first of all, sorry for the unspecific question, I couldn't think of a specific enough title, if there is any suggestion, I can edit the title.
This question seems simple enough, but I can't seem to solve it? Two of the sections are clearly a right based triangular pyramid missing a chunk of a similar triangular pyramid. So if we continue the line through P, the side profile would look like this, and if we label PV x we can get the other lengths through similar triangle shenanigans.
Then, $\frac{1}{1-x}\cdot 1 \cdot 1 /3 - \frac{x}{1-x}\cdot x \cdot x / 3 = 1/3$ and rearranging would give $x^2-1=0$, which is nonsensical. Could someone please point out my mistake?
*Question is from 2019 Senior Division Australian Mathematics Competition
Let $PV = x$, then extending the triangular pyramid chunk below the cube, it will form the full pyramid at $z = -h$, where $z = 0$ is the horizontal base of the cube.
From similar triangles, we have the following ratios
$ \dfrac{h}{x} = \dfrac{1 + h}{1} $
From which $ h = (1 + h) x $
$ h(1 - x) = x $
and
$ h = \dfrac{x}{1 - x} $
The volume of the pyramidal chunk (inside the cube) is
$ V = (1 - \bigg(\dfrac{h}{h+1}\bigg)^3 ) V_0 $
Where V_0 is the volume of the all the pyramid, which is one third of the base area times the total height
$V_0 = \dfrac{1}{3} \dfrac{1}{2} (1 + h) $
Now $V = \dfrac{1}{3} $, so
$ (1 - \bigg(\dfrac{h}{h+1}\bigg)^3 ) \dfrac{1}{2} (1 + h) = 1 $
Now, $1 + h = \dfrac{1}{1 - x} $ and $ \dfrac{h}{h+1} = x $
So,
$ \dfrac{ 1 - x^3 }{1 - x} = 2 $
And this simplifies to
$ 1 + x + x^2 = 2 $
So that
$ x^2 + x - 1 = 0$
whose postive solution is
$ x = \dfrac{-1 + \sqrt{5}}{2} $
So the correct answer is $(E)$