Area of interior triangle of pyramid normal to a side length

[member 428835]

Homework Statement

Given a pyramid who's base is an equilateral triangle with corner taper $\psi$, what is the cross-sectional area of a plane that is normal to one of the interior corners (not normal to the altitude of the pyramid) $z$ distance from the corner taper?

Relevant Equations

A = 1/2 bh for a triangle

This isn't homework, but I figured it's fine if I make it a HW problem and post here (if not, please let me know).

Let $z^*=0$ be the vertex of the pyramid, and let $z^*$ run the altitude. It's easy to show the area of the base normal to the altitude is $A = 4 \left.z^*\right.^2 \tan(30^\circ)\tan(\psi)^2$. However, if we now let $z$ define the length along one of the corner vertices, I'm not sure how to calculate the area of the plane within the pyramid normal to $z$.

If we lay the pyramid on a side so that $z$ is going in the flat direction, I believe the taper angle would be $2\psi$ from $z=0$, making me think the height of the interior triangle would be $z\tan(2\psi)$, but I'm having issues thinking about that interior angle that used to be $30^\circ$. Any help?

If this is confusing, let me pose the problem differently: given a pyramid with an equilateral triangular base $ABC$ and vertex $V$, define $O$ to be the point on base ABC such that VO is the altitude (height) of the pyramid. $\angle CVO= \psi$, the taper angle of the pyramid. Now let the point $z$ lie on $\overleftrightarrow{VC}$ such that $\angle AzV = \angle BzV = 90^\circ$. What is the area of triangle $AzB$?

 

[anuttarasammyak]

What is the area of triangle AzB?

Say it S
$$\frac{S\cdot VC}{3}=V$$
where V is volume of the pyramid.

 

[docnet]

Now let the point z lie on VC↔ such that ∠AzV=∠BzV=90∘. What is the area of triangle AzB?

Then z and O are the same point and the area of triangle AzB is 1/3 the area of triangle ABC.
these two conditions are inconsistent with the geometry of the problem you first stated

 

[member 428835]

I think I've been consistent with the notation, but in case not, here's a picture.

Cool, so $ABC \cdot VO = S\cdot Vz + S\cdot zC = S\cdot (Vz+zC) = S \cdot VC$, which is three times the total volume. Thanks so much @anuttarasammyak !