# Martini Glass - Extension

This is my extension to the very interesting question on the martini glass from 538.com by the Riddler as posted here earlier by MP Droid.

Recap of original configuration.

A martini glass has the shape of an inverted right circular cone with sides of unit length. The glass is positioned upright and filled with martini up to a level $p(<1)$ on each side. See diagram below.

Extension: The extension is as follows:

(1) Refer to the first diagram above. As in the original question, when tilted such that the glass is just short of overflowing, what is the angle of tilt $\alpha$ from the vertical?

Now assume that the martini glass is covered with a lid such that the martini does not overflow when tilted more than $\alpha$.

(2) Refer to the second diagram above. If the martini glass is further tilted such that the right side (now the bottom) of the glass is parallel to the water level, i.e. horizontal,

• what is the length $u$ of the left side of the glass?
• what is the angle of tilt $\beta$ from the vertical?
• what is the shape of the top surface of the martini?

(3) Refer to the third diagram above. If the martini glass is further tilted such that the top surface of the martini touches the apex of the inverted cone of the glass,

• what is the length $v$ of the right side?
• what is the angle of tilt $\gamma$?
• what is the shape of the top surface of the martini?
• Why isn't the answer accepted? Commented Dec 14, 2020 at 15:43

(1) In this case the surface forms an ellipse with major axis $AB$ (see diagram below) and semi-minor axis $MN$. Let $\beta$ be the angle between $AV$ and the horizontal plane. This is linked to your angle $\alpha$ by $\beta=\pi/2-\theta-\alpha$. By the sine rule we have $AB:\sin2\theta=q:\sin\beta$, that is: $$AB=q{\sin2\theta\over\sin\beta}.$$ And in the circle of diameter $FG$ we have: $$MN^2=GM\cdot FM=BE\cdot CD=q\sin^2\theta, \quad\text{that is:}\quad MN=\sqrt{q}\sin\theta.$$ The volume of the cone of vertex $V$ and base the ellipse $ABN$ is then $$volume={\pi\over3}{AB\over2}MN\cdot AH= {\pi\over3}q{\sin2\theta\over2\sin\beta}\sqrt{q}\sin\theta\cdot \sin\beta= {\pi\over3}\sin^2\theta\cos\theta\ q^{3/2}.$$ Equating that to the original volume ${\pi\over3}\sin^2\theta\cos\theta\ p^3$ we get: $$q=p^2.$$ This also entails $MN=p\sin\theta$: the semi-minor axis of the elliptic surface is the same as the radius of the original circular surface (a result also obtained in the old thread). To obtain tilt angle $\beta$ we can plug $AB=\sqrt{1+q^2-2q\cos2\theta}$ (from the cosine rule) into the first formula above, to get: $$\sin\beta={p^2\sin2\theta\over\sqrt{1+p^4-2p^2\cos2\theta}}.$$
(2) In this case the surface of the liquid forms a parabolic segment, with axis $AB=AE=1-u$ (see diagram below). Tilt angle is simply $\beta=\pi/2-\theta$. In the circle of diameter $DE$ we have $$BM^2=BD\cdot BE=2AC(2OE-2AC)=4u(1-u)\sin^2\theta.$$ The area of the parabolic segment is $4/3$ the area of triangle $AMN$, hence: $$area={4\over3}AB\cdot BM={8\over3}\sin\theta\sqrt u (1-u)^{3/2}.$$ The area of any horizontal section intersecting $VA$ at a distance $x$ from $V$ is given by the same expression, with $u$ replaced by $x$. We then obtain the volume by integrating such horizontal slices: $$volume=\int_0^u{8\over3}\sin\theta\sqrt x (1-x)^{3/2}dx ={\sin\theta\over9}\left( (14u-8u^2-3)\sqrt{u(1-u)}+3\arcsin\sqrt{u} \right).$$ Equating that to the original volume we can, in principle, compute $u$. But this won't give, in general, a simple closed expression.