A Fistful of Brain Teasers

Don Atkinson posted:

Slightly more refined at 10mins : 54 secs past midnight.

I'll try to post details later. But it's not an elegant, pure maths style approach. More like a sledge-hammer !

Is the correct answer - well done.

The first time I did this problem I brutalised it with calculus which takes about two pages - yuck. There is a relatively simple approach which gives the same answer in a few lines if you look at it from the correct point of view. (That's a clue of anyone apart from Don is playing along). 

The maximum rate of change is highlighted in yellow on the abstract from my excel spreadsheet.

The time (minutes past the hour) associated with β = 65.46 is got by dividing by 6 (each minute on a clock face is 6 deg from the previous) giving 10.91 minutes or, 10 minutes and 54.6 seconds past midnight.

But as you can see, it's a bit rough and ready !

Ian G. posted:

Don Atkinson posted:

Slightly more refined at 10mins : 54 secs past midnight.

I'll try to post details later. But it's not an elegant, pure maths style approach. More like a sledge-hammer !

Is the correct answer - well done.

The first time I did this problem I brutalised it with calculus which takes about two pages - yuck. There is a relatively simple approach which gives the same answer in a few lines if you look at it from the correct point of view. (That's a clue of anyone apart from Don is playing along). 

When i first saw your teaser, I thought "rate of change" requires a formula for length AND some calculus.

And as you said, (after 2 pages) "yuck"

So this morning the old excel came out of hibernation together with the sledge hammer...........

Ian G. posted:

Don Atkinson posted:

Slightly more refined at 10mins : 54 secs past midnight.

I'll try to post details later. But it's not an elegant, pure maths style approach. More like a sledge-hammer !

Is the correct answer - well done.

The first time I did this problem I brutalised it with calculus which takes about two pages - yuck. There is a relatively simple approach which gives the same answer in a few lines if you look at it from the correct point of view. (That's a clue of anyone apart from Don is playing along). 

"relatively simple" and "correct point of view" have got me intrigued !

Don, It's me that has a nice glass of Riesling in hand tonight so I'll post a diagram tomorrow that makes it clear ! 

Enjoy the Riesling ,

Lazely looking at my clock diagram, angle C looks remarkably similar to a right angle.

And sqrt 3 looks remarkably similar to the length BC joining the tips of the hands at the time I mentioned earlier.

Needs checking out.................

Don, yes it is a right angle. Now to see why, imagine you are sitting in the minute hand and hence from your new viewpoint  the hour hand is going anti-clockwise at the difference in rotation speeds of the two arms. It's moving away from you fastest when you're looking down the tangent to the hour hands 'circle' - hence the right angle. Hope that makes sense. 

Ian 

Here's a description of what I mean, 

Neat, Very neat. And, as you said "relatively simple"

Here's an easy one, doable even on a wine drinking evening... 

Many large airports have moving walkways, which are like conveyor belts for people. Two airport passengers conduct an experiment: One passenger walks from one end of the moving belt to the other in the direction of the belt’s motion and then back again against the belt’s motion. The other passenger (walking at the same speed) simply walks along the corridor next to the moving belt and back again. Who returns to the starting point first?

Immediate gut reaction is both the same, but then I thought no, it’ll depend on travellator speed, as if it is > or equal to walking speed the person would never get back to the start. Then if the belt is zero speed, the two will be the same - so whatever the belt speed, that one will be something longer than the same, so the person not on the travellator gets back first.

Good thinking IB, if the belt length is L, the belt speed is v and the walking speed is w can you  find a formula to back up your conclusion?

It’s a bit like our calculations to allow for the effect of wind.

The fastest return journey in always without wind.

As soon as a steady wind is present, the return journey or round trip (triangle, square, circle etc) is always more time consuming.

Ian G. posted:

Good thinking IB, if the belt length is L, the belt speed is v and the walking speed is w can you  find a formula to back up your conclusion?

Not that it needs backing up, as that is definitive!

Dustance (each way) = y

Walk speed = y

Travellator speed = z

Time normal = 2.L/w

Time travellator if v<w = L/(w+V)+L/(w-v) 

If v> or = w then never get back to start.

Don Atkinson posted:

It’s a bit like our calculations to allow for the effect of wind.

The fastest return journey in always without wind.

As soon as a steady wind is present, the return journey or round trip (triangle, square, circle etc) is always more time consuming.

Winds can of course change ... tailwind both ways?

Innocent Bystander posted:

Don Atkinson posted:

It’s a bit like our calculations to allow for the effect of wind.

The fastest return journey in always without wind.

As soon as a steady wind is present, the return journey or round trip (triangle, square, circle etc) is always more time consuming.

Winds can of course change ... tailwind both ways?

That’s why I stated “steady wind”

Don - Missed the previous windy problem and I haven't been diligent in following this thread :-) 

IB 

If anyone  was not persuaded by your definitive ( and correct) logic he could try 

Time normal = 2.L/w

Time travellator if v<w = L/(w+V)+L/(w-v)     =  2L/w* w^2/(w^2-v^2) = 2L/w * 1/( 1 - v^2/w^2 )  

so Time travellator = Time normal / ( 1 - v^2/w^2 ) and so long as v<w   then ( 1 - v^2/w^2 ) <1 so 

Time travellator >  Time normal

Ian G. posted:

Don - Missed the previous windy problem and I haven't been diligent in following this thread :-) 

IB 

If anyone  was not persuaded by your definitive ( and correct) logic he could try 

Time normal = 2.L/w

Time travellator if v<w = L/(w+V)+L/(w-v)     =  2L/w* w^2/(w^2-v^2) = 2L/w * 1/( 1 - v^2/w^2 )  

so Time travellator = Time normal / ( 1 - v^2/w^2 ) and so long as v<w   then ( 1 - v^2/w^2 ) <1 so 

Time travellator >  Time normal

 

 

 

 

Hi Ian,

Quite a long time ago I posted a couple of problems dealing with aircraft flying with/against the wind.

But it's easy enough to demonstrate that providing the wind is steady and the a/c flies at a constant true airspeed, a "round trip" in any steady wind will take longer than the same trip in nil wind.

Don Atkinson posted:

Quite a long time ago I posted a couple of problems dealing with aircraft flying with/against the wind.

But it's easy enough to demonstrate that providing the wind is steady and the a/c flies at a constant true airspeed, a "round trip" in any steady wind will take longer than the same trip in nil wind.

yep, it's the same problem in a different guise. 

Another simple one, otherwise there is a risk of only IanG and Eoink taking part !

A firm sells stationery in packages as follows :-

A notebook and two biros cost 48p; a notebook and 4 pencils cost 58p, a notebook, a pencil and 3 felt-tipped pens cost 64p.

How much should you expect to pay for a package containing a notebook, a pencil, a biro and a felt-tipped pen ?

The circle has centre O, and TX is a tangent touching the circle at X

The area of the triangle OXT is …… ?

....... yes, I know, it's dead easy !

30sq.cm