2}{3}a^3\)

The volume of the two pyramids is: .\(\displaystyle \frac{4}{3}a^3\) cubic units.

 

[daon2]

Thanks for your insightful answer.

I tried to visualise the problem for an hour so if I understood you well the volume would be

V = \int_{-a}^{a} 2 * (a^2 - t^2) dt

because at z = t the square would have a side length of sqrt(2 * (a^2 - t^2))

then V = (8*a^3)/3

Is this right ?

Thanks

The way I visualize the solid, I think you are correct. In the case of a pyramid, I believe the vertices would trace out straight lines rather than circles.

 

[maxhk]

The way I visualize the solid, I think you are correct. In the case of a pyramid, I believe the vertices would trace out straight lines rather than circles.

Thanks for your answers.

I think the correct form is two pseudo-domes (domes with square cross sections ) rather than two pyramids as explained in the problem
" .... is a square whose vertices lie on these circles ...."

So the volume is (8 * a^3) /3

The volume of two opposite pyramides with common base is (4 * a^3) /3 but this is another question ...

Thanks a lot for your both excellent contributions.