This week’s puzzle was related to me by Jay Goldman:
I have two glasses, one containing precisely 12 fluid ounces of coffee, the other containing precisely 12 fluid ounces of tea.
I pour one fluid ounce from the coffee cup into the tea cup, and then thoroughly mix the contents. I then pour one fluid ounce from the (mostly) tea cup back into the coffee cup, and then thoroughly mix the contents.
Is there more coffee in the tea cup, or more tea in the coffee cup?
There is a total of 12 fluid ounces of coffee and 12 fluid ounces of tea between 24 fluid ounces in two cups.
If that was too short...
There is a total of $V$ fluid ounces of tea and $v$ fluid ounces of coffee in the tea cup. $\frac{v}{V+v}$ of that is poured back into the coffee cup. $\frac{V}{V+v}$ remains in the tea cup. The amount of tea in the coffee cup or coffee in the tea cup is thus $$\frac{Vv}{V+v}$$
The amount of coffee in the tea cup is the same as the amount of tea in the coffee cup.
In this case, $\frac{12}{13}$ fluid ounces.
I have two glasses that can hold a maximum volume of 24 fluid ounces. Initially, one glass contains precisely 12 fluid ounces of coffee, while the other contains precisely 12 fluid ounces of tea.
Your goal is to dilute the amount of coffee in the “coffee cup” by performing the following steps:
Pour some volume of tea into the coffee cup.
Thoroughly mix the contents.
Pour that same volume out of the coffee cup (i.e., into the sink), so that precisely 12 fluid ounces of liquid remain.
After doing this as many times as you like, in the end, you will have 12 ounces of liquid in the coffee cup, some of which is coffee and some of which is tea. In fluid ounces, what is the least amount of coffee you can have in this cup?
If 6 fluid ounces of tea is poured into the coffee cup, 4 fluid ounces of coffee and 2 fluid ounces of tea will be poured out.
8 fluid ounces of coffee and 4 fluid ounces of tea remain in the coffee cup, and 6 fluid ounces remain in the tea cup.
If 6 fluid ounces of tea is poured into the coffee cup, $\frac{8}{3}$ fluid ounces of coffee and $\frac{10}{3}$ fluid ounces of tea will be poured out.
This yields $$\boxed{\frac{16}{3} \approx 5.3333 \text{ fluid ounces of coffee}}$$
If 4 fluid ounces of tea is poured into the coffee cup, 3 fluid ounces of coffee and 1 fluid ounce of tea will be poured out.
9 fluid ounces of coffee and 3 fluid ounces of tea remain in the coffee cup, and 8 fluid ounces remain in the tea cup.
If 4 fluid ounces of tea is poured into the coffee cup, $\frac{9}{4}$ fluid ounces of coffee and $\frac{7}{4}$ fluid ounces of tea will be poured out.
$\frac{27}{4}$ fluid ounces of coffee and $\frac{21}{4}$ fluid ounces of tea remain in the coffee cup, and 4 fluid ounces remain in the tea cup.
If 4 fluid ounces of tea is poured into the coffee cup, $\frac{27}{16}$ fluid ounces of coffee and $\frac{37}{16}$ fluid ounces of tea will be poured out.
This yields $$\boxed{\frac{81}{16} \approx 5.0625 \text{ fluid ounces of coffee}}$$
In order to minimize coffee at the end, a maximal amount of coffee must be poured out. However, this maximal amount is relative to the volume that is poured out, not 12 ounces.
Let $n$ be the number of pairs of pours.
$\frac{12}{n}$ is the number of fluid ounces of tea poured into the coffee cup, and thus, the number of fluid ounces of the coffee-tea concotion poured out.
The proportion of coffee remainining in the coffee cup after each pair of pours is
\begin{align*} & \dfrac{12}{12+\frac{12}{n}} \\ =& \dfrac{1}{1+\frac{1}{n}} \\ =& \dfrac{n}{n+1} \\ \end{align*}Since there are $n$ pours,
\begin{align*} Min =& \lim_{n \to \infty} 12\left(\frac{n}{n+1}\right)^n \\ =& \dfrac{12}{\lim_{n \to \infty} \left(\frac{n+1}{n}\right)^n} \\ =& \dfrac{12}{\lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n} \\ \end{align*}After an infinite number of infinitesimal pours, the minimum value of coffee remains,
$$\boxed{\frac{12}{e} \approx 4.4146 \text{ fluid ounces}}$$