Brain Teasers ? or 50 Years On........... ?

Peter,

I have just measured the fold length on my scaled drawing. It scales close to 3.4 units.

Cheers, Don

Don Atkinson posted:

Peter,

I have just measured the fold length on my scaled drawing. It scales close to 3.4 units.

Cheers, Don

Hi Peter, winky, et al

I have now done the number-crunching and get ...........3.399 units

I used Cartesian co-ordinates with circles radius 2, and centers at (2,2) and (8/3, 4)

It wasn't elegant, the quadratic for solving "x" was 90x² - 420x + 256 = 0

Some of the subsequent numbers in the Pythag part suggested there might be an "elegant" solution somewhere eg x² = 10.399999 and y² = 1.15555 giving a square on the hypotinuse = 11.55555. All suggesting that my scale drawing 3.4 might be THE answer ! 

I think I deserve an egg sandwich after all that.........

Cheers

Don

When I get time, i'll type up my solution, the manuscript version is far too revealing of careless errors

First draft, warts and all...................

Don

I went through my scribbles and notice a howler ! Basically I was simplifying and decided that 36/4 = 12 rather than 9 !!

I adopted the same approach as you. Only subtle differences was that I used the upper circle to be centred at (0,0) and also scaled up to a circle diameter of 6 units to avoid the fractions.

When I scale down to a diameter of 2 units I get intersection points as (-1.912 , -0.5847) and (1.179 , -1.615). This gives a fold length of 3.399. 

So I agree that your scale drawing might be more accurate 

Well done,

Peter

ps - Any update on the crusty problem ??

I keep on wondering if there is a "neat" solution based on two intersecting lines within a circle.

Lines AB and CD with ABCD = points on the circumference and which cross at X

AX*XB = CX*XD

One of the lines (AB) would be a diameter and the other (CD) would be the chord required such that CX = XD

But meanwhile, i'm happy with the "Bulldozer" approach giving 3.399

Don

You are absolutely right !

You end up with a right angle triangle. The hypotenuse is the radius (2) and one of the sides is the half the distance between the circle centres ( (40^0.5)/2). The third side of the triangle is equal to 1/2 * fold length.

Again I get an answer of 3.399

Regards,

Peter 

Whoops - I meant to say the 1/2 distance between the circle centre is ((40^0.5)/6).

The exact answer works out to be (104 ^ 0.5) / 3 which is 3.399

Back to JR Hardee's sandwich and using the various clarifications of the texts..............and fond memories more than 50 years ago of my school days............which I hope are still right .......

"If a point P is always equidistant from a fixed point (the center of the sandwich) and a fixed straight line (the edge of the sandwich) then the locus of the set of points P is called a Parabola."

Basically, thefixed point is called the Focus and the straight line is called the Directrix. The Parabola is symetrical about the line through the Focus which is perpendicular to the Directrix. This line is called theAxis. The point where the Parabola crosses the Axis is called the Vertex.

In Cartesian co-ordinates, if the axis is horizontal along the x axis, with the focus at (a, o) and the line x = −a for the directrix, then the equation of the parabola can be written as y² = 4ax.

So my line NM is part of a parabola, as others have already said.

Hopefully, I am on the right track now. Does any of this ring bells ?

Next step, when I can wrack my old brain, is to do a bit of integration and geometry. (which I note JR has side-stepped for the benefit of others ) but Peter has already done

What does this work out to in terms of percentage of the sandwich that is eaten? The published solution (intersecting parabolas) says 21.9%.

JRHardee posted:

What does this work out to in terms of percentage of the sandwich that is eaten? The published solution (intersecting parabolas) says 21.9%.

JR, it's going to take a while for me to recall the integration..............

Excuses ?

• I leave for work 07:15 and get home 20:15.
• Mrs D is in Canada so I have to cook and do all our domestic chores and keep in touch.
• The telly's broke (see Home Theatre), the water pipe is broke (but fixed at last - see other post)
• The Leave campaign won - I voted Remain
• I ate sandwiches all last week trying to figure out how close I could get to the crust - and I don't even like sandwiches...............

But on the bright side........... Sophiebear and myself managed to sort out winky's problem.......

Nice sandwich problem BTW !

OK, I did a bit of integration.

The area of a segment of a parabola (between the Vertex and a normal to the axis) is :-

• 2/3 of the chord x the maximum thicknes of the segment

The parabola in question has a focal length = 1.0 and therefor an equation y² = 4x

The diagonal of the sandwich can be described as a straight line y = 1-x

Solving for the intersections gives

• x = 3 - 2 √2 = 0.171573
• y = 1 - x       = 0.828427

Segment area = 0.667 x 0.171573 x 0.828427   Eq (1)

Triangular area between the diagonals and the axis = ½ x 0.828427 x 0.828427   Eq (2)

Area of sandwich that can be eaten = Eq (1)  +  Eq(2)

Ratio of the area eaten to the whole sandwich = 0.218951

My basic calculation.

The integration to get 2/3*w*h took a bit of effort........

My general summary.

Phew !

sophiebear0_0 posted:

Whoops - I meant to say the 1/2 distance between the circle centre is ((40^0.5)/6).

The exact answer works out to be (104 ^ 0.5) / 3 which is 3.399

Neat, very neat.

I'm guessing that is the solution winky had in mind, rather than the "Bulldozer"

Swimming (against the tide ?)

I plan to cross a river that is 1 mile wide to the point directly opposite on the other bank.

I can swim at 2.5 mph and I can walk at 4 mph. The river current is 2 mph.

I realise I can cross in one of two basic ways.

1. Head somewhat upstream so that my resultant track is straight across.
2. Head directly towards the opposite bank and then walk back along the bank from the point downstream to which the current has carried me.

Which is the quickest way, and by how much ?

Fools rush in, but I don't think it's either one. You are proposing that the crosser either walk and swim or swim and walk, unless I'm over-simplifying. My guess is that the fastest way is to swim slightly upstream, cutting a little bit off your walk, because you will be carried downstream to a lesser extent. The amount to which you swim upstream is determined by your relative speeds on land and at sea.

JRHardee posted:

Fools rush in, but I don't think it's either one.You are proposing that the crosser either walk and swim or swim and walk, unless I'm over-simplifying. My guess is that the fastest way is to swim slightly upstream, cutting a little bit off your walk, because you will be carried downstream to a lesser extent. The amount to which you swim upstream is determined by your relative speeds on land and at sea.

Not quite.

Option (1) is that I head (point) upstream sufficient such that my track across the river takes me directly to my destination on the opposite bank. No walking involved. Just swimming.

Option (2) is that I head (point) directly across the river at my destination and keep heading directly at the opposite bank (not my destination) despite being carried downstream by the current. When I reach the opposite bank, I am well down stream from my intended destination and so have to walk upstream along the bank to reach my destination which is directly opposite my starting point.

Hope this makes the task clear ?

The diagram shows a system of pulleys driven by belts

The circumference of the rim of the outer pulley is exactly twice that of the inner pulley and there is no slip between belt and pulley.

If pulley A rotates at 100 rpm, how fast will pulley E rotate ?

(see next page for picture of pulleys............)

Pulley system referred to on previous page

The diagram shows a system of pulleys driven by belts

The circumference of the rim of the outer pulley is exactly twice that of the inner pulley and there is no slip between belt and pulley.

If pulley A rotates at 100 rpm, how fast will pulley E rotate ?

Don Atkinson posted:

JRHardee posted:

Fools rush in, but I don't think it's either one.You are proposing that the crosser either walk and swim or swim and walk, unless I'm over-simplifying. My guess is that the fastest way is to swim slightly upstream, cutting a little bit off your walk, because you will be carried downstream to a lesser extent. The amount to which you swim upstream is determined by your relative speeds on land and at sea.

Not quite.

Option (1) is that I head (point) upstream sufficient such that my track across the river takes me directly to my destination on the opposite bank. No walking involved. Just swimming.

Option (2) is that I head (point) directly across the river at my destination and keep heading directly at the opposite bank (not my destination) despite being carried downstream by the current. When I reach the opposite bank, I am well down stream from my intended destination and so have to walk upstream along the bank to reach my destination which is directly opposite my starting point.

Hope this makes the task clear ?

JR,

I confined the available options in the question to two. This was so as to keep the solution simple.

Of course, there is wide range of possible options in addition to the two presented.

You have now got me wondering what the minimum Origin to  Destination  time might be ?

I have a feeling it's one of the two options in the question, but..........

Don Atkinson posted:

If pulley A rotates at 100 rpm, how fast will pulley E rotate ?

 

100 rpm

Don Atkinson posted:

Swimming (against the tide ?)

I plan to cross a river that is 1 mile wide to the point directly opposite on the other bank.

I can swim at 2.5 mph and I can walk at 4 mph. The river current is 2 mph.

I realise I can cross in one of two basic ways.

1. Head somewhat upstream so that my resultant track is straight across.
2. Head directly towards the opposite bank and then walk back along the bank from the point downstream to which the current has carried me.

Which is the quickest way, and by how much ?

I'm thinking a 1 mile wide river will be laminar flow for most of its width. In which case the speed of water flow will vary across it's width.

fatcat posted:

Don Atkinson posted:

Swimming (against the tide ?)

I plan to cross a river that is 1 mile wide to the point directly opposite on the other bank.

I can swim at 2.5 mph and I can walk at 4 mph. The river current is 2 mph.

I realise I can cross in one of two basic ways.

1. Head somewhat upstream so that my resultant track is straight across.
2. Head directly towards the opposite bank and then walk back along the bank from the point downstream to which the current has carried me.

Which is the quickest way, and by how much ?

I'm thinking a 1 mile wide river will be laminar flow for most of its width. In which case the speed of water flow will vary across it's width.

The speed of the river is 2mph. It is uniform across the entire cross-section.

I can swim at precisely 2½mph for as long as it takes.

i can walk at precisely 4mph for as long as it takes, even if my clothes are soaking wet and i'm tired and hungry and wondering how far I can chomp through my sandwiches towards the crust........

Don. I dread to think what Scotty of the starship enterprise would say to that.

Option 2, swimming and walking will be the quickest, by 4 minutes. I think.