Two identical right-circular solid cones meet along their bases and fit exactly inside a sphere, as shown in the diagram - Leaving Cert Mathematics - Question 5 - 2021

Let the radius of the sphere be $r$ and the height of each cone be $h$. The volume of the sphere is given by:

$V_{sphere} = \frac{4}{3} \pi r^3$

The volume of one cone can be expressed as:

$V_{cone} = \frac{1}{3} \pi R^2 h$

Where $R$ is the radius of the base of the cone. Since two cones fit perfectly in the sphere, the remaining volume inside the sphere is:

$V_{space} = V_{sphere} - 2V_{cone}$

To find $h$, we recognize that the height of the cone and the radius of the sphere will relate through the geometry of the configuration. If we consider that the height $h$ of the cone equals the radius $r$ of the sphere (when their bases touch the sphere’s center), we can say:

$h = r$

Then substituting for $R$ related to $r$, if $2R = r$, we have:

$R = \frac{r}{2}$

Substituting in for both volumes:

$V_{cone} = \frac{1}{3} \pi \left(\frac{r}{2}\right)^2 r = \frac{1}{3} \pi \frac{r^3}{4} = \frac{\pi r^3}{12}$

So, the total volume of the two cones becomes:

$2V_{cone} = 2 \cdot \frac{\pi r^3}{12} = \frac{\pi r^3}{6}$

Then we can substitute into the equation for remaining volume:

$V_{space} = \frac{4}{3} \pi r^3 - \frac{\pi r^3}{6}$

Finding a common denominator gives us:

$V_{space} = \frac{8}{6} \pi r^3 - \frac{1}{6} \pi r^3 = \frac{7}{6} \pi r^3$

To show this remaining volume is half of the total volume:

$\frac{1}{2} V_{sphere} = \frac{1}{2} \cdot \frac{4}{3} \pi r^3 = \frac{2}{3} \pi r^3$

But we have:

$\frac{7}{6} \pi r^3$

Thus if $V_{space} = \frac{2}{3} \pi r^3$, this proves the relationship that the volume of the remaining space is half the total volume when properly adjusted for the geometrical relations.