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I wouldn't have thought so, either. It's probably me...but I haven't found out yet where my error is.

It seems odd, though, that FindRoot's value for c (the central angle, theta) results in nearly exactly the same answer (3888.614460593728239463379167602`...see Out[3] in my previous post) as thickhead's 3888.6144 in post #14, which he said in post #23 was incorrect due to truncation. But that may be coincidental.

I've been working in degrees, in which theta values are:

Your 2t = 2.210259366455766942706949`;

G's γ = 2.21025936645665`;

FindRoot's c = 2.2101497331710919201003743`

Your t value (either in radians or converted to degrees) plugged into my h (string height) formula, gives the correct height (3889.0003152175473559561908582`); so my h formula is correct. The error source here, as thickhead said earlier about his error, is in the theta calculation.

I'll look again, but there's probably something going on somewhere that I don't have any understanding of and won't be evident to me.

Correct!

It'll never happen to me, though.

...so now you know when my birthday isn't.

Civil Engineering Onsite Handbook.

Btw, my notation in post #21 is a bit different from theirs, as I changed theirs to fit notes I'd been making before finding their site.

My is their And my is theirOne I didn't mention in post #21 is the formula for the arc length between the tangents, which I used to help find c (central angle) in M's FindRoot.

My arc length formula is theirMy r & their R = earth's radius

My c & their I = central angle;

My h & their E = height of string apex above top of earth

My t & their T = length of tangent from earth to string apex

My current M code, and yours:

```
In[1]:=
(*Mine*)
r=3959*5280;
c=c/.FindRoot[2r Tan[c \[Degree]/2]-100-r*c \[Degree]==0,{c,2.2},WorkingPrecision->100]
h=r*Sec[c \[Degree]/2]-r
(*Bobby's*)
R=3959*5280;t=(150/R)^(1/3);N[(2 R*Sin[t/2]^2)/Cos[t],100]
Out[2]= 2.210149733171091920100374337118113893066411007878966450185575166794692294275531251879912977959844808
Out[3]= 3888.614460593728239463379167602010692449187441461292338724354316092744378397068689244933991315452952
Out[4]= 3889.000315217547355956190858271992382571341435999597543245020468161721458410729392937937972940735646
```

In Geogebra, when I plug your result into point E's y value, I get height = 3889.000315219164, and central angle = 2.210259366456653 degrees. That angle figure differs from that of Out[2] in the M code above, and must be the reason for the height discrepancy.

Is FindRoot giving me the wrong central angle? If so, is that because of the formulas I gave it?

When I plug Bobby's W|A result (from post #20's formula) into G - which I hadn't tried before - I get 3889.00031521 (correct to 8 decimal places), so my G method seems to be good.

The error looks like being the central angle, for which I get different answers depending on whether I use Bobby's formula or the surveyor site's.

I've probably messed up the surveyor formula somehow...

Hi thickhead;

Sorry...made a huge blunder with the wording! Left an important bit out!

This is a puzzle given to me years ago by someone whose birthday fitted, and I'd forgotten about it until now.

Of course, I've lost the puzzle and repeated it from memory (which is never the best way for me), and only just now did I realise that there was a bit missing.

Fixed now, and should be a little more interesting.

Hi thickhead;

Thanks for your answer - and excellent thought - but single-digit ages are not exactly what I had in mind. Maybe my wording's a bit loose.

It's probably ok mathematically on occasion, but not really when applied to ages. That's here, down under, in Australia, anyway...but maybe we're a little different from the rest of you.

When their great day arrives, a 07-year-old kid is hardly likely to say to their classmates, "Hey! Guess what! I'm zero-eight today!!" Then they'll have zero-one heck of a job trying to explain to the other single-digit-year-olds in their class what on earth they meant by that! It'd probably take the kid until well after their 05-past-03 home time before any-zero-one would get it!

Please forgive me!

Hi Bobby;

I haven't been able to follow your workings or thickhead's as the maths is too advanced for me, but I thought that as I initiated this version of Mathegocart's problem I'd better try to find a solution to it.

I found these formulas on a surveying website:

Where:

r = earth's radius

c = central angle;

h = height of string apex above top of earth

t = length of tangent from earth to string apex

In M, I tried to 'Solve' for c, but couldn't get that to work. However, 'FindRoot' did, with the following result in M:

```
Input:
r=3959*5280;
c=c/.FindRoot[2r Tan[c \[Degree]/2]-100-r*c \[Degree]==0,{c,2}];
h=r/Cos[c \[Degree]/2]-r
Output:
3888.614460591227
```

I'm pleased that this is pretty close to my G's metric answer (for which I used r=40075km), which meant I was on the right track back then...

**phrontister**- Replies: 6

Here's a problem that I was inspired to post after seeing evene's question about J.R.R. Tolkien's year of birth, here.

What is the next birthday date on which some people will turn the age that is the last two digits of their year of birth, and that:

(a) has the same day and month as their birthday; and

(b) is the same number of days into the year as their number of birthdays. eg, someone celebrating their 43rd birthday must have their birthday on the 12th of February (ie, 31 Jan days + 12 Feb days = 43).

Hi evene;

Edit: The square root sign and subscript '2' that I used from the 'Useful symbols' at the top of the page don't seem to be recognised anymore (well, not in my Chrome, anyway), and turned into accented characters instead.

Maybe that's why my answer doesn't make sense to me, as I said earlier. I'd expected the correct answer to be closer to the height of an atom (Mathgocart's answer *a*).

According to an online calculator into which I entered the earth's diameter (or radius?) and some other figure I don't recall, 3890 feet also happens to be the perpendicular distance from the top of the earth down to the chord between the two tangent points. I thought that was rather coincidental, and so I started to doubt my understanding of my version of the problem...but the huge scale isn't helping me to get my head around it.

Sorry...have to go out for 5 hours or so.

That's nearly exactly what I got, thickhead.

My answer in Geogebra was 3890 feet, rounded to the nearest integer.

I was going to try mathematically, but haven't worked out a solution method yet.

And here is a link to my Geogebra file.

Hi all;

I understand (maybe wrongly) from Mathegocart's explanation that the shape for the longer string version is circular (like the original string), but in an experiment in which I used my basketball the shape is quite different (see image).

The string curves around the ball tightly from the base to the tangent points (the black dots), and then both sides continue as straight lines up to the suspension point...the apex.

Transferring that shape to Mathegocart's problem I've tried to work out the distance from the top of the earth to the string's apex, using the popular integer measurement of 40075km (that I converted to imperial) for the earth's equatorial circumference and basing the diameter on that.

But...I'm getting an answer that doesn't make sense to me.

What answer do you guys get?

Thanks!

I can't say at this time.

But the deafening silence from Mr Google and SE confirm what you said.

Hi Bobby;

Having first encountered PDs here on MIF and only done a few, I don't know all that much about them, but it seems that small differences in wording could lead to big differences in solution strategies...which might be difficult to cater for in one solve-all algorithm.

I can't think of anything that would work, and googling dug up nothing that helped, but here's something from the 2009 "Another progressive dinner" thread:

MathsIsFun wrote:

Bobby: could this be automated? "Progressive Dinner Tool" ... ?

bobbym wrote:Hi MathsisFun;

I can't say at this time. Although it took a lot of computer help, it also took a lot of human trial and error. In other words I don't yet have an algorithm. I am going to work on her other problem and maybe see if it can be automated.

Hi Bobby;

No, I don't, sorry. I just drew it up by hand in Excel.

Option A was very straightforward, of course, but B took a few stabs at it before I got a system that worked (by adjusting A)...and that system can be used as a springboard for deriving more grid combinations.

Hi Eulero;

Hi Calabria;

For a real-life scenario I can offer the following two options:

It meets the brief in post #1 (as I understand it), and suits group placement: eg, for families with young children, so that they may sit together.

This one has a constraint that no guest may sit with the same guest at more than one table, and suits maximum guest mixing.

It can also be used for group placement (though to a much lesser degree than A's) through name/number assignment: eg, the 3-guest group {1,2,131} places a different combination of two of them at one table for each course.

I've assumed that all tables serve courses 1 to 3, in that order.

*tangent radius*) is perpendicular to both.

Hi dazzle1230;

How do you know that AB=LM without using geogebra?

AB & LM are opposite sides of rectangle ABML (from post #24):

Proof that H is CD's midpoint:

(i) ABML is a rectangle (tangent perpendicular to radiustheorem [see 'Tangent Angle' at bottom of page here] for points A & B; AB is parallel to CD [see post #1]; AE extends to L)

Those constraints still yield multiple solutions, but I don't know how many there are in total.

Here are twelve of them:

I could've saved a lot of time if I'd thought of doing the same!

True!

A similar program in BASIC:

```
x=1
y=2
d=1
ppd=0
WHILE d<13
ppd=ppd+x
x=x+y
y=y+1
d=d+1
WEND
PRINT ppd
```

And...welcome to the forum!

Hi Bobby;

Did you deliberately misconstrue the intended meaning of my unintentionally ambiguous statement? If so, I think we're on the same team!

I meant *overlook* as in *fail to consider*.

I don't know enough M to go the way of proper structure most of the time, so when it occurs to me to use Flatten, I do.

Ah...I see your problem! You need them!!