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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!!

I'm expert at initiating 'first time' events, like:

- asking a whitegoods service technician why my machine is making an xyz noise: "Sorry, sir, but we've never heard of our machines making an xyz noise before." (maybe I should practise up on making xyz noises over the phone, or send them an mp3 recording of said noise);

- asking Customer Service if there's a fix for this or that: "Oh, we've never been asked about this or that before. Hang on a minute". That's when you discover that your understanding of the length of time a minute takes is way off the mark!

You have your specs on inside out!

*in* those posts, not my search word. I sent MIF all the facts that I remembered...

The search word was *incircle*, which gave 25 results.

I was interested in 6 of them, 4 of which (years 2015 & 2016) opened up fine; but 2 (or was it 1? - I don't recall now) of the older ones (year 2014) opened up the 'forbidden access' page.

My symbolic version of the factorial method has about 15 fewer keypresses than yours.

Yes, I knew about censored words, but this one threw me because all I did was click on a search result.

But I clicked it several times, just in case there was some hiccough. If I'd only clicked once (or twice? or thrice?) I may not have been locked out.