Math Is Fun Forum

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Hi Kerim;

Well, not really, unless the "I would like" in Nick's "I would like everyone to meet at least once" is only a desire, not a requirement.

01 doesn't meet 08, 15 & 22
08 doesn't meet 01, 15 & 22
15 doesn't meet 01, 08 & 22
22 doesn't meet 01, 08 & 15
02 doesn't meet 09, 16 & 22
09 doesn't meet 02, 16 & 22
etc, etc.

Ditto for 01, 08, 15 & 22 in your post #16.

Also, in post #16, guest 02 only meets 01, 03, 04, 05, 06, 07, 08, 15 & 22, and fails to meet any of the other 18 guests.

Just looking quickly, 03 does much the same as 02.

I didn't check further...

Sorry Bob, but I've got no real clues on how to go about solving that one.

I don't know how to solve progressive dinners with formulas (bobbym had some success, with Mathematica).

And so I tried with Excel (like I've done successfully with a couple in the past), but no such luck this time.

No, not at all.

I just love doing number puzzles, with or without computer aided software.

For me, tackling a challenging number puzzle is simply a nice way to relax.

Hi ktesla39;

No, it's not Python.

Mathematica has its own [proprietary] language, and you need to have the program (not free) to run the code.

I only have a basic knowledge of programming in it, as any Mathematica coder would quickly see if they saw my code.

I'm more familiar with BASIC, and was able to use some of that knowledge in Mathematica, which can run adjusted BASIC-like code, but with reduced efficiency compared to its own language.

Hi Bob;

Thanks for your method...I'll have to look at it tomorrow (bed now).

But before I go, here's my method:

Hi;

Whether I'm logged in or out, I get this message: "You do not have permission to access this page."

Hi Kerim;

Thanks for trying for a numerical proof of my graphical solution!

Xlnt! Excel Solver agrees with me on the AE+BG sum! That sum is also the length of FB, which is √28 ≈5.2915026222 (see my post #17 drawing).

However, your area (1.484614945) is nearly twice the smallest area that I got (≈0.742307489).

When G is anywhere on FB, AE+BG is always at its smallest sum (which I discovered by T&E), meaning that CEG's area can vary between its smallest (≈0.742307489) and its largest (≈6.9282032303).

In my previous post (#17), I showed how to accurately position G on FB at the location where CEG's area is smallest: draw a circle, radius √(12/5.25), on B, intersecting with FB at G. CEG's area grows if G moves along FB from that initial location, as the video in post #13 shows (G moving off FB does the same).

Well, that's odd!

Both images display on my desktop, my laptop, my TV (which has internet connection), and my mobile phone (yes, I finally got one of those clever bricks...sadly, I had to ditch my trusty 3G Samsung GT-B2710 when our 3G network shut down about 6 weeks ago).

Here they are from my Google Drive account, but in 'hide' boxes, because of their huge display size (I could've shrunk them, I suppose):

I hope you can see my Google Drive images!

Btw, can you see my images in posts #2 (3 pics), #5 & #9, and my videos when you click on their links in posts #7, #9 & #13?

I'm sorry, Kerim, but I can't follow much of your explanation.

The main reason for that is my low maths standard. My last year of school was 4th-year high, after which I went straight to work.

However, I enjoy doing maths puzzles, and have learnt some things along the way.

Good work on getting a solution!

Is my 'solution' (the one in the two shaded boxes in my previous post) anywhere near yours?

In the next day or two I'll have a closer look at what you did, and maybe I'll see something there to help me progress towards a solution too.

Thx - Phro

Ah, yes...that makes sense now.

What was also throwing me (apart from the 'overline' and 'm', neither of which I'd seen before), was your 0.5/2. I was using 1/4 (same thing).

Not to mention the multiplication dot, and writing the square longhand instead of with '^2'.

I gave up too quickly with trying to fathom your style out, but then returned to it later because I knew that you knew what you were doing, and that I should persevere.

Hi Bob;

I took some measurements when triangle CEG is smallest (from my observation) that don't occur at other CEG positions, and have shown them in the image below:

Btw, FB is the line along which G travels (from B to F), and is the "constant-slope trajectory" I referred to in post #4.

Here's a link to an animation video I made of CEG travelling its full distance along the x-axis: Triangle area. It uses the image details from post #5.
Note: The two bold-font boxes in the bottom right-hand corner are dynamic, continually updating as CEG moves along its path along the x-axis. The area box changes constantly...but the AE+BG box remains unchanged, confirming the accuracy of the FB trajectory slope.

I've had a good look to see if I could improve on my observation 'solution' technique, but no luck so far.

Edit: I've just realised that if my drawing actually is of the smallest CEG, then it's probably no coincidence that AEC & CGB are right angles, giving CE and CG their shortest route to AE and BG respectively. Any other angles there would stretch the size of CEG larger than what it is in the image.

Hi Bob;

I went 'P&O' discovery cruising again this morning, and saw that AEC and CGB are both right angles for minimum CEG.

I zoomed in at Geogebra's maximum level, adjusted CEG's current position a tiny bit, and the standard curved symbol for both angles magically changed into right-angle symbols.

As you can see from my image, I couldn't quite get the angles to actually be 90°, but that was close enough for Geogebra.

Maybe this info can help with what you're working on...