Brain Teaser No 3
Posted by: Don Atkinson on 11 March 2007
I know I started Brain Teaser No 1 about 5 years ago. ISTR another with No 2 in the title, so hopefully this is not duplicating somebody elses Brain Teaser No 3.....
Flight Around the World
A group of aeroplanes is based on a small island. Each plane holds just enough fuel to take it half way around the world. Any amount of fuel can be transfered from the tank of one aeroplane to another aeroplane whilst the planes are in flight. The ONLY source of fuel available to these aeroplanes is on this small island. Assume that there is no time lost when refueling, either in the air or on the ground.
What is the smallest number of aeroplanes required to ensure the flight of one aeroplane around the world on a great circle, assuming that all areoplanes have the same constant groundspeed and rate of fuel consumption and that all aeroplanes return safely to their island base.
Cheers
Don
Flight Around the World
A group of aeroplanes is based on a small island. Each plane holds just enough fuel to take it half way around the world. Any amount of fuel can be transfered from the tank of one aeroplane to another aeroplane whilst the planes are in flight. The ONLY source of fuel available to these aeroplanes is on this small island. Assume that there is no time lost when refueling, either in the air or on the ground.
What is the smallest number of aeroplanes required to ensure the flight of one aeroplane around the world on a great circle, assuming that all areoplanes have the same constant groundspeed and rate of fuel consumption and that all aeroplanes return safely to their island base.
Cheers
Don
quote:Originally posted by Ian G.:
Ah - 89mins 42sec ?
no no silly boy, skelp, - 92mins 42secs
If v1 and v2 are the speeds of the two walkers and vt the speed of the train in m/s, then the length of the train is (vt-v1)*20 and (vt+v2)*18 from the two passings. Equating gives vt = 10*v1 +9*v2.
It is 618 seconds between the COMPLETION of the two passings during which time the first walker has covered v1*618 metres.
So the distance between the walkers at the instant the second passing is complete is
D = vt*618 - v1*618
= (10*v1 + 9*v2)*618 - v1*618
= 9 * (v1+v2)*618
= 5562 * (v1+v2)
Since their closing speed is also v1+v2 it will take them 5562 seconds or 92mins and 42 secs to meet.
I'll probably chage my mind again soon - watch this space.
.........and for the benefit of Mr Cavendish.....
yes, I am aware that the average 12 year-old would be able to use his simple algebra to knock both the above out in less than 5 minutes....
But we don't have too many 12 year-olds on the forum at present.....and most 52 year-olds will now have 40 years of congealed grease to overcome before getting the old brain into gear and running smoothly.........or have to recall where they filed their Bostock & Chandler
Cheers
Don
yes, I am aware that the average 12 year-old would be able to use his simple algebra to knock both the above out in less than 5 minutes....
But we don't have too many 12 year-olds on the forum at present.....and most 52 year-olds will now have 40 years of congealed grease to overcome before getting the old brain into gear and running smoothly.........or have to recall where they filed their Bostock & Chandler
Cheers
Don
How embarassing then - unlike an average 12 year old I've spent at least 10mins on the escalator problem and haven't found the 'key' yet. 
I do know how many bricks there are in the wall tho'
Ian
I do know how many bricks there are in the wall tho'
Ian
I managed to miscalculate on the bricks thing
but yes, 900 and 80.
but yes, 900 and 80.
ok, here's a little geometry one so Don can join in.
Archimedes worked out upper and lower bounds for the value of pi by circumscribing two circles on a hexagon - one 'on the outside' through all the 6 vertices and one 'on the inside', just touching to all the 6 edges.
What are the upper and lower values he came up with.
Ian
(I pose this not 'cos it is hard but because this idea is so deeply beautiful I need to share it True genius.)
Archimedes worked out upper and lower bounds for the value of pi by circumscribing two circles on a hexagon - one 'on the outside' through all the 6 vertices and one 'on the inside', just touching to all the 6 edges.
What are the upper and lower values he came up with.
Ian
(I pose this not 'cos it is hard but because this idea is so deeply beautiful I need to share it
True genius.)