Math Is Fun Forum

Hi Bob;

Your wife would have had an argument on that topic with bobbym, who said this about that! (which mathsyperson later commented on)

Anyway, I did as you did, and resorted to 'cheating' by using W|A to solve a horrible equation.

The equation was for the sum of the lengths of the arcs in the M, R and L circular sectors = 2πr...and W|A said that r≈15.5175.

Sorry, sonu1997, but I'll decline your invitation.

I suggest you provide some more info for general discussion here, which could generate more interest in the thread and maybe even an outcome.

Btw, here's my code:

My clunky general code just uses Mathematica as the vehicle...

EDIT: Included a True/False check in my second code

Hi sonu1997 & irspow;

I also tried, and failed...but that doesn't mean much!

I wonder if the equation's been entered correctly, though.

The reason I ask is that in case the trailing constant (which I'll call 'c') happens to be wrong, I tried a range of alternatives for c...with some success:

Using Mathematica, I found 31 solutions for {x,y,z}, subject to the following constraints:

1. x = between -100 & 100
2. y = between -100 & 100
3. c = between 100 and 999, containing at least two digits of the OP's '114' in their correct positions (in case c contains just one incorrect digit).

Three examples of my solutions:
(a) x = 7, y = ±48, z = 46, c = 118
(b) x = 10, y = ±69, z = 66, c = 164
(c) x = 5, y = ±4, z = 6, c = 514

Hi Bob;

I found a proof for 0<x<0, but it doesn't quite satisfy 'always'.

Hi Bob;

Nice to have another go at something.

I see that your last post came before I finished an edit to my previous post (#5), and so you might like to have a look at that one.

I wrote a small, clunky, BASIC program to get my results, and can't think of a cleverer approach.

It's interesting that all my results for the lowest n have n=131...which looks like a pattern to me.

But it all seems rather frivolous, and so I think I'll leave it there.

Hi Nasir;

I got exactly the same as you did (including the two given questions), and, like you, didn't include numbers with repeated digits (in the spirit of the puzzle).

I initially considered omitting questions whose incorrect copy was also one of the original questions - which would reduce the number of questions (including the two given questions) to 14 - but decided against that because Hungry Horace wouldn't have noticed the 'repeats' as he hadn't recorded the originals.

Here's a link to the puzzle: Sides Reversed Is Puzzle

Hi Bob;

Sorry, but for brevity's sake I omitted some steps and comments...which are:

1. I drew LM, which from the 3:4:5 triangle rule on ΔLKM = 5.

2. Then I drew KC = 5 (from being parallel to LM, and KL being parallel to BM).

3. CM = 3 (from the 3:4:5 triangle rule on ΔCMK).

4. As the opposite sides of LKCM are parallel, LKCM is a parallelogram.

5. Also, LKCM's 4 angles and side lengths are constant because of the right angles at K and M, and so LKCM retains its 'fixed-dimension shape' while it rotates around its hinge point L...a critical feature enabling me to draw the conclusion I did about AB and BC changing in length by equal amounts.

I'll have to look at your proof tomorrow coz I'm off to bed now, but thanks for your work!

Hi Bob;

The Escher Waterfall drawing (see image) uses certain visual techniques to show water flowing uphill from the wheel to the upper aqueduct...which water can't do.

Likewise, the Wiki drawing uses incorrect visual sizes of A's intersection angles and BC's intersection lengths, and shows AOD as a crooked line...which it can't be.

Conclusion: Escher's and Wiki's drawings are illusions, each representing something that is untrue as being true.

Conversely, my drawing in post #12 shows that LC is actually a straight line, with M lying on it.

Hi Bob;

Wow...that's quite some proof!

However, the trig's too advanced for me to understand it...but I'm happy to wait for a simplified solve if you can find one.

I'm glad, though, that your proof confirms that my solution is correct.

Btw, I don't think that AKM is a straight line (see my Geogebra image below).

Hi Bob;

The only dimensions given by the OP are those of 3:4:5 triangles, and therefore I strongly suspected that the solution would involve such triangles. (barring red herrings, that is!)

So I drew the LABC rectangle, which led to the image below (Geogebra) and my post #2 solution.

That's probably not a proper proof as such, but I've failed to find one of those things, though I've spent some time trying.

And I can't find any other solution...

Hi Bob;

I received my Excel file ok, thx.

Only had time to have a quick look at it, and all I did was check for duplications in the columns by using the UNIQUE function that's part of the new Excel Dynamic Arrays family.

Example: Entering '=COUNTA(UNIQUE(B2:B47))' in B48 returns '46' there, confirming that your 46 column B entries are unique. Similarly with Column G.

I'll have to take a closer look at earlier posts to try and understand this thread before examining further. Previously I've only skimmed through the posts. 

Might take me a bit of time, though!

Hi Bob,

Aren't 2 & 6 the same? Or are my cataracts playing up?

I thought it wouldn't hurt to have someone else check carefully, too.

I'm onto you there!

I used Mitch's spreadsheet layout from post #4.

Couldn't get the right answer with MIF's calculator. Tried quite a few options that made sense to me, but not to the calc!

I got a bit further with Mathematica, but need Bobby to tidy up some things for me cos I just don't know enough!
Edit 24/4/2021: Got my Mathematica code to work, and it found & printed the 46...just not in the Excel format.

Edit 23/4/2021: I overlooked a combo in Excel (making 46 instead of 45), and have fixed the image.