Math Is Fun Forum

Hi all;

I've added post #4 to the thread that ktesla39 gave the link to.

In that post I included steps of how I used Imgur to display the image that appears there.

Done on my smartphone, a Samsung Galaxy A35.

Hi Paul;

I don't know the reason for that happening here on MIF , but this is what I do to get the approx size I want for my display:

1. Crop out what I don't need displayed (using M$ Paint, or some such). Cropping your image reduced it from 800px x 1778px to 743px x 530px, which removed about 72% of the original area!
2. Upload the cropped version to Imgur.
3. Enter Imgur's BBCode image link into my MIF post, and before pressing Submit, click Preview to assess image size.
4. If I want to change the image size, then immediately before the full stop of the image extension (eg, .jpeg, .png) I insert one of the following letters:
     s (Small square)
     b (Big square)
     t (Small thumbnail)
     m (Medium thumbnail)
     l (Large thumbnail)
     h (Huge thumbnail)

I copied your image, cropped it, and uploaded it to my Imgur.

To resize that image from original to medium, its original link https://i.imgur.com/8adTo66.jpg would become https://i.imgur.com/8adTo66m.jpg (note the inserted 'm' before the extension's full stop).

Here's the cropped image at original size:

And here's the image shrunk to medium size:

And here are the displays for all 7 size options (including the original) based on the cropped rectangular image I uploaded to Imgur:

You'll notice that the left and right sides of the 'small square' and 'big square' images are cropped. Unfortunately, Imgur does that automatically on rectangular images whose width exceeds their height, so if such cropping matters, either:
(a) don't use the 'small square' or 'big square' options on the rectangular image, or
(b) convert the rectangular image to a square (eg, by adding false top &/or bottom with image software).

I've used Imgur since 2017, and think it gives a very good file host service.
They've changed their webpage layout to a somewhat confusing one that I haven't totally sussed out, but (lucky for me) on one of their pages (I forget which one) they give an option to use their old style...which I select straight away!

I don't know of a better alternative file host.

I use a PC &/or laptop. I've only had a 'smart'phone for a few months and am still learning it, so if you're on one of those things and have Qs about what I've said, we may need someone else (Bob?) to help out.

Hi punhub85;

Yes...the adverbial forms of the two adjectives you mentioned ('facecious' & 'abstemious') fit: 'faceciously' & 'abstemiously'.

For hyphenated words, there's also the adverb 'half-seriously' (from the adjective 'half-serious').

Collins Dictionary (a British dictionary) has 'halfseriously' but not 'halfserious', only the hyphenated 'half-serious'!

The unhyphenated 'halfseriously' is valid in Scrabble games where British words are used.

I've done the following in Excel (spreadsheet...by hand, coz I don't know how to code progressive dinners):

Guests are numbered 1-28, and are colour-coded green (1-7), peach (8-14), blue (15-21), pink (22-28).

All guests attend all 4 courses and all 4 houses (the image shows all guests 1-28 in each of the courses and in each of the houses). They do this in groups of 7 (of which there are 16, all unique).

No one attends...
(a) the same house more than once, nor
(b) the same course more than once.

Everyone...
(a) fails to meet many of the other guests, and
(b) meets several guests more than once (the colour-coding helps identify that).

Same here...

Echo_23's last post (#13) appends a paragraph that doesn't appear in paulb203's post #1...

Hi ktesla39;

I revised my code to be much more Mathematica-like:

Code:
Length[Cases[Table[Divisors[i++],{i,999}],{_,_,_}]]

Output:
11

Hi ktesla39,

Good idea...I tried it with Mathematica, and here's my code and solution:

It tested all integers in that range, and for each integer that had only 3 divisors it printed the solution number, the integer number, and the divisors. When done, it gave the total number of solutions.

Edit:

Hi Bob;

Yes, I got that answer too...with a 27-character Mathematica formula using Sum & Mod that also runs in WolframAlpha.

Hi Kerim;

Yes, I have the same impression.

That's why I wrote these:
Post #2 - "or maybe there's some info missing"
Post #4 - "the OP's problem doesn't say anything about the sought area size"
Post #7 - "I would've thought that, ideally, there'd only be one location for the smallest AE+BG, and that we'd only need to report the area of CEG at that location.
               So maybe the OP's wording in post #1 is missing some information we need?"

The only satisfactory goal (for when AE+BG is smallest) has to be to find either:
(a) the largest CEG area; or
(b) the smallest CEG area.
The number of different CEG areas in between (a) & (b) is infinite.

I'm opting for (b), as (a) is too easy. But maybe it's supposed to be an easy puzzle?!

Unfortunately, the OP hasn't returned to the thread to clarify...

Here's an update to my post #9, showing the interesting length relationships a bit differently.

It includes a formula I hadn't thought of before:

That makes it easy to position G accurately on FB: ie, draw a circle on B, with radius a, and use the Intersect tool on FB and the circle to create G on FB at the exact right spot (so no more moving G along FB in search of the exact right spot!)

And hey presto, everything falls easily into place: the right angles, some other angles, the GB-length relationships, the smallest AE+BG and smallest CEG area...!

But I'm as close to finding a non-graphical solution as ever!

Hi Bob;

Here's an update to my post #9...this time showing formulas:

Video link: Triangle area (smallest) - formulas

The figures in the shaded boxes are governed by the formulas.

But...I still can't prove
1. that the smallest AE+BG only occurs with G on FB (which I think it does); and
2. that the smallest area of triangle CEG only occurs as shown in the image (which I think it does).

Edit: Just spotted something: triangles FGC and AEC are congruent. I wonder if that helps with anything...

Hi Bob;

Have worked it out...but it looks like you were testing me by deliberately misplacing the decimal point!

Here's my formula (with a = AE):