Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

Post #3 by Alexander_Thomas contains a gambling link...

Hi Checker 555!

I went to the Checkers game, and also got a pop-up that prevented play.

However, my pop-up has a 'New Game' button at the bottom, and clicking that makes the pop-up disappear and starts a new game.

The pop-up also has a 'Close' button next to the 'New Game' button, but clicking that only closes the pop-up. To start a game you'd then have to click the 'New Game' button above the playing board.

I hope your pop-up's the same as mine!

Hi tony123;

Hi Bob...the guitarist is Gerhard Gschossmann.

And here's the link to the recording I used: Gerhard Gschossmann - 'Summertime' (George Gershwin) guitar solo fingerstyle

That version is played at the song's correct tempo, with good feeling and expression...which makes my Geogebra soundtrack sound rather chipmunkeyish.

Here's a vocal clip of the song: GEORGE & IRA GERSHWIN 'Summertime' HAROLYN BLACKWELL ('Porgy & Bess'). Very moving!!

Hi Bob,

I added an instrumental soundtrack to my video, and I'm sure you'll recognise the tune. This piece is played by a brilliant guitarist, in an amazing finger-picking style that is his own arrangement. I think you'll like it!

The original @ 62 seconds is much slower than my soundtrack, which I sped up to squeeze the whole of the first verse into the 46-second Geogebra recording.

Bob wrote:

But I'm no nearer a proof.

I hope you can find it!

Hi Bob,

Here's a little video of an animation of the range of non-symmetrical scenarios obtained by running Point B along line AP:

Non-symmetrical range (video)

Edit: Sorry, I lied. The range includes the symmetrical scenario I mentioned in my previous post. The symmetry appears 3 times: at the start of the video, at the 14-second mark, and at the end of the video. Each time, N is smack in the middle of the square and the 4 vertices are 67.5°.

Hi Bob,

I've returned to this puzzle again, hoping to get somewhere with it this time...

I think I've found a proof (with the help of Geogebra), but only in the case of symmetrical shapes as per this image:

The symmetry only occurs when N is *precisely* in the centre of square APDO.

I've typed my workings onto the drawing, and I hope it's all clear enough.

Btw, 12 for AP, PD & BF (all equal, per post #1) is just an arbitrary figure I chose to get some numbers into my drawing so I could use Geogebra.

The hide box below has two more configurations, each obeying tony123's rules, with Geogebra showing that the enclosed angle is 45° in both cases:

For the last 2 drawings I just slid Point B up/down the AP axis (the stops were random locations). The interconnectivity between components (eg, 'Perpendicular Line', 'Point on Object') enabled the whole BCFE figure to move and change shape automatically while still obeying all OP rules and the 45° angle requirement.

I have no idea how to prove a non-symmetrical solution, though squillions (maybe limitless?) of such solutions exist (that is, as I understand it)!!

Hi Bob,

I suspect that making a square with those extensions will help!

I've redrawn the diagram:

It also 'happens' that PO runs through N...and, of course, PO divides right angles APD and DOA into 45s.

...and there are umpteen similar triangles (but I haven't seen anything yet).

Hi tony123,

Another one of your puzzles that I can't solve!

I drew it up in Geogebra (see image). That gave me some angle pairs I hoped might help, but they haven't triggered anything...

Thanks, coolboyx12.

I resign!

...not that I think there's anything wrong with the thoughts I've presented!

Hi ganesh & coolboyx12;

I've redone my work (using Excel tables this time), with calculations as per the options named in the hide buttons below:

Both options include scenarios for insurance premiums charged for 3 days and for 4 days.

Thanks, coolboyx12.

So I've looked at ganesh's method, and have some comments about it:

A. Depreciation:

The problem doesn't mention depreciation, nor its percentage, but ganesh's method includes a yearly fixed depreciation rate of 10%, applied to 3 years.

1. There seems to be no generally-used standard regarding the amount of vehicle depreciation: instead, the depreciation percentage depends on many factors, such as the make and model of the vehicle, its age and its use (eg, private or business).

2. Vehicle depreciation rates are not necessarily fixed (eg, diminishing rates).

3. If we assume that depreciation applies, it's important to know when it takes effect: eg, whether it's at the beginning, or at the end, of each year (often a vehicle's value depreciates immediately a person takes possession of it). If applied at the start of each year, it should be applied in each of the 4 years, not just the first 3 (the damages claim was in year 4).

B. Insurance premiums:

There are 4 insurance terms ('After 3 years he got a claim of damages'), but ganesh's method only charges for 3.

Another thing unclear to me is the huge discrepancy between the initial worth of the vehicle (Rs. 450,000) and the subsequent 'claim of damages' (Rs. 750,000).

To me, it implies that the initial worth (unadjusted over time) was simply used as a basis for premium calculation, and that the insurance policy actually covered a higher amount.

Back again...

What exactly does the phrase 'he got a claim of damages' mean?

My first understanding was that he'd received a bill from a repairer to fix Rs. 750,000 worth of damages, but I see now that it could mean that he'd lodged an insurance claim and been paid that amount by the insurer.

I have other Qs about the wording rattling around in my head, but they may change depending on the answer I get to the above.

Hi coolboyx12;

Edit: On re-reading the question, I've come to the conclusion that I don't understand the question at all! So maybe you should ignore what I wrote in the hide bar.

I'll think about it some more, but there seems to be some critical info missing...

Hi!

I did it like this:

D = original number of production days (10).

Chairs: final 16 is > original 8, resulting in increased production days of:

(a) D x 16/8.

Carpenters: final 10 is > original 2.6, resulting in reduced production days of (a) x 2.6/10.

The resulting sum:

Sorry, I didn't see the 2 previous posts until now...but I'll post mine anyway.

Hi tony123;

Too tough for me, so I tried trial and error in Excel and got this:

Hi Bob;

Can anyone find a simple edit that makes the solution unique?

Here's a simple edit to Clue 5 that produces a unique solution by replacing 'To the right of' with 'Next':

5. Next to the ship carrying cocoa is a ship going to Marseille.

Clues 7, 11 & 14 also use 'next'.

Hi jiaoqian, and welcome to the Forum!

I did this puzzle many years ago, and, like you, could not resolve the French/Brazilian cocoa/tea issue.

That means there are two solutions, which, for me, is disappointing.

I've done many of these types of logic puzzles, and the Ships Puzzle is one of the very few I've come across that has multiple solutions...

Hi Keckman,

Keckman wrote:

There is six number in divide operation and you get seven right numbers of pi. But is there even better ones? I guess there is not same kind rational number under one trillion.

Well, according to my research, there's a very big jump from your 355/113 to the next one (103993/33102): your Pascal program might be quite busy for a while.

I initially thought of trying brute force like you did, but changed my mind after I googled 'oeis pi fractions approximations 22/7 355/113 104348/33215' (as a single-line search using your fraction, imcute's and the 22/7 I know)...which introduced me to a minefield of info about this topic!

I started here: OEIS (sequence of convergents to Pi): A002485 (numerators) & A002486 (denominators).

Those two links (and a StackExchange post) quoted the following Mathematica formula for obtaining a sequence of increasingly accurate fraction approximations: Convergents[Pi, x].

That interested me as I have Mathematica. I chose 10 for x (for an output of a sequence of 10 fractions) and ran the code in my program and also online at WolframAlpha, with the following output in both:

3, 22/7, 333/106, 355/113, 103993/33102, 104348/33215, 208341/66317, 312689/99532, 833719/265381, 1146408/364913.

The 2nd fraction in the sequence shows the well-known 22/7, the 4th is yours from post #1, and the 6th is imcute's from post #2.

Links to a couple of other sites I looked at:

Wikipedia: Continued fraction expansion of π and its convergents.

Wolfram MathWorld: Pi Continued Fraction.

Hi;

I thought I'd try Excel's Goal Seek, and managed to solve this thread's cubic and a couple of others I tested it on:

The two extra cubics have 3 roots each, and to find them I entered into column D the initial x values that I'd predetermined and stored in Column C, those values being:

(a) a high initial value to try to find the highest x;

(b) a low initial value to try to find the lowest x;

(c) half the sum of the resulting highest & lowest x values from (a) & (b) to find the remaining x (not sure if that's a good idea generally, but it worked for these equations).

Then I ran Goal Seek (individually for the equations in rows 2, 3, 5-7 & 10-12) with the respective Goal Seek boxes filled from columns E & D as per the image example.

Btw, to increase Goal Seek's precision (quite a necessary step for good accuracy with this method), I first set 'Maximum Change' in Excel's File/Options/Formulas/'Enable iterative calculation' to 0.000000000000001, which gave accuracy to 14 decimal places for this thread's cubic: i.e., x ≈ -2.09144638072228 (the last few digits aren't displayed in the image).

The above cubics are the only ones I've tried this Goal Seek method on...

Hi imcute;

imcute wrote:

5x^3+10x^2=-2

then you factor out 5x^2

x^2(x+2)=2/5

The resulting equation after factoring out 5x² is missing something...

**phrontister**- Replies: 1

Hi Bob;

I thought I'd keep this going, even though the other thread is closed.

We've done a fair bit of work on this and maybe it's worth trying my latest finding, so that, if we're successful (hope against hope!), we have the solution for anyone who needs it.

Sorry to bother you further after the disappointing direction this episode has taken and just leave it if you'd rather, but maybe still keep my thread open in case someone else takes an interest in it later.

Cheers,

phro

This is what I wrote on the other thread:

Bob wrote:

I'm using Android. Tried all sorts of image sharing but have yet to get the bcc version. Elsewhere on the help it says bcc is not an option for mobiles

This is from a Quora post:

"Unfortunately there is no option to upload photos to imgur using your phones web browser because they want you to download their app. But some mobile browsers (like Chrome) have an option for you to view the desktop version of a website. If you select this option, it will load imgur as it would on your desktop PC and you can upload images that way."

Here's a link to some info on enabling/disabling the desktop version view on a range of browsers: How to View Desktop Version of Any Site on Mobile.

Maybe the desktop view gives the ability to copy the url of the actual image (ie, one with an image extension) instead of an imgur web page?

Hi Bob;

Bob wrote:

I'm using Android. Tried all sorts of image sharing but have yet to get the bcc version. Elsewhere on the help it says bcc is not an option for mobiles

This is from a Quora post:

"Unfortunately there is no option to upload photos to imgur using your phones web browser because they want you to download their app. But some mobile browsers (like Chrome) have an option for you to view the desktop version of a website. If you select this option, it will load imgur as it would on your desktop PC and you can upload images that way."

Here's a link to some info on enabling/disabling the desktop version view on a range of browsers: How to View Desktop Version of Any Site on Mobile

Maybe the desktop view gives the ability to copy the url of the actual image (ie, one with an image extension) instead of an imgur web page?

harpazo1965 wrote:

Please copy and paste the following link. Tell me if you can see the image.

https://imgur.com/gallery/PXM8kTT

Yes, I can...as part of the following Imgur web page (a screen capture):

However, while your link works for me now, I've found from your other posts that your web page links only work for a limited time, if at all.

Bob wrote:

Worryingly, I don't remember a thing from that thread.

Oh dear!