Archive for Math

Building a Picnic Table

Someone writes

Jonathan and/or [redacted],

I’ve got a basic geometry question for either of you, that I need a little help with…

I am working on making a picnic table with [redacted] and we wanted to design our own legs for the table which are angled and cross. I have been able to work out all relationships between heights, lengths, widths, angles. etc. and now have two equations with two unknowns that will allow me to solve for the correct angle at which to cut the wood for the legs. Unfortunately I have forgotten how to solve basic cos, sin, and tan when variables are involved. The two equations are (all units are in inches, not that it matters):

$\displaystyle \textrm{sin} (\alpha)$ $\textstyle =$ $\displaystyle \frac{5.5}{x}$ (1)
$\displaystyle \textrm{tan} (\alpha)$ $\textstyle =$ $\displaystyle \frac{29}{35.5-x}$ (2)

How do I solve for $x$ and $\alpha$? Also, what is the answer ūüôā

Thanks,
[redacted]

Because

\begin{eqnarray*}\textrm{tan} (\alpha) & = & \frac{\textrm{sin} (\alpha)}{\text......rac{\textrm{sin} (\alpha)}{\sqrt{1 - \textrm{sin}^{2} (\alpha)}}\end{eqnarray*}

we have

$\displaystyle \frac{29}{35.5-x}$ $\textstyle =$ $\displaystyle \frac{\frac{5.5}{x}}{\sqrt{1 - \left(\frac{5.5}{x}\right)^{2}}}$ (3)

Simplifying and solving by computer algebra, we get

\begin{eqnarray*}\frac{3243}{841} x^2 + \frac{8591}{841} x - \frac{1017005}{3364} & = & 0\end{eqnarray*}

and

\begin{eqnarray*}x & = & \frac{ \pm 1276\sqrt{2071} - 8591}{6486}\end{eqnarray*}

Only the positive choice is a root of Eqns. (1) and (2), giving

\begin{eqnarray*}x & = &7.62835577107986... \\\alpha & = & 0.805235953662952... \\& = & 46.1366216570791... \textrm{(degrees)}\end{eqnarray*}

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The Biggest Box Fedex Will Take

My friend Megan Goering asks what is the largest volume box you can send through Fedex given that length + twice the width + twice the height must be less than 165 inches. It might seem odd to care about only the volume instead of the dimensions, but it makes sense if you’re shipping walnuts, balloons or bows for Christmas presents. Anything pourable, that is.

Here’s my solution. Say that the dimensions are $x$, $y$ and $z$. Then we can write the problem as

\begin{eqnarray*}\textrm{max } & & xyz \\\textrm{subject to} & &x + 2y + 2z \le 165 \\& & x,y,z \ge 0 \\\end{eqnarray*}

First, observe that if $x + 2y + 2z < 165$, then the product $xyz$ can be increased by increasing any one of the dimensions. Therefore, the maximum will occur when $x + 2y + 2z = 165$. Thus, we can eliminate $z$ because

$\displaystyle z$ $\textstyle =$ $\displaystyle \frac{165 - 2y -x}{2}$ (1)

Here is the problem

\begin{eqnarray*}\textrm{max } & & xy\left(\frac{165 - 2y -x}{2} \right)\\\textrm{subject to} & &x + 2y \le 165 \\& & x,y \ge 0 \\\end{eqnarray*}

The constraints impose that the maximum be achieved in a triangular area with one vertex at the origin and the other two each on one of the $x$ and $y$ axes. Notice that if we use a point $(x,y)$ on the border of this triangular region, the product $xyz$ will be equal to 0, so this is clearly not the maximum. Therefore, the maximum occurs on the inside of the triangle. This is important, because it means that the maximum is some sort of “bump” which means the problem can be solved directly with calculus.

We want to maximize $xy\left(\frac{165 - 2y -x}{2} \right)$. The choice of $x$ and $y$ that achieves the maximum is unaffected by multiplying that product by a positive constant, so use this opportunity to get rid of the $\frac{1}{2}$. Let $f = xy(165 - 2y -x)$. Then

\begin{eqnarray*}f & = & 165xy - 2xy^{2} -x^{2}y \\\frac{\partial f}{\partial...... -2xy \\\frac{\partial f}{\partial y} & = & 165x - 4xy -x^{2}\end{eqnarray*}

Thus, in order for the maximum to be achieved, we need both derivatives to be equal to 0. Thus, we need

\begin{eqnarray*}165y - 2y^{2} -2xy & = & 0 \\165x - 4xy -x^{2} & = & 0\end{eqnarray*}

Which is, conveniently enough, two equations in two variables, so we can probably solve this. First simplify as

\begin{eqnarray*}165 - 2y -2x & = & 0 \\165 - 4y -x & = & 0\end{eqnarray*}

Then as

\begin{eqnarray*}2y + 2x & = & 165 \\4y + x & = & 165\end{eqnarray*}

And finally

\begin{eqnarray*}x & = & 55 \\y & = & \frac{55}{2}\end{eqnarray*}

and we get

\begin{eqnarray*}z & = & \frac{55}{2}\end{eqnarray*}

from Equation (1).

Thus the maximum volume you can send using this Fedex rule is $55 \times \frac{55}{2} \times \frac{55}{2} = \frac{166375}{4}$ cubic inches, or about 24 cubic feet. This many walnuts weighs over 300 kg, so presumably the problem was motivated by something lighter.

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Final Olympic Medal Race Results

A few days ago, when I wrote a post proposing a metric to rank countries in the 2008 Olympics, it wasn’t yet over. Now that we have final results, let’s see how we rank:

China   100%
United States   94%
Russia  60%
Britain 43%
Australia       38%
Germany 36%
South Korea     29%
France  29%
Italy   23%
Japan   21%

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Who is Winning the Olympic Medal Race?

Several rankings have been proposed as to how to determine which country is winning the Olympic medal race.¬† Here I’m only going to look at the medals in isolation.¬† Folks have suggested dividing the medals by the countries’ populations, by the GDP or by the number of athletes.¬† You could factor in the sport popularity, fourth place rankings or time zone differences.¬† You could even rank the countries by how well they exceeded expectations on online betting sites.¬† I would like to see that.¬† But it’s too complicated to do here.

Back to the medals alone.  That means three numbers per country.

Clearly,
1) Using the total number of medals leaves out some important information.  3 golds is obviously better than 3 bronzes.
2) Using the number of golds leaves out some important information.  3 golds and 3 bronzes is obviously better than 3 golds.

I’m not the first one to suggest weighting the numbers.¬† It’s progress, but it punts on how to do it.¬† I’ve seen 4, 2 and 1. I’ve also seen 5, 3 and 1.¬† It’s pretty easy to make up numbers, but it’s more satisfying to have a rationale.¬† Here I propose the <i>distinguishing power</i> rationale.¬† There’s only one gold medal.¬† There’s also only one silver medal, but somebody else did better, so silver has half the distinguishing power of gold.¬† Similiarly, bronze has 1/3 of gold.¬† The actual weights don’t matter as long as the ratio is right.¬† 6, 3 and 2 are the smallest integers that express the distinguishing power ratios.

And finally the rankings (today), as a percentage of the leader:

China   100%
United States   93%
Russia  49%
Britain 43%
Australia       36%
Germany 31%
South Korea     27%
Japan   24%
France  24%
Italy   20%

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Permutable Dates

Gregorus seems to have already posted about this, but I can post my solution. We came up with the puzzle of finding the 8-digit number (with zeros allowed at the beginning) which has the largest possible number of permutations that are possible dates. Some numbers like 99,999,999 don’t have any. Others like 01,234,567 have a lot (starting with 01/23/4567, and most of those digits can be swapped around).

I wrote my program in the Bash shell language and it has two lines.

Line One:

perl -e 'use  Date::Manip; my $d = ParseDate("01/01/0001");while(UnixDate("%Y") < 10000) {print UnixDate($d,"%m%d%Y\n"); $d = DateCalc($d, "+ 1 day");}' >> all-dates.txt


This just makes all the possible dates, one day at a time. Perl’s Date::Manip module is notoriously slow, and it took almost two hours to make all 3,652,059 dates in this millenium (not quite actually: the last year of the millenium has a five digit year). As for the delay, I rationalized that I had optimized my code for “programmer time”.

Line Two:

perl -ne 'print join("", sort (split("", $_)))' all-dates.txt | sort | uniq -c | sort -n


This sorts the digits in each date to make ordered digit clusters, then orders the clusters of digits by how many times they appear.

The results? 00,123,578 will make 2,472 dates! Of course, it is a 20,160-way tie, because any permutation of 00,123,578, such as 57,800,123 will make the same 2,472 dates.

Unfortunately, I can’t make any dates near to us from 00,123,578. The closest I can get on both sides is 07/08/2135 and 03/20/1875. Can anyone do better?

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Clockwork Powers

Here’s another problem I’ve been looking at lately:

Find a number $x$, such that if $j$ is any digit from 0 to 9, $x^{j}$ has $j$ as its first digit past the decimal point. For instance, 63.18 works for the first power $(63.18^{\b{1}} = 63.\b{1}8)$ and the third power $(63.18^{\b{3}} = 252196.\b{3}89432)$, but it doesn’t work for the second $(63.18^{\b{2}} = 3991.\b{7}124)$ where the first digit is 7 instead of 2.

Hint: The smallest such number is less than 10.

To prove that the problem is possible to solve quickly, I’ll provide the solution to a more complicated version. Suppose that the first two digits of $x^{j}$ must match $j$ instead of just the first one. This means that $x^{3}$ has “03” just to the right of the decimal point all way to up to $x^{99}$ having “99” just after the decimal point. Then the smallest solution is the following number:

95.01063709417263364386569653065071891390859441647675346849
01552762611728856071543459219055118914594166160224386992990
49592128996983422861693601042135900702684741868297684742429
253949832737577847233354…

The smallest solution’s digits actually never end so I had to cut it off somewhere. I choose the cut off point to be the first place where the remaining digits still worked. In other words, if you truncated the 4 off of the end, the number would be changed too much and would fail to meet all the conditions. Furthermore, I was forced to round up (because anything smaller doesn’t work).

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The Monkey Problem

I’ve been forwarded a classic problem, in bold below.

The Monkey Problem

A rope over the top of a fence has the same length on each side, and weighs one-third of a pound per foot. On one end hangs a monkey holding a banana, and on the other end a dead-weight equal to the weight of the monkey. The banana weighs 2 ounces per inch. The length of the rope in feet is the same as the age of the monkey, and the weight of the monkey in ounces is as much as the age of the monkey’s mother. The combined ages of the monkey and its mother are 30 years. One-half the weight of the monkey, plus the weight of the banana is one-fourth the sum of the weights of the rope and the dead-weight.

The monkey’s mother is one-half as old as the monkey will be in when it is three times as old as its mother was when she was one-half as old as the monkey will be when it is as old as its mother will be when she is four times as old as the monkey was when it was twice as old as its mother was when she was one-third as old as the monkey was when it was as old as its mother was when she was three times as old as the monkey was when it was one-fourth as old as it is now.

How long is the banana in inches?

In fact, I’ve seen this problem back when I was kid, though I’ve never tried to solve it, mostly because that last paragraph looked tedious and as though it would require a deeply nested set of multiplied and offset variables. It turns out however, to be simpler than I imagined. Here we go….

A rope over the top of a fence has the same length on each side, and weighs one-third of a pound per foot.

Everything else in the problem uses ounces, so let’s make it consistent. We’ll make up variables as go along. First we need rope weight and rope length.

\begin{eqnarray*}RW = RL * 1/3 * 16\end{eqnarray*}

On one end hangs a monkey holding a banana, and on the other end a dead-weight equal to the weight of the monkey.

Add in variables for Dead Weight and Monkey Weight.

\begin{eqnarray*}DW = MW\end{eqnarray*}

The banana weighs 2 ounces per inch.

Add in variables for banana weight and banana length.

\begin{eqnarray*}BW = 2 * BL\end{eqnarray*}

The length of the rope in feet is the same as the age of the monkey

Rope Length and Monkey Age.

\begin{eqnarray*}RL = MA\end{eqnarray*}

… and the weight of the monkey in ounces is as much as the age of the monkey’s mother.

Add in MMA for Monkey Mother Age.

\begin{eqnarray*}MW = MMA\end{eqnarray*}

The combined ages of the monkey and its mother are 30 years.

\begin{eqnarray*}MA + MMA = 30\end{eqnarray*}

One-half the weight of the monkey, plus the weight of the banana is one-fourth the sum of the weights of the rope and the dead-weight.

\begin{eqnarray*}1/2 * MW + BW = 1/4(RW + DW)\end{eqnarray*}

The monkey’s mother is one-half as old as the monkey will be in when it is three times as old as its mother was when she was one-half as old as the monkey will be when it is as old as its mother will be when she is four times as old as the monkey was when it was twice as old as its mother was when she was one-third as old as the monkey was when it was as old as its mother was when she was three times as old as the monkey was when it was one-fourth as old as it is now.

\begin{eqnarray*}MMA = \frac{1}{2} ( 3 ( \frac{1}{2} ( 4 ( 2 ( \frac{1}{3} ( 3 ( \frac{1}{4} MA)))))))\end{eqnarray*}

How long is the banana in inches?

\begin{eqnarray*}\textrm{Find } BL\end{eqnarray*}

No doubt these equations are easy to solve by hand, but why settle? We can write them in a form that can be solved by computer. Right now, I prefer the mathematical software Sage . Formatted for Sage, the equations look like this:

var('RW, RL, DW, MW, BW, BL, MA, MMA')
equations = [
RW == RL * 1/3 * 16,
DW == MW,
BW == 2 * BL,
RL == MA,
MW == MMA,
MA + MMA == 30,
1/2 * MW + BW == 1/4 * (RW + DW),
MMA == 1/2 * (  3 * ( 1/2 * ( 4 * ( 2 * ( 1/3 * ( 3 * ( 1/4 * MA)))))))]
s = solve(equations, RW, RL, DW, MW, BW, BL, MA, MMA, solution_dict = True) 
print s

And the output is:

18\}\end{eqnarray*}

Notice the banana weighs 11.5 ounces and the monkey only weighs 18 oz., so something is little implausible about the problem, but as mathematicians, that’s not our problem…

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