famous problems and famous mathematicians

Solution to a “nice analysis question for RMO practice”

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The question from a previous blog is re-written here for your convenience. 

Question:

How farthest from the edge of a table can a deck of playing cards be stably overhung if the cards are stacked on top of one another? And, how many of them will be overhanging completely away from the edge of the table?

Solution:

The figure below shows how two and three cards can be stacked so that the mass of cards is equal on either side of the vertical line passing through the corner of table’s edge in order to just balance them under gravity:

the set of first two cards are arranged as follows (the horizontal lines represents the cards):

xxxxxxxxxxxxxxxx\line(5,0){170}

\line(5,0){150}xxxxxxxxxxxxxxxxxxx

the set of three cards are arranged as follows:

xxxxxxxxxxxxxxxxxxxxxx\line(5,0){150}

xxxxxxxxxxxxx\line(5,0){150}

\line(5,0){150}xxxxxxxxxxxxxxxxxxxxxx

We can see that the length of the overhand is a harmonic series of even numbers multiplied by the length of one card, L.

Overhand distance is (\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \ldots + \frac{1}{52 \times 2})L for 52 cards.

It may be noted that the series if continued to infinity leads to H_{\infty}^{E}.

That is, H_{\infty}^{E}=\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \frac{1}{8} + \ldots

This series is known to diverge as proved below:

First consider, H_{\infty}=1+ \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8}+ \ldots, which is, greater than

1+ \frac{1}{2} + \frac{1}{4} + \frac{1}{4} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8}+ \ldots, which is greater than 1+ \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \ldots. Hence, H_{\infty} diverges as we go on adding 1/2 indefinitely.

Now, let H_{E}=\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \frac{1}{8} + \ldots = \frac{1}{2}(1+ \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \ldots)=\frac{1}{2}H_{\infty}

Since H_{\infty} diverges, H_{E}also diverges.

Hence, the “overhang series” also diverges.

This means that the cards can be stacked indefinitely and the overhang distance can reach infinity. However, this will happen very slowly as shown in the table below:

\begin{array}{cc} n^{E} & H_{n}^{E}\\ 2 & 0.5 \\ 10 & 1.46 \\ 100 & 2.59 \\ 1000 & 3.74 \\ 10000 & 4.89 \\ 100000 & 6.05 \end{array}

Computing the number of cards that completely overhang off the table needs information about the overhang distance for different number of cards. As shown below in the figure, four cards are required to have one card completely away from the edge of the table. This is because (\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \frac{1}{8}=1.0417 >1).

(the set of four cards are arranged as follows:)

xxxxxxxxxxxxxxxxxxxxxxxxxxxx\line(5,0){150}

xxxxxxxxxxxxxxxx\line(5,0){150}

xxxxxxxxxx\line(5,0){150}

xxxxx\line(5,0){150}

We can see that the length of the overhang is a harmonic series of even numbers multiplied by the length of one card, L:

Overhang distance = (\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \ldots + \frac{1}{52 \times 2})L for 52 cards

It may be noted that the series if continued to infinity, leads to H_{\infty}^{E}

H_{\infty}^{E} = \frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \frac{1}{8} + \ldots

This series is known to diverge. This means that the cards can be stacked indefinitely and the overhang can reach infinity. However, this will happen very slowly as shown in the table below:

\begin{array}{cc} n^{E} & H_{\infty}^{E} \\ 2 & 0.5 \\ 10 & 1.46 \\ 100 & 2.59 \\ 1000 & 3.74 \\ 10000 & 4.89 \\ 100000 & 6.05 \end{array}

Computing the number of cards that completely overhang off the table needs information about the overhang distance for different numbers of cards. As shown in the above schematic figures of cards with overhangs, four cards are required to have one card completely away from the edge of the table. This is because

(\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \frac{1}{8})=1.0417>1

For the second card to overhang completely, leaving the first card (and hence one half) that is already completely overhung, it is now necessary that

(\frac{1}{4} + \frac{1}{6} + \frac{1}{8} + \ldots + \frac{1}{2n})>1, or

(\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \frac{1}{8} + \ldots + \frac{1}{2n} )>1+ \frac{1}{2}

where n needs to be found out. By generating some more data, we can find the value of n to be 11.

For third overhanging card, we need

(\frac{1}{6} + \frac{1}{8} + \ldots + \frac{1}{2n})>1 or

(\frac{1}{2} + \frac{1}{4} + \frac{1}{6}+\frac{1}{8}+ \ldots + \frac{1}{2n})>1+\frac{1}{2} + \frac{1}{4}

Thus, for m completely overhanging cards, we find n such that H_{2n}^{E} > 1+ H_{2(m-1)}^{E}

The table below shows these values wherein we see an approximate pattern of arithmetic progression by 7.

\begin{array}{cccc} m & n & m & n \\ 1 & 4 & 11 & 78 \\ 2 & 11 & 12 & 85 \\ 3 & 19 & 13 & 92 \\ 4 & 26 & 14 & 100 \\ 5 & 33 & 15 & 107 \\ 6 & 41 & 16 & 115 \\ 7 & 48 & 17 & 122 \\ 8 & 55 & 18 & 129 \\ 9 & 63 & 19 & 137 \\ 10 & 70 & 20 & 144 \end{array}

By examining the pattern in the table, we can get a simple rule to estimate the number of completely overhanging number of cards m, with an error of utmost one, for n cards stacked.

m = round(\frac{n}{7.4})=round(\frac{10n}{74}).

Reference:

Popular Problems and Puzzles in Mathematics by Asok Kumar Mallik, IISc Press, Foundation Books.

Hope you enjoyed the detailed analysis…

More later,

Nalin Pithwa

 

Pick’s theorem: a geometry problem for RMO practice

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Pick’s theorem:

Consider a square lattice of unit side. A simple polygon (with non-intersecting sides) of any shape is drawn with its vertices at the lattice points. The area of the polygon can be simply obtained as B/2+I-1 square units, where B is the number of lattice points on the boundary; I = number of lattice points in the interior region of the polygon. Prove this theorem.

Proof:

Refer Wikipedia 🙂 🙂 🙂

https://en.wikipedia.org/wiki/Pick%27s_theorem

Cheers,

Nalin Pithwa.