Brain Teasers ? or 50 Years On........... ?
Posted by: Don Atkinson on 02 June 2015
50 Years on…….
50 years ago, I was doing what many 18 year olds are doing this week and over the next few weeks……………….their A-Levels.
Mine were Pure Maths; Applied Maths; Physics and Chemistry. We also had a new subject called The Use of English.
About 10 years ago I started a few “Brain Teaser” threads on this forum. One or two people complained that many of the so-called Brain Teasers were no more than A-Level maths dressed up. That was true of a few teasers, but most were real teasers, especially the ones like “The Ladder” posted by Bam and also the one about the maximum number of 1cm diameter spheres that can be packed into a 10x5x5 cm box.
Any way, never mind Brains or Teasers, I guess one or two other Forumites are also looking back 50 years and would be delighted to tease their brains with calculus, probability, spherical geometry, geometric progressions, Newton’s Laws of Motion ……………………….no ? Then probably best if you drink your weekly 21 units tonight and wake up in the Music Room tomorrow to recover from the nightmare !
First one to follow shortly, and please, please add your own favourites !!
Before I forget...........the circumference of an ellipse "c"................to save you looking it up in wikipedia 
C = 2Пa [1 – (1/2)^2(e^2) – (1.3/2.4)^2((e^4)/3) – (1.3.5/2.4.6)^2((e^6)/5) …..]
a = semi-mjor axis
e = eccentricity
..........so can we get a bigger lorry park ?
First express "a" as a=a0 /(1-e^2)^(1/4),with a0^2=A/pi,where A is the park area.
You get
C = 2Пa0 [1 + (3/64)e^4+..)+..]
So you cannot get a shorter fence (at least for small eccentricity).
The ancient Greeks knew that the shorter perimeter is obtained with a circle. But the rigorous proof dates from the early twentieth century.
Before I forget...........the circumference of an ellipse "c"................to save you looking it up in wikipedia
C = 2Пa [1 – (1/2)^2(e^2) – (1.3/2.4)^2((e^4)/3) – (1.3.5/2.4.6)^2((e^6)/5) …..]
a = semi-mjor axis
e = eccentricity
..........so can we get a bigger lorry park ?
First express "a" as a=a0 /(1-e^2)^(1/4),with a0^2=A/pi,where A is the park area.
You get
C = 2Пa0 [1 + (3/64)e^4+..)+..]
So you cannot get a shorter fence (at least for small eccentricity).
The ancient Greeks knew that the shorter perimeter is obtained with a circle. But the rigorous proof dates from the early twentieth century.
Yes, a circle encloses the largest area for the least amount of fencing.
I think that for all rectangular areas, a square encloses the largest area for the least amount of fencing. i.e. there is no rectangle aspect ratio that beats a square.
However, we have seen already that when one side of the lorry park is already provided and the fence only has to enclose the other three sides, then a rectangle 1:2 encloses the largest area. I am thinking of a rectangle as an an "elongated square"
I had assumed Steved had in mind that likewise, some sort of "elongated circle" might be more efficient than the semi-cirlce that we know is more efficient than the rectangle.
Before I forget...........the circumference of an ellipse "c"................to save you looking it up in wikipedia
C = 2Пa [1 – (1/2)^2(e^2) – (1.3/2.4)^2((e^4)/3) – (1.3.5/2.4.6)^2((e^6)/5) …..]
a = semi-mjor axis
e = eccentricity
..........so can we get a bigger lorry park ?
First express "a" as a=a0 /(1-e^2)^(1/4),with a0^2=A/pi,where A is the park area.
You get
C = 2Пa0 [1 + (3/64)e^4+..)+..]
So you cannot get a shorter fence (at least for small eccentricity).
The ancient Greeks knew that the shorter perimeter is obtained with a circle. But the rigorous proof dates from the early twentieth century.
Yes, a circle encloses the largest area for the least amount of fencing.
I think that for all rectangular areas, a square encloses the largest area for the least amount of fencing. i.e. there is no rectangle aspect ratio that beats a square.
However, we have seen already that when one side of the lorry park is already provided and the fence only has to enclose the other three sides, then a rectangle 1:2 encloses the largest area. I am thinking of a rectangle as an an "elongated square"
I had assumed Steved had in mind that likewise, some sort of "elongated circle" might be more efficient than the semi-cirlce that we know is more efficient than the rectangle.
But this elongated rectangle is half a square. This is coherent with the square being the optimal shape without a wall.
Could it be
(2L)^2 – (2L-d)^2 +d^2 / (2L)^2
Before I forget...........the circumference of an ellipse "c"................to save you looking it up in wikipedia
C = 2Пa [1 – (1/2)^2(e^2) – (1.3/2.4)^2((e^4)/3) – (1.3.5/2.4.6)^2((e^6)/5) …..]
a = semi-mjor axis
e = eccentricity
..........so can we get a bigger lorry park ?
First express "a" as a=a0 /(1-e^2)^(1/4),with a0^2=A/pi,where A is the park area.
You get
C = 2Пa0 [1 + (3/64)e^4+..)+..]
So you cannot get a shorter fence (at least for small eccentricity).
The ancient Greeks knew that the shorter perimeter is obtained with a circle. But the rigorous proof dates from the early twentieth century.
Yes, a circle encloses the largest area for the least amount of fencing.
I think that for all rectangular areas, a square encloses the largest area for the least amount of fencing. i.e. there is no rectangle aspect ratio that beats a square.
However, we have seen already that when one side of the lorry park is already provided and the fence only has to enclose the other three sides, then a rectangle 1:2 encloses the largest area. I am thinking of a rectangle as an an "elongated square"
I had assumed Steved had in mind that likewise, some sort of "elongated circle" might be more efficient than the semi-cirlce that we know is more efficient than the rectangle.
But this elongated rectangle is half a square. This is coherent with the square being the optimal shape without a wall.
Good point !
Still trying to get my head around this "disc overlaps more than two squares" thingy.
What I am having difficulty visualising is, why are others looking at two side-adjacent squares ? Assuming that's what (2L*L) implies. ?
Perhaps I am reading winky's question completely differently to other people ?
I don't know about you, but I certainly haven't read the question correctly.
Still trying to get my head around this "disc overlaps more than two squares" thingy.
What I am having difficulty visualising is, why are others looking at two side-adjacent squares ? Assuming that's what (2L*L) implies. ?
Perhaps I am reading winky's question completely differently to other people ?
I don't know about you, but I certainly haven't read the question correctly.
Changed my mind, I did read the question correctly, but obviously doesn't mean I'm correct.
I've considered an area of 4 squares of length L and calculated the probability of not landing on more than 2 squares.
Still trying to get my head around this "disc overlaps more than two squares" thingy.
What I am having difficulty visualising is, why are others looking at two side-adjacent squares ? Assuming that's what (2L*L) implies. ?
Perhaps I am reading winky's question completely differently to other people ?
I don't know about you, but I certainly haven't read the question correctly.
Changed my mind, I did read the question correctly, but obviously doesn't mean I'm correct.
I've considered an area of 4 squares of length L and calculated the probability of not landing on more than 2 squares.
The chess board is infinite.
The "corners" of the "area of 4 squares" (assuming you have the 4 squares forming a larger square 2L*2L, are the intersections of another group of 4 squares ad infinitum.
Still trying to get my head around this "disc overlaps more than two squares" thingy.
What I am having difficulty visualising is, why are others looking at two side-adjacent squares ? Assuming that's what (2L*L) implies. ?
Perhaps I am reading winky's question completely differently to other people ?
I don't know about you, but I certainly haven't read the question correctly.
Changed my mind, I did read the question correctly, but obviously doesn't mean I'm correct.
I've considered an area of 4 squares of length L and calculated the probability of not landing on more than 2 squares.
The chess board is infinite.
The "corners" of the "area of 4 squares" (assuming you have the 4 squares forming a larger square 2L*2L, are the intersections of another group of 4 squares ad infinitum.
But the disc can only land on a maximum of 4 squares.
I think my formula is correct but only if d > L/2.
Probably need to tweek it a little.
I'm thinking if the centre of disc lands in shaded area it will be in contact with less than 3 squares.
A new sketch
If the centre of the disc lands in the unshaded area it will touch more than 2 squares.
A new formula
((2L)^2 – (2L-d)^2 +d^2 - 4d(L-d))
-----------------------------------------
(2L)^2