Cutting a pyramid in half

Let $x = FB$ , then the coordinates of points $G, H$ are

$G = ( 12 - \frac{1}{2}x , \frac{1}{2}x , \frac{1}{2}x \tan \theta )$

$H = (12 - \frac{1}{2}x , 12 - \frac{1}{2}x , \frac{1}{2}x \tan \theta)$

where $\theta$ is the angle between the base and a triangular face.

The volume is,

$V = \frac{1}{2} (x)(12 - x) (\frac{1}{2}x \tan \theta) + \frac{2}{3} (x) (\frac{1}{2}x) (\frac{1}{2}x \tan \theta)) = \frac{1}{2}\frac{1}{3}( 12^2 )(10)$

Multiply through by $12 / \tan \theta$ , and noting that $\tan \theta = \frac{5}{3}$ ,then the equation becomes,

$3 x^2 (12 - x) + 2 x^3 = 12 (240)(3/5) = 1728$

or

$x^3 - 36 x^2 + 1728 = 0$

Solving this cubic equation gives the result of $x = 7.832$

Let $x = FB$ and $y = HG$ and draw the following segments as follows:

Since $AB = 12$ and $FB = x$ , $y = HG = KG = AF = AB - FB = 12 - x$ .

The lower half of the cut solid is made up of on triangular prism $GMOHLN$ (in blue) and two congruent pyramids $HLNCI$ and $GMOBF$ (in red), and the two pyramids can be combined to make a single pyramid that is similar to the original pyramid. Since they are similar and the original pyramid has a side length of $12$ and a height of $10$ , the red pyramid has a height of $h = \frac{10}{12}x = \frac{5}{6}x$ .

Since the original pyramid is cut in half by volume, $\frac{1}{2}V_{\text{orig pyr}} = V_{\text{red pyr}} + V_{\text{blue prism}}$ , or $\frac{1}{2} \cdot \frac{1}{3}12^210 = \frac{1}{3}x^2h + \frac{1}{2}xhy$ , or $240 = \frac{1}{3}x^2(\frac{5}{6}x) + \frac{1}{2}x(\frac{5}{6}x)(12 - x)$ , which rearranges to $x^3 - 36x^2 + 1728 = 0$ , and has a solution of $x \approx \boxed{7.832}$ for $0 < x < 12$ .