Giza-IQ Test - Solution 1

Today, instead of proposing more puzzles, I have decided to begin giving you solutions.

This is the solution to the first of the two problems I posted on July 11.

   

1.01    In the figure the height is not 2/3 of the width, but it doesn’t matter.

We indicate with rs the radius of the sphere and with hc and rc the height and radius of the cone. Further, we can set rc = 1 without any loss of generality. We can then write:

BE = hc - rs = 4/3 - rs

DE = rs

AB = sqrt(hc²+rc²) = sqrt(16/9 + 1) = sqrt(16+9) / 3 = 5/3

AC = rc = 1

The angles EDB and ACE are right angles, and the triangle ABC is similar to EBD. We can therefore write:

BE/DE = AB/AC

If we substitute the lengths of the segments, we obtain:

(4/3 - rs) / rs = 5/3

which, resolved in rs, gives us the radius of the sphere:

r = 1/2

Vc = pi * rc² * hc / 3 = pi * (4/3) / 3 = 4 * pi / 9

Vs = (4/3) * pi * r³ = pi * (4/3) * (1/8) = pi / 6

Vs/Vc = pi / 6 * 9 / (4 * pi) = 3/8