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Thanks, Bob...glad it works for you.

But the OP said:

harpazo1965 wrote:

Thanks but I don't have a computer or a laptop to do all that fancy stuff you talked about.

What device did you try it with, Bob? I'm hoping it was a smartphone, because, although I don't have one of those and can't perform the test, I thought my method would've been non-fancy enough to even work on a smartphone.

They have the capacity to perform the kind of copy/paste I referred to, don't they?

Hi Bob;

I use a PC, and so maybe my experience differs from how the OP's device works (or can work) with Imgur, but this is what I've observed and got:

The OP's address isn't to the image itself...it's only to the Imgur web page with the image on it (amongst other things).

An image address has an image format extension, such as PNG, JPEG...which isn't the case with the OP's 'https://imgur.com/a/MMz2Kir'.

Solution:

I copied the OP's url into my browser's address bar, right-clicked the image that appears on the resulting Imgur page, selected 'Copy image address' from the drop-down menu, and then pasted that address into MIF (in 'img' tags nested inside 'hide' tags)...which gives me:

You'll see that the image address is very different from that of the OP's.

Maybe my method would work for the OP on his device...

J : Jij (Dutch, meaning 'you')

Z : Zzz (sleep)

Hi CurlyBracket;

* Question 6: How many children do they have?*

They have 3 children.

Yes, that would do it.

That also works in the event of there being joint winners, who would then be named as such.

Hi CurlyBracket;

Yes, I like that method.

Just a comment re the puzzle wording:

I know what you meant with it, but, in a strict interpretation of the wording, answering the second question could have been a bit awkward if there was a tie...and there are 3 scenarios with at least one winner in which a tie occurs.

Here's a link to a video I created from my Geogebra file.

Marquis De L'Hôpital's pulley puzzle

The animation is from the Slider function that I linked to BC values, and is set to oscillate between BC=70 & BC=72, with 0.00001 increments.

I didn't capture all the increments, of course(!), although it can be done.

Changing BC values moves Point C along Circle A's circumference, impacting on other objects' measurements...including the length of ED.

The oscillation shows ED approaching and reaching its max of ≈51.90438 at BC≈71.322, continuing past it until BC=72, then returning past it until BC=70, and finally back again to ED's max of ≈51.90438.

Jeremy Desmond wrote:

Relying on trigonometry I used my calculator to work out the various altitudes of the weight depending on different angles of BAC, and I repeated the laborious calculation to gradually home in on the maximum result. I got the answer to be 51.90438 to 5 decimal places. This method gives the right answer but in the most boring and prosaic manner possible. It misses out on the beauty and elegance inherent in math.

Oh well...I'm sure I'll get over it!

Here are some interesting links about L'Hôpital's Pulley Problem:

Simple Geometric Solutions to De L'Hôpital's Pulley Problem

Optimization – L'Hôpital's Pulley Problem

Proof of length of line segment x

Bob wrote:

Triangle BCE is isosceles...

Shouldn't that read CBE, Bob?

But maybe placing the letter of the vertex angle of an isosceles triangle in the middle is just a habit of mine...similar idea to the order in naming angles.

*Edit: I deleted my answer, which was just the number. I posted it very prematurely without giving any help at all, despite this being the Help Me! forum. Sorry about that!* >blush<

Hi Jeremy & Bob,

The maths is beyond me, but I wanted to try graphically, drew this up in Geogebra, and got ED ≈ 51.90438 to 5 decimal places.

I moved C (radius 100 - BC) along circle A (radius 40) to get the largest ED (approx).

Is that anywhere near the mark?

Hi CurlyBracket;

** Question 5: The Woodcutting Competition**:

I think that there was no tie, and Third won, the felling ratio being Third:Second:First = 5:4:3.

*EDIT: Included my solution method...*

Hi Median Joe;

I opened my Excel file in LibreOffice Calc, and it ran perfectly...including the Goal Seek function.

Sorry, Median Joe, but the Goal Seek image had some errors in column D. Fixed now.

Hi Median Joe;

Here's an image of the worksheet for my solution using Excel's What-If Analysis (Goal Seek).

Goal Seek's settings:-

Set cell: $A$11

To value: 44

By changing cell: $B$3

As I mentioned in D3, the solution can be obtained by changing B3 until A11 becomes 44. Doing that manually doesn't take long at all.

Hi Median Joe,

...I don't quite understand this line :

At that point, Ann was (3/2)(M−k)−(M−A).Not quite sure why you subtract M - A?

I'm confused. Did you deliberately omit the brackets?

Anyway, if (M−A) weren't subtracted, M & A's ages would be identical *at that point* (which they're not). Mary's age *at that point* is stated by the 'bright spark' as being (3/2)(M−k).

I used the same method: see post #4's *"Note: Their age difference at each stage is a constant, which in the first stage is established as being 2c (ie, 3c - c)."*

Hi Median Joe;

I didn't think of trying that Edna method, but your Maple code sparked my interest.

I don't know Maple (nor have it), but the following works in WolframAlpha here and in Mathematica, with the same result as yours: Solve[{m+a==44,m==2(a-x),m-x==(a+z)/2,a+z==3(m-y),m-y==3(a-y)},{m,a,x,y,z}]

I should be in bed!

Hi Median Joe;

Yes, I also got the answer 27.5.

I'd forgotten about the algebraic method I showed in these two threads I opened back in 2009 - Rita's age & Edna's age - and instead went straight to Excel's Goal Seek to solve your puzzle.

Below is my solution to your puzzle, adapting the algebraic method I posted here:

============================================================

Let c = Ann's age when Mary was 3 times as old as Ann.

*Note: Their age difference at each stage is a constant, which in the first stage is established as being 2c (ie, 3c - c).*

```
Mary Ann
when Mary was 3 times as old as Ann was 3c c
when Ann is 3 times as old as Mary was 9c
when Mary was half as old as Ann will be 4.5c 2.5c
Mary is twice as old as Ann was 5c 3c
```

So Mary's present age is 5c and Ann's is 3c, from which 5c + 3c = 44 (their given combined present age).

Solving: c = 5.5

∴ Mary is 27.5 (5*c = 5*5.5) & Ann is 16.5 (3*c = 3*5.5).

============================================================

The following equation is based on the above info, and evaluates to c = 5.5:

I haven't looked at your solution yet, nor your "Not quite sure why you subtract M - A?" question, but I will...

Hi Median Joe;

I'd suggest structuring from right to left.

That's the method I used for my equations, and I think it worked coz my solution makes sense at each stage (well, to me it does, anyway).

I cheated by using Excel's Goal Seek for the final solve computation, but I'm working on adapting my equations to obtain an algebraic solution to replace Goal Seek's 'what-if analysis' approach.

Struggling, though, and I may not get there!

Hi Bob & CurlyBracket;

I solved it this way:

A = B^2 - 108

B = A^2 - 158

Solution 1:

Substituting B into the first equation resulted in A^4 - 316A^2 - A + 24856 = 0 ... which I couldn't solve longhand, only online (eg, WolframAlpha) or with CAS (eg, Mathematica, Excel).

Answer: A = 13

Solution 2:

Trial & error: Evaluating the first 2 equations with A = 13 yields B = 11, evaluating with B = 11 yields A = 13, and figures in either direction from those yield increasingly large errors.

Answer: A = 13

Well, yes...it is now, but it began 30/10/2021 as an apparently genuine maths 'problem' (see post #2, which I edited to include the original 'problem' that the OP since replaced with the chair ad, possibly on 26/11/2021).

I said 'problem', because the level of simplicity of the OP's 'problem' raises the suspicion that the post was a bogus springboard to be used later, as it now seems to have been, for placing the ad via an edit to the original post.

A moderator should look at this...

Hi CurlyBracket;

Yes, that's my reasoning also (see post #2), and my m=RIGHT((2016-ax-by)/3+x+y) expression is an attempt to state it algebraically.

I probably cheated by employing spreadsheet's RIGHT function to extract the last digit of 708, but used that to keep the expression small (though non-standard).

CurlyBracket wrote:

...how did you get this expression?

m=RIGHT(x+y+z-(ax+by+cx-2016)/3)

Sorry...was a bit hasty there.

This is better:

m=RIGHT((2016-ax-by)/3+x+y)

Is that what you got?

I've fixed my previous post.

Hi CurlyBracket;

Nice one!

I hope I got it right!

*Edit 5/5/2022: Improved my solution (see post #4)*

U : Unau *n* a two-toed sloth

V : Vav *n* sixth letter of the Hebrew alphabet

X : Xerox *v* to produce a copy (of a document, etc) using a machine employing a xerographic copying process

All three are valid Scrabble™ words (English): Scrabble™ Official Word Check

Q : Qajaq (Inuit) *n* a kayak

Q : Qulliq (Inuit) *n* a type of oil lamp used by Inuit people

Both are valid Scrabble words: Scrabble™ Official Word Check

That link is to the Collins online Scrabble word checker that uses the official tournament word list, including over a quarter of a million permissible words. It's endorsed by Mattel and the World English-Language Scrabble™ Players’ Association.