Daily Challenges

[Paddy]

Just now, Sword said:

Your pictures are far more elegant than my words! I couldn't be bothered taking all those photos!

 

[1949threepence]

1 hour ago, Diaconis said:

Paddy, please put me out of my misery, I've got it down to four but three seems impossible, thx

Think of the top row as coins 1 2 & 3 from left to right, and then the bottom row as coins 4, 5 & 6 from left to right:-

Move 1 - move coin 4 to rest under and between coins 5 & 6

Move 2 - move coin 5 leftwards to rest underneath coins 1 & 2

Move 3 - move coin 1 to between coins 5 & 4

voila !!!

  

[Sword]

1 hour ago, Diaconis said:

Maybe this one will elicit a response.

Referring to the layout below, the problem is to draw a line connecting the 3's, a separate line connecting the 2's and a separate line connecting the Aces.

You can't cross any line and you cant go out of the layout or cross over any card.

 

 

[Diaconis]

Paddy, i get this

instead of this

 

[Paddy]

OK - here is a serious challenge for a mathematician. It has been bugging me for years and I don't know the solution:

You have a perfectly circular field and a goat. You tether it to a fixed point on the edge of the field - how long does the chain need to be so that it can eat precisely half the area of the field? (You can ignore the reach of the goats neck.)

 

[Paddy]

4 minutes ago, Diaconis said:

Paddy, i get this

instead of this

 

Sorry if I confused you - the orientation of the ring is irrelevant. Sorry!

 

[Diaconis]

4 minutes ago, Paddy said:

Sorry if I confused you - the orientation of the ring is irrelevant. Sorry!

 

😅The second one is possible in 4

[Peckris 2]

On 4/1/2020 at 2:28 PM, DaveG38 said:

This is how it works. Point to one door and ask each robot the question 'if I ask you if this is the door to safety would you say yes.'  Assume the door I point to is the safe one. The truthful robot will simply say yes, he would say this because it is true. The lying robot says to himself, I know that the door pointed to is the safe one, but because I'm a liar I would say no. But then the rest of the question asks whether he would say yes to his own lying answer. Logically, the robot being a liar will directly contradict his earlier 'no' answer so he says yes to the question, since he is lying about his own earlier lie. Hence both robots indicate which is the correct door and so you can escape. Works the other way round for the wrong door.

That's not in the original riddle.

[Peckris 2]

39 minutes ago, Paddy said:

OK - here is a serious challenge for a mathematician. It has been bugging me for years and I don't know the solution:

You have a perfectly circular field and a goat. You tether it to a fixed point on the edge of the field - how long does the chain need to be so that it can eat precisely half the area of the field? (You can ignore the reach of the goats neck.)

 

That's a very difficult one! The solution will be a pointed oval (like an American football) but I have no idea how to calculate half the circle in that shape!

[Diaconis]

An ice cube floats in a glass of water.

Problem: lift it out with a piece of string. Must not touch the cube with fingers, or knot the string in any way.

 

[Sword]

3 minutes ago, Diaconis said:

An ice cube floats in a glass of water.

Problem: lift it out with a piece of string. Must not touch the cube with fingers, or knot the string in any way.

 

Push the string onto the ice cube. The string will gradually cut into the ice. Water refreezes on top of string. Lift.

[Paddy]

11 hours ago, Paddy said:

OK - here is a serious challenge for a mathematician. It has been bugging me for years and I don't know the solution:

You have a perfectly circular field and a goat. You tether it to a fixed point on the edge of the field - how long does the chain need to be so that it can eat precisely half the area of the field? (You can ignore the reach of the goats neck.)

 

With no one volunteering a solution I decided to see if the answer was online anywhere - and it is. Here is a comprehensive solution on wikipedia along with several further problems and solutions on the same theme:

https://en.wikipedia.org/wiki/Goat_problem

I didn't know it was quite THAT complicated!

 

[Sword]

14 hours ago, Diaconis said:

And finally Esther

The Queen can move any number of squares horizontally, vertically, or diagonally but must cover all 9 of the shaded corner squares in 4 moves. An old chestnut in another guise.

I started with the queen at a corner if that's allowed.

 

[Sword]

1 hour ago, Paddy said:

With no one volunteering a solution I decided to see if the answer was online anywhere - and it is. Here is a comprehensive solution on wikipedia along with several further problems and solutions on the same theme:

https://en.wikipedia.org/wiki/Goat_problem

I didn't know it was quite THAT complicated!

 

I was thinking of doing it by a different way. The area of sector + the areas of the 2 segments. Rather curious if we can still remember the A level maths done decades ago. But on second thought, I would rather make some lunch instead.

 

[Rob]

4 minutes ago, Sword said:

I was thinking of doing it by a different way. The area of sector + the areas of the 2 segments. Rather curious if we can still remember the A level maths done decades ago. But on second thought, I would rather make some lunch instead.

 

I was doing something similar by trying to add two segments with similar chord length but differing radii. It all went t*ts up when it dawned that  I had a second variable (the angle) in the equations - at which point I realised I had forgotten about integrals.

 

[Sword]

Yes, I too realised the angle would be an additional variable. The first idea to come to mind is to express sin (angle) in terms of length and radius. But then it will get messy and rearranging to get the ratio of length/radius might be difficult or not feasible. 

[blakeyboy]

The solutions on Wiki are under   'Recreational Mathematics'.

I glanced at them and had to go and lie down.

[1949threepence]

18 hours ago, Sword said:

Push the string onto the ice cube. The string will gradually cut into the ice. Water refreezes on top of string. Lift.

That pre-supposes the ambient temperature is below freezing point. If it is, then at say -18 degrees, all the water in the glass will freeze within a few hours, and your piece of string will be trapped in it, together with the original ice cube. Or even at just 0 degrees, eventually the entirety of the water in the glass will freeze.

Above freezing point, the water will remain as water, and the ice cube will melt at a rate commensurate with how far above freezing point the surrounding air is. It will be shrinking as it melts, and therefore not able to trap a piece of sting within it.     

[Sword]

6 minutes ago, 1949threepence said:

That pre-supposes the ambient temperature is below freezing point. If it is, then at say -18 degrees, all the water in the glass will freeze within a few hours, and your piece of string will be trapped in it, together with the original ice cube. Or even at just 0 degrees, eventually the entirety of the water in the glass will freeze.

Above freezing point, the water will remain as water, and the ice cube will melt at a rate commensurate with how far above freezing point the surrounding air is. It will be shrinking as it melts, and therefore not able to trap a piece of sting within it.     

I am not totally certain as I have not tried this before. Ice melts more easily as you put pressure on it. Hence the string can cut into the ice. Although the ambient temperature is more than zero degrees, the ice itself can still be -10 degrees. Hence, the water on top of the string will refreeze. 

[blakeyboy]

7 hours ago, Rob said:

I was doing something similar by trying to add two segments with similar chord length but differing radii. It all went t*ts up when it dawned that  I had a second variable (the angle) in the equations - at which point I realised I had forgotten about integrals.

 

I remember the South African Police being asked about integration in the late 1970's, in those dark racist days there.

The Police Chief said there was nothing to worry about, since:  "our boys are up all night integrating suspects"...

 

Must get a snack - all this talking has given me quite an apartheid.....

[Sword]

15 hours ago, Sword said:

 I am not totally certain as I have not tried this before. Ice melts more easily as you put pressure on it. Hence the string can cut into the ice. Although the ambient temperature is more than zero degrees, the ice itself can still be -10 degrees. Hence, the water on top of the string will refreeze. 

Diaconis, can we have the official answer please. 

[Sword]

What is greater than God,
more evil than the devil,
the poor have it,
the rich need it,
and if you eat it, you'll die?

[Diaconis]

33 minutes ago, Sword said:

Diaconis, can we have the official answer please. 

lay the string over the ice cube, shake plenty of salt on top of the cube. In a few moments the cube will freeze to string and can be lifted. Also works with a match or toothpick.

[Sword]

That's cheating somewhat. You didn't say you can use additional substances!

[bagerap]

2 hours ago, Sword said:

What is greater than God,
more evil than the devil,
the poor have it,
the rich need it,
and if you eat it, you'll die?

Nothing