I can't vouch for his behavour but his use of English is truly shocking :-)))

Christopher_MMember

431 replies

Christopher_MMember

I can't vouch for his behavour but his use of English is truly shocking :-)))

Christopher_MMember

*behaviour* ;-)

Don AtkinsonMember

Christopher_M posted:*behaviour* ;-)

Who is confusing who ?

Adam MeredithMember

Don Atkinson posted:Christopher_M posted:*behaviour* ;-)

Who is confusing who ?

Is the space before the '?' for the missing 'm'?

Pedant that.

Don AtkinsonMember

mmmmmm...... I wondered who would spot that and how long it would take.........

Don AtkinsonMember

Adam Meredith posted:Don Atkinson posted:Christopher_M posted:*behaviour* ;-)

Who is confusing who ?

Is the space before the '?' for the missing 'm'?

Pedant that.

I would have said “ idiosyncratic” if I thought I could spell “idiosyncratic” and was sure it was the right word to use

Don AtkinsonMember

Don Atkinson posted:

Nobody looks back a page when reading the Forum, so I thought i'd bring this and the next one forward...

Don AtkinsonMember

Don Atkinson posted:Now for something a little more challenging

Four CirclesIts (pretty) obvious that the maximum number of different sized circles (excluding infinitely large circles that have become straight lines!) that can be made to 'just touch' one another is four. (each and every circle just touches the other three).

In the attached diagram above, the three circles (Yellow), (Green) and (Blue) have radii of 1, 2 and 3 respectively.What are the radii of the two 'fourth' circles (shown in red in the diagram below), that can be made to just touch each of Yellow, Green and Blue. (note the two 'fourth' circles will not touch each other, ‘cos that would break the maximum rule!!)

Oh! BTW, there is a standard way and an elegant way to solve this little problem.

As I said above, No need to look back !!

OK.

1247 divided by 43 equals 29

The other missing numbers are 86 and 387.

Don AtkinsonMember

sjbabbey posted:OK.

1247 divided by 43 equals 29

The other missing numbers are 86 and 387.

Nicely done SJB

That just leaves the Red Circles (for the moment) - anyone ? (Frank is excluded)

sophiebear0_0Member

Don

I wouldn't like a nice problem to go to waste. So with a "little" help from Mr Google I went in pursuit of the elegant solution for the circles problem.

I get the following:

Radius of inscribed (inner) circle: 6/23

Radius of circumscribed circle (outer): 6

The alternative geometry-based solution looks pretty messy - and I soon ran out of patience !

Cheers,

Peter

Don AtkinsonMember

sophiebear0_0 posted:Don

I wouldn't like a nice problem to go to waste. So with a "little" help from Mr Google

I went in pursuit of the elegant solutionfor the circles problem.I get the following:

Radius of inscribed (inner) circle: 6/23

Radius of circumscribed circle (outer): 6

The alternative geometry-based solution looks pretty messy - and I soon ran out of patience !

Cheers,

Peter

I hope you found the "elegant" solution interesting.

It's surprising what we all forget as we grow older and become more specialised in our work and hobbies.

Using 1/r to represent "Curvature" is something I find most people have forgotten about !

Christopher_MMember

Don Atkinson posted:Using 1/r to represent "Curvature" is something I find most people have forgotten about !

Always assuming they knew it in the first place :-)

Don AtkinsonMember

Christopher_M posted:Don Atkinson posted:Using 1/r to represent "Curvature" is something I find most people have forgotten about !Always assuming they knew it in the first place :-)

eeK !!!.......Chris, what sort of school did you go to ! .......(probably had to settle for Harrow rather than Eaton ?) but at least NOW you know

And the circles with radii 1, 2 and 3 work out in a very satisfying way when using what I think is a rather elegant formula. Keep everything in vulgar fractions (none of this modern decimalisation) and they virtually cancel themselves out. All very neat.

Don AtkinsonMember

sophiebear0_0 posted:Don

I wouldn't like a nice problem to go to waste. So with a "little" help from Mr Google I went in pursuit of the elegant solution for the circles problem.

I get the following:

Radius of inscribed (inner) circle: 6/23

Radius of circumscribed circle (outer): 6

The alternative geometry-based solution looks pretty messy - and I soon ran out of patience !

Cheers,

Peter

Hi Peter,

There's nothing wrong with using the internet. Most of us would have to get a text book out to brush up on geometry if we tried it from scratch. Using the internet sensibly is just the same as reading a book !

Don AtkinsonMember

A couple of easy ones again, before the next "hard one"

**Dishing up the Prize Money**

Nine people, Adam, Bruce, Chris, Dave, Erik, Frank, Graham, Huge and Ian, share a prize of £450.

Bruce gets £1 more than Adam. Chris gets £1 more than Bruce. Dave gets £1 more than Chris and so on.

How much of the £450 does Ian get ?

Don AtkinsonMember

**How Far is it to Salisbury?**

It’s coming up to Xmas and Naim is running a special delivery service to get product to dealers in time for the big day. Naim have two delivery vans that shuttle between Salisbury and their busiest dealer.

Each day, one van leaves Salisbury bound for the dealer, whilst at the same time the other van leaves the dealer, bound for Salisbury. Both vans follow the same route and pass each other (giving the famous 'Naim Salute') 30 miles from the dealer's. At their destinations each van takes 15 minutes to unload/reload. On the return journey, following the same route, the vans pass each other (giving the famous 'Naim Salute') 15 miles from Salisbury.

How far is it from Salisbury to the dealer's ?

sjbabbeyMember

Ian gets £54

Innocent BystanderMember

Agreed: 1+2+3+4+5+6+7+8=36, Bruce=414+1... Ian=414+8=4

450-36=414 /9=46 for Adam ....46+8=54 for Ian

Don AtkinsonMember

Last edited by Don Atkinson

Don Atkinson posted:

How Far is it to Salisbury?It’s coming up to Xmas and Naim is running a special delivery service to get product to dealers in time for the big day. Naim have two delivery vans that shuttle between Salisbury and their busiest dealer.

Each day, one van leaves Salisbury bound for the dealer, whilst at the same time the other van leaves the dealer, bound for Salisbury. Both vans follow the same route and pass each other (giving the famous 'Naim Salute') 30 miles from the dealer's. At their destinations each van takes 15 minutes to unload/reload. On the return journey, following the same route, the vans pass each other (giving the famous 'Naim Salute') 15 miles from Salisbury.

How far is it from Salisbury to the dealer's ?

Thanks to SteveD for holding back on the solution

The answer is 75 miles.

We just need the arithmetic now !!

(perhaps I should have made it clear that the vans travel at steady speeds and that deceleration/acceleration are either negligable or included in the 15 minute turnround times. ie, you can ignore it !

stevedMember

How far is it to Salisbury

Firstly, thanks Don for remembering my response the first time you set a similar puzzle! Given that no-one has replied, I will try and explain:-

This is a classic puzzle, in that it is solved more by deductive insight than by mathematical equations. It is a version of the Sam Loyd Ferryboat puzzle, first published over 100 years ago by American chess expert and recreational mathematician Sam Loyd.

When they first meet, the "dealervan" has traveled 30 miles, and together both vans have traveled "X" (X being the distance between Salisbury and Dealer).

When they next meet, both vans together have travelled 3X. Therefore at that point, each vehicle has traveled 3 time as far as when they first met, meaning that dealervan will have traveled 90 miles in total at that point. That 90 miles includes 15 miles on the reverse route, so the distance between Dealer and Salisbury is 75 miles (ie 90 less 15).

I must confess to have tried and failed to solve this "mathematically".

Don AtkinsonMember

Thank you Steve, that is a neat, clear and concise explanation. Beautiful.

I expect one or two others will be pleased (possibly relieved ?) that you have posted the solution.

Don AtkinsonMember

Last edited by Don Atkinson

steved posted:How far is it to Salisbury

Firstly, thanks Don for remembering my response the first time you set a similar puzzle! Given that no-one has replied, I will try and explain:-

This is a classic puzzle, in that it is solved more by deductive insight than by mathematical equations. It is a version of the Sam Loyd Ferryboat puzzle, first published over 100 years ago by American chess expert and recreational mathematician Sam Loyd.

When they first meet, the "dealervan" has traveled 30 miles, and together both vans have traveled "X" (X being the distance between Salisbury and Dealer).

That 90 miles includes 15 miles on the reverse route, so the distance between Dealer and Salisbury is 75 miles (ie 90 less 15).When they next meet, both vans together have travelled 3X. Therefore at that point, each vehicle has traveled 3 time as far as when they first met, meaning that dealervan will have traveled 90 miles in total at that point.I must confess to have tried and failed to solve this "mathematically".

Hi Steve,

I have highlighted that part of the solution that I find most people have difficulty grasping, and which I consider is at the heart of the solution.

Don AtkinsonMember

**Xmas Sales !**

A shop was selling six bespoke shirts in their sale at £15, £22, £30, £26, £16 and £31.

Five of the shirts were sold to two customers, the second customer spending twice as much as the first.

Which shirt wasn’t sold ?

Nick from SuffolkMember

The £26 one

Don AtkinsonMember

Hi Nick,

Was that a good guess ? (Obviously not !,)

Trial & error ? Or some sort of formula ?

Well done

Cheers, Don

EloiseMember

Don Atkinson posted:Hi Nick,

Was that a good guess ? (Obviously not !,)

Trial & error ? Or some sort of formula ?

Well done

Cheers, Don

You’re as bad as my maths teacher Don... “always show your working young lady or you will score zero”.

Don AtkinsonMember

Eloise posted:Don Atkinson posted:Hi Nick,

Was that a good guess ? (Obviously not !,)

Trial & error ? Or some sort of formula ?

Well done

Cheers, Don

You’re as bad as my maths teacher Don... “always show your working young lady or you will score zero”.

I did it by trial & error. A lucky guess is improbable - I tried a couple. But by eliminating one a time and trying the remaining combinations, I got a fairly rapid insight and solution.

My “well done” was a 9/10 for Nick. Telling us it was a good guess would get 10/10 !

But, yes, I am as bad as your maths teacher.....

Nick from SuffolkMember

I simply answered the question. Spent a few minutes looking for an algebraic-type solution . Instead used some logic to dismiss certain combinations, which meant that brute force was only needed on a small handful of combinations. There is, therefore, no 'workings'.

However, 2 x (22 + 16) = (15 + 30 + 31)

Don AtkinsonMember

Nick from Suffolk posted:I simply answered the question. Spent a few minutes looking for an algebraic-type solution . Instead used some logic to dismiss certain combinations, which meant that brute force was only needed on a small handful of combinations. There is, therefore, no 'workings'.

However, 2 x (22 + 16) = (15 + 30 + 31)

Looks like you did much the same as me Nick !

I wrote down all the combinations of two-shirt purchaces - that was was quick and easy.

A bit of idle browsing and semi-logic followed.

No magic formula. At least, not that I noticed !

Don AtkinsonMember

**The Point of No Return !**

The term “Point of No Return” is now more commonly referred to as the “Point of Safe Return”. It seems a bit more reassuring in the modern-day aviation world of Health & Safety. Nevertheless, the words mean the same.

There are still a few destinations that are remote and isolated such that a diversion to a nearby alternative is impractical, but non-the-less, on many flights, the economics are such that fuel (including reserves) is often only sufficient for the outbound trip and not the round trip. So the Point of Safe Return needs to be identified. Shortly before reaching the Point of Safe Return, the weather and runway availability at destination will be updated and a decision made as to whether to continue or turn around.

In very simple terms, ie assuming the aeroplane flies at a constant speed and relative to the aeroplane the wind blows from a steady direction at a steady speed, the Point of Safe Return is easy to calculate.

For example, A to B is 240 nm (nautical miles). The aeroplane flies at a steady 120 kt (knots) and has sufficient fuel for 3 hours (plus reserves). In nil-wind conditions it will reach its Point of Safe Return after 1 ½ hours. At this time it will have travelled 180 miles from A, and be 60 nm short of B. It would have 1 ½ hours of fuel available for the homebound leg.

If the wind is constant throughout, such that there is a steady 30 kt tailwind on the outbound leg (*). And a steady 30 kt head wind on the homebound leg (*), whereabouts is the Point of Safe Return in terms of distance from A and distance short of B ?

(*) 120 + 30 = outbound groundspeed; 120 - 30 = homebound groundspeed

Nick from SuffolkMember

Don Atkinson posted:Nick from Suffolk posted:I simply answered the question. Spent a few minutes looking for an algebraic-type solution . Instead used some logic to dismiss certain combinations, which meant that brute force was only needed on a small handful of combinations. There is, therefore, no 'workings'.

However, 2 x (22 + 16) = (15 + 30 + 31)

Looks like you did much the same as me Nick !

I wrote down all the combinations of two-shirt purchaces - that was was quick and easy.

A bit of idle browsing and semi-logic followed.

No magic formula. At least, not that I noticed !

No need to write down all the combinations of 2. No combination of 3 can involve only a single odd number. That reduces the combinations to check by a lot.

Don AtkinsonMember

Nick from Suffolk posted:Don Atkinson posted:Nick from Suffolk posted:However, 2 x (22 + 16) = (15 + 30 + 31)

Looks like you did much the same as me Nick !

I wrote down all the combinations of two-shirt purchaces - that was was quick and easy.

A bit of idle browsing and semi-logic followed.

No magic formula. At least, not that I noticed !

No need to write down all the combinations of 2.

No combination of 3 can involve only a single odd number.That reduces the combinations to check by a lot.

That's where, in my case, **"A bit of idle browsing and semi-logic followed". **

168 nautical miles from A i.e. 72 nautical miles short of B

168/150 = 1.12 hours

168/90 = 1.8777 hours

total = 2.9977 hours

Nick from SuffolkMember

Or 168 3/4 nautical miles (1 and 1/8 hours), which gets rid of those pesky 7.992s.

Don AtkinsonMember

sjbabbey posted:168 nautical miles from A i.e. 72 nautical miles short of B

168/150 = 1.12 hours

168/90 = 1.8777 hours

total = 2.9977 hours

That's pretty close sj, and **certainly close enough for all practical purposes** which would need to allow for the turn and a few other practical factors but...........(at the risk of Eloise intervening )

...........we are only dealing with the simple mathematics of the distance to the Point of Safe Return and hence the times will add up to 3 hours **exactly**, and the distance(s) are therefore not quite so clean-cut as 168 & 72, although they are mathematically fairly elegant

Don AtkinsonMember

Nick from Suffolk posted:Or 168 3/4 nautical miles (1 and 1/8 hours), which gets rid of those pesky 7.992s.

ah-ha !

Neat one Nick ! you beat me to it whilst I was drafting the above rely.

Don AtkinsonMember

.............and **if**, in the original question, **the wind was reversed**, so there was a 30 kt headwind outbound and a 30 kt tailwind homebound, the Point of Safe Return would be...........?

Nick from SuffolkMember

Exactly the same, the problem is symmetric

Don AtkinsonMember

Nick from Suffolk posted:Exactly the same, the problem is symmetric

Yep. Spot on.

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