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

Solution 1:

Consider this diagram, representing a side view of the martini glass with "tipped" so that the beverage reaches a maximum distance $x := |\overline{OX}|$ and minimum distance $y := |\overline{OY}|$ from cone vertex $O$:

Let $O^{\prime\prime}$ be the midpoint of $\overline{XY}$, so that it is the center of the represented ellipse; thus, $|\overline{O^{\prime\prime} X}| = |\overline{O^{\prime\prime} Y}| = a$ is the ellipse's "major radius".

Two observations follow quickly:

• With $H$ the foot of the perpendicular from $O$ to $\overline{XY}$, so that $h := |\overline{OH}|$ is an altitude of $\triangle XOY$ (and congruent to the altitude of the beverage cone itself), $$a h = |\triangle XOY| = \frac{1}{2} x y \sin 2\theta \tag{1}$$

• With "horizontals" $\overline{XX^\prime}$, $\overline{YY^\prime}$, and $\overline{X^{\prime\prime}Y^{\prime\prime}}$ as shown above, we see that $\overline{O^{\prime\prime} X^{\prime\prime}}$ and $\overline{O^{\prime\prime}Y^{\prime\prime}}$ are necessarily midpoint segments of respective triangles $\triangle XYX^\prime$ and $\triangle XYY^\prime$; therefore, $$|\overline{O^{\prime\prime}X^{\prime\prime}}| = \frac{1}{2} |\overline{XX^\prime}| = x \sin\theta \qquad\text{and, likewise}\qquad |\overline{O^{\prime\prime}Y^{\prime\prime}}| = y\sin\theta \tag{2}$$

Also, we note that $b$, the ellipse's "minor radius", is the length of a segment perpendicular to $\overline{X^{\prime\prime}Y^{\prime\prime}}$ at $O^{\prime\prime}$ that meets a (semi-)circle with diameter $\overline{X^{\prime\prime}Y^{\prime\prime}}$:

This is the classical construction of the geometric mean of two values, and we have $$b^2 = |\overline{O^{\prime\prime}X^{\prime\prime}}| |\overline{O^{\prime\prime}Y^{\prime\prime}}| = x y \sin^2 \theta \tag{3}$$

Now, since the volume of a cone whose base is an ellipse with radii $a$ and $b$, and whose height is $h$, is given by $$V = \frac{1}{3}\cdot(\text{area of base})\cdot(\text{height}) = \frac{\pi}{3} ab h \tag{4}$$ we have $$V = \frac{\pi}{3}\cdot a h \cdot b = \frac{\pi}{3}\cdot \frac{1}{2}x y \sin 2\theta \cdot \sqrt{ x y \sin^2\theta} = (xy)^{3/2}\cdot\frac{\pi}{6}\sin\theta\sin 2\theta \tag{$4^\prime$}$$

That is, for fixed $\theta$, we can say more simply:

$$V^2 \;\;\text{is proportional to}\;\; (xy)^3 \tag{$\star$}$$

This being so ... If a certain volume of beverage in a "tilted" glass reaches extreme distances $x$ and $y$ from the vertex, and the same volume of beverage in a "level" glass reaches distance $z$ (in all directions) from the vertex, then we clearly have

$$x y = z^2 \tag{$\star\star$}$$

The "level" distance is the geometric mean of the "tilted" extremes! I suspect that, given the tidiness of this relation, there should be a shorter route to it.

Edit to add ... As @hypergeometric has made the observation that (for constant volume) the ellipse's minor axis and the side-view area remain constant, I'll mention how that these facts arise from my analysis. Simply combine $(\star)$ with $(3)$ and $(1)$ to get:

• $V$ is proportional to $b^3$.
• $V^2$ is proportional to $|\triangle XOY|^3$.

Then, the constancy of $V$ implies the constancy of the other values.

Solution 2:

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.