Area of interior triangle of pyramid normal to a side length

AI Thread Summary
The discussion revolves around calculating the area of a triangle within a pyramid, specifically triangle AzB, defined by a vertex and a point along the altitude. The area of the base normal to the altitude is established as A = 4z^2 tan(30°) tan(ψ)^2. The taper angle and height of the interior triangle are explored, leading to the conclusion that the area of triangle AzB is one-third the area of the base triangle ABC when z and O coincide. The participants clarify the geometric relationships and confirm the consistency of their calculations. The conversation emphasizes the geometric properties and relationships within the pyramid structure.
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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##?
 
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joshmccraney said:
What is the area of triangle AzB?
Say it S
\frac{S\cdot VC}{3}=V
where V is volume of the pyramid.
 
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joshmccraney said:
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
 
I think I've been consistent with the notation, but in case not, here's a picture.
IMG_3761.jpg

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 !
 
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