538.com's Puzzle of the Overflowing Martini Glass - How to compute the minor and major axis of an elliptical cross-section of a cone FiveThirtyEight.com Riddler Puzzle / May 13
The puzzle goes like this;
"It’s Friday. You’ve kicked your feet up and have drunk enough of your martini that, when the conical glass () is upright, the drink reaches some fraction p of the way up its side. When tipped down on one side, just to the point of overflowing, how far does the drink reach up the opposite side?"
I came up with an an answer of $ p^2 $, based on the following (probably incorrect) solution;

$
\begin{aligned}
\overline{AE} &= \overline{AD} &&= 1 \dots \text{length of one side}\\
\overline{AC} &= p &&=\text{initial distance of liquid  along one side from the bottom}\\
\overline{AB} &= q &&= \text{distance when glass is tipped over to the other side}\\
h_p &= p \cos\theta &&= \text{height of upright triangle} \\
b_p &= 2p \sin\theta &&= \text{base of upright triangle} \\
h_q &= \overline{AE} \sin(2\theta) &&= \text{height of tipped-over triangle} \\
&= 2\sin\theta\cos\theta\\
b_q &= q &&= \text{base of tipped-over triangle} \\
&&&\\
\triangle(ACF) &= \left( \displaystyle \frac{h_p b_p}{2} \right) &&= p^2\sin\theta \cos\theta\\
\triangle(ABE) &= \left(\displaystyle\frac{h_q b_q}{2}\right) &&= q\sin\theta\cos\theta&&\\
\end{aligned}
$
$\begin{aligned}
\text{As   } & \text{volume of cone AEB} &&= \text{volume of cone ACF,}\\ 
  & \triangle(ACF) &&= \triangle(ABE) \dots \text{(leap of faith?)}\\ 
& q &&= p^2 \\ 
\end{aligned}
$
So the questions are;
a) Is it valid to assume that equal volumes imply equal vertical cross-sections?
b) How to derive the major and minor axes of the elliptical cross section $BE$ (so we can state the equation for the tilted volume in terms of $\overline{AB}$, and ultimately state $\overline{AB}$ as a function of $p$)?

Diagram below shows side and aerial views of upright and tilted martini glass. 

For the upright glass, the shape of the martini liquid is an inverted right circular cone, with height $p\sin\theta$ and circular base of area $A_1$. Volume of martini is given by:
$$\begin{align}
V&=\frac 13\cdot A_1 \cdot p\cos\theta\\
&=\frac 13\cdot \pi (p\sin\theta)^2\cdot p\cos\theta\\
&=\frac 13\cdot \pi p\sin\theta\cdot \underbrace{\left(\frac 12 \cdot p^2\cdot \sin 2\theta\right)}_{\text{area of sideview cross-section}}
&&\cdots (1)
\end{align}$$
For the tilted glass, the shape of the martini liquid is an inverted ellipsoidal cone, with height $h$ and ellipse base with area $A_2$ and semi-major and semi-minor axes of $a$ and $b$ respectively.  Volume of martini is given by:
$$\begin{align}
V&=\frac 13\cdot A_2 \cdot h\\
&=\frac 13\cdot \pi ab\cdot h\\
&=\frac 13 \cdot \pi b \cdot \underbrace{\left(\frac 12 \cdot 2a\cdot h\right)}_{\text{area of sideview cross-section}}\\
&=\frac 13 \cdot \pi b \cdot \underbrace{\left(\frac 12 \cdot q\cdot 1\cdot \sin 2\theta\right)}_{\text{area of sideview cross-section}}\\
\end{align}$$
Also, when glass is tilted, top surface of martini liquid changes from a circle to an ellipse, elongating along direction of tilt, with width remaining unchanged, i.e semi-minor axis of ellipse is equal to radius of original circle, or $b=p\sin\theta$. Hence
$$V=\frac 13 \cdot \pi p\sin\theta \cdot \left(\frac 12 \cdot q\cdot 1\cdot \sin 2\theta\right)\qquad\cdots (2)$$
Volume remains the same:   
$(1)=(2)$ :
$$\begin{align}
\frac 13\cdot \pi p\sin\theta\cdot \left(\frac 12 \cdot p^2\cdot \sin 2\theta\right)
&=\frac 13 \cdot \pi p\sin\theta \cdot \left(\frac 12 \cdot q\cdot 1\cdot \sin 2\theta\right)\\
\underbrace{\frac 12 \cdot p^2\cdot \sin 2\theta}_{\text{area of upright sideview cross-section}}&=\underbrace{\frac 12 \cdot q\cdot 1\cdot \sin 2\theta}_{\text{area of tilted sideview cross-section}}\\\\
\color{red}{q}&\color{red}{=p^2}\qquad\blacksquare
\end{align}$$
The above also shows that the area of the sideview cross-section remains the same before and after tilting. 

Special Note:
In the name of science and mathematics, I ordered myself a martini!
Pics shown below.  Toothpick measures diameter of circle and minor axis of  ellipse, and shows that they are both the same.  Olive marks centre of circle/ellipse respectively.

Upright martini glass

Tilted martini glass

Edit
This section has been added to address the point raised by @DavidK about the assumption on the ellipse semi-minor axis.
Recap:
By equating volumes, we have
$$\begin{align}
V_\text{upright}&=V_\text{tilted}\\
\frac 13\cdot \pi (p\sin\theta)^2 \cdot (p\cos\theta)&=\frac 13\cdot \pi ab \cdot h\\
p\sin\theta\cdot \left(\frac 12\cdot 2p\sin\theta\cdot p\cos\theta\right)&=b\cdot \left(\frac 12 \cdot 2a\cdot h\right)\\
p\sin\theta\cdot \left(\frac 12 p^2 \sin 2\theta\right)&=b\cdot \left(\frac 12\cdot q\cdot 1\cdot \sin 2\theta\right)\\
p^3\sin\theta&=bq&&(1)
\end{align}$$
Now consider the side view of the tilted glass (see diagram below).

From the diagram it is clear that
$$\begin{align}
r&=\frac 12 (1+q)\sin\theta\\
d&=\frac 12 (1-q)\sin\theta\\
b&=\sqrt{r^2-d^2}\\
&=\sqrt {q} \sin\theta&&(2)
\end{align}$$
From $(1), (2)$, 
$$\begin{align}
p^3\sin\theta&=\sqrt{q}\sin\theta\cdot q\\
&=q^{3/2}\sin\theta\\
\color{red}q&\color{red}{=p^2}\qquad\blacksquare\end{align}$$
Substuting the result in $(2)$ gives
$$b=p\sin\theta$$.

Additional Note:
From additional analysis it can be seen that, for a conical glass, 
constant volume implies constant lateral width, i.e. $b=p\sin\theta$, and vice versa. 
