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

steved posted:

As a break from looking at cones, here is a genuine practical problem from a party last night.

• The host had invited 16 guests, and the host had prepared a team quiz which had 4 rounds.
• The host wanted 4 teams of 4 people for each round.
• For each successive round, teams were to be reorganised so that each member had new team-mates - ie their new team had no member with whom they had played previously.
• The objective was to ensure that, by the fourth round, everyone had "played" with everyone else.
• All went well for the first 3 rounds, but for the fourth round we couldn't get a solution which avoided a number of "repeat members".
• The host looked at me, as the supposed maths expert, to confirm that it should have been possible to achieve her objective. I was unable to come up with a solution which worked, but neither was I able to explain why it was impossible.

Any suggestions gratefully received!

Steve D

Hi Steve,

I'm glad "all went well for the first 3 rounds"  because I don't think there are too many solutions even to the first 3 rounds.....

A preliminary look at this suggests there might only be one possible solution to get through the first 3 rounds with different members of each team, with no "repeat members" whatsoever !

I'm not entirely certain about this. I do have a solution for the first 3 rounds. However, if I make even 1 single transposition, the 3rd round then involves a couple of "repeat members". In other words, it looks like even the 3rd round only has 1 possible solution.

However, for the fourth round, I don't think it is at all possible to field four teams with no repeat members.

Proving it ?.........................

Hopefully there aren't any "repeat members" in any of the teams in the first three games.....

.....we're off to the cinema, so i'll check more carefully when I get back !

Time to try and explain the results, this is not a very detailed proof; I would certainly lose markets on lack of working.

Let's start by taking a single cone of half apex angle A.

We then surround the cone with three planes, and as tangents to the surface of the cone and two planes at right angles to the other two planes.

We place the cone so it is centred at coordinate 0,0,0 (x,y,z axis)

The cone is directed along z axis such that the radius of the cone's circle for a given value of z  is R = z tan A

Considering this circle we then wish to define the tangent a a point at angle Q from the x axis such that R = x cos Q= y sin Q.

The intercept of this tangent gives R / cos Q for the x axis and R / sin Q for the y axis.

This gives us two vectors in the tangential plane for any given value of angles Q and A.

Using cross products we define the equation of the planes as:

x cos Q + y sin Q - z tan A = 0

If we assume for one plane Q =0 we have the equation of the plane

x - z tan A = 0

The angle between these planes can be found by using the angle between the normal to the planes. You can find the proof for these online....

this gives us the angle between the planes (P)

cos P = (cos Q + tan^2 A) / sqrt ((cos^2 Q +sin^2 Q + tan^2 A) x (1+tan^2))

for P= 90 degrees we have cos Q = -tan^2 A

Next, we need to find the angle between the outer planes which are both 90 degrees to the central plane.

so we can slot in angle P and -P into the formulas for the planes above, this gives us...

x cos Q +y sin Q - z tan A = 0 and x cos -Q + y sin -Q - z tan A = 0 = x cos Q - y sin Q - z tan A

the angle between them (S) can be described as

cos S = (cos^2 Q - sin^2 Q - tan^2 A)/sqrt ((cos^2 Q +sin^2 Q + tan^2 A) x (cos^2 Q +sin^2 Q + tan^2 A))

cos S = (2cos^2 Q -1 + tan^2 A) / (1+ tan^2 A)

but we know cos Q = -tan^2 A so...

cos S = (2tan^4 A - 1 + tan^2 A) / (1 + tan^2 A)

this is where I was pulling my hair out assuming that S was 60, but actually need to use S = 120, we therefore know cos = -0.5 and we can then solve he quadratic for tan^2 A

2 tan^4 A - +1.5 tan^2 A -0.5 = 0

so tan^2 A = (-1.5 +- sqrt ( 2.25+4))/8 = 0.25

so A = arctan (0.5) or 53.1301 etc

Sorry for the lack of diagrams but hopefully the explanation is understandable.

A Merry Chrstmas to you all.

Matthew

Thanks Don, I am very happy to report back to the host that she was asking the impossible (also impossible was her expectation that 16 people could have a "friendly" quiz without falling out over disallowed answers to some of her quiz questions........)

Steve D

Don Atkinson posted:

steved posted:

As a break from looking at cones, here is a genuine practical problem from a party last night.

• The host had invited 16 guests, and the host had prepared a team quiz which had 4 rounds.
• The host wanted 4 teams of 4 people for each round.
• For each successive round, teams were to be reorganised so that each member had new team-mates - ie their new team had no member with whom they had played previously.
• The objective was to ensure that, by the fourth round, everyone had "played" with everyone else.
• All went well for the first 3 rounds, but for the fourth round we couldn't get a solution which avoided a number of "repeat members".
• The host looked at me, as the supposed maths expert, to confirm that it should have been possible to achieve her objective. I was unable to come up with a solution which worked, but neither was I able to explain why it was impossible.

Any suggestions gratefully received!

Steve D

Hi Steve,

I'm glad "all went well for the first 3 rounds"  because I don't think there are too many solutions even to the first 3 rounds.....

A preliminary look at this suggests there might only be one possible solution to get through the first 3 rounds with different members of each team, with no "repeat members" whatsoever !

I'm not entirely certain about this. I do have a solution for the first 3 rounds. However, if I make even 1 single transposition, the 3rd round then involves a couple of "repeat members". In other words, it looks like even the 3rd round only has 1 possible solution.

However, for the fourth round, I don't think it is at all possible to field four teams with no repeat members.

Proving it ?.........................

............I'll let Matthew have a go first................looks like he's nicely warmed up for an "impossible" bit of maths

For some reason, putting actual names in the list made selecting teams a bit easier !

Probably easier to spot any lingering mistakes as well............

Matthew T posted:

Time to try and explain the results, this is not a very detailed proof; I would certainly lose markets on lack of working.

Let's start by taking a single cone of half apex angle A.

We then surround the cone with three planes, and as tangents to the surface of the cone and two planes at right angles to the other two planes.

We place the cone so it is centred at coordinate 0,0,0 (x,y,z axis)

The cone is directed along z axis such that the radius of the cone's circle for a given value of z  is R = z tan A

Considering this circle we then wish to define the tangent a a point at angle Q from the x axis such that R = x cos Q= y sin Q.

The intercept of this tangent gives R / cos Q for the x axis and R / sin Q for the y axis.

This gives us two vectors in the tangential plane for any given value of angles Q and A.

Using cross products we define the equation of the planes as:

x cos Q + y sin Q - z tan A = 0

If we assume for one plane Q =0 we have the equation of the plane

x - z tan A = 0

The angle between these planes can be found by using the angle between the normal to the planes. You can find the proof for these online....

this gives us the angle between the planes (P)

cos P = (cos Q + tan^2 A) / sqrt ((cos^2 Q +sin^2 Q + tan^2 A) x (1+tan^2))

for P= 90 degrees we have cos Q = -tan^2 A

Next, we need to find the angle between the outer planes which are both 90 degrees to the central plane.

so we can slot in angle P and -P into the formulas for the planes above, this gives us...

x cos Q +y sin Q - z tan A = 0 and x cos -Q + y sin -Q - z tan A = 0 = x cos Q - y sin Q - z tan A

the angle between them (S) can be described as

cos S = (cos^2 Q - sin^2 Q - tan^2 A)/sqrt ((cos^2 Q +sin^2 Q + tan^2 A) x (cos^2 Q +sin^2 Q + tan^2 A))

cos S = (2cos^2 Q -1 + tan^2 A) / (1+ tan^2 A)

but we know cos Q = -tan^2 A so...

cos S = (2tan^4 A - 1 + tan^2 A) / (1 + tan^2 A)

this is where I was pulling my hair out assuming that S was 60, but actually need to use S = 120, we therefore know cos = -0.5 and we can then solve he quadratic for tan^2 A

2 tan^4 A - +1.5 tan^2 A -0.5 = 0

so tan^2 A = (-1.5 +- sqrt ( 2.25+4))/8 = 0.25

so A = arctan (0.5) or 53.1301 etc

Sorry for the lack of diagrams but hopefully the explanation is understandable.

 

A Merry Chrstmas to you all.

Matthew

Hi Matthew, great Brain Teaser and many thanks for managing to sort out the proof, good job.

By way of summarizing the overall picture,

the initial six cones, with their "Altitudes" horizontal, formed a "circle", (or rather a Hexagon when viewed from above), with their Apex points just touching in the centre. These cones had  a base radius R and sloping sides 2R giving Base/Side angle 60 degrees and Apex 60 degrees.

When laid on  their sides on a flat surface (neutral stability situation) cones with a base radius R need an Altitude 2R (rather than sloping sides 2R) in order to form a "circle", (or rather a Hexagon when viewed from above), with their Apex points just touching in the centre. the Apex angle is 53.13 degrees.

Neat ! [2R base x 2R sloping side] versus [2R base x 2R Altitude] I like it !!

Don.

The way see it, the altitude of the cone isn't 2R. It's greater than 2R.

If you look at the lower sketch I posted the other day, line AB is the altitude, its length only appears to be 2R (2) when viewed from above.

I must load autocad onto my PC and see if I can draw it.

Hi Frank,

Yes, I thought about using Autocad or Sketchup but felt this might be cheating...................

Assuming Matthew's calculation is correct (he set the puzzle and I'm pretty sure he knew the answer when he set it - he just had to recall how to prove it) then Matthew derives a solution for in which A = arctan 1/2

This means that the Altitude (or vertical height) of the cone is twice the Base radius.

In your drawing, you have used a dimension of 1.0 for the radius and you have introduced a dimension of 2.0 which you say represents AB when viewed from above. However, this dimension of 2.0 is somewhat arbitrary. The common line between two cones does not intersect the base of these cones at the end of a common diameter with next adjacent cones. It's close !, but not exact. (at least, as far as I can tell !)

It's quite a difficult problem to visualise, never mind solve. And as I said above, i'm trusting Matthew's solution.

One or two "interesting" points emerge, now that Matthew has revealed his solution to the "perfect" cone problem.

Below I have copied a larger chunk of the spreadsheet I produced when trying to find a solution. Having started with the cone Altitudes horizontal, I then rotated them downwards in one degree intervals until Apex Angle and Rotation were equal. This produced the wrong answer because I didn't properly take into account the Cone/Cone contact line. But.....

....the table does show that at 30 degree rotation, a cone with "Altitude = Base Diameter" and "Apex Angle = 53.13 degrees" fits the Bill, so to speak. Not the right answer, but IMHO, interesting never-the-less.

Created a cone having the same base diameter and height. Multiplied and positioned on a virtual plane, six of them fit together perfectly.

Drawing them seems to indicate the line of contact between the cones isn’t in the position I expected it to be.

Nice drawings Frank !

It needs Cones where the Altitude and Base Diameter are the same, as you have now found.

Like I said above -"The common line between two cones does not intersect the base of these cones at the end of a common diameter with next adjacent cones. It's close !, but not exact. (at least, as far as I can tell !)"

I find the solution to Matthew's problem very satisfying. And those drawings are pretty cool as well.

If two men each take the other's mother in marriage, what would be the relationship between their sons ?

x² - x² = x² - x²

x(x - x) = (x + x)(x - x)

x(x - x) = (x + x)(x - x)

x = (x + x)

x = 2x

x = 2x

1 = 2       QED

Not sure if this belongs here, or in the Best Jokes thread ?

4 times. 3 times to roll over the circumference plus 1 time to orbit the big circle.

Don Atkinson posted:

x² - x² = x² - x²

x(x - x) = (x + x)(x - x)

x(x - x) = (x + x)(x - x)

x = (x + x)

x = 2x

x = 2x

1 = 2       QED

Not sure if this belongs here, or in the Best Jokes thread ?

Logical fallacy is the factoring, multiplication and then division by zero (X-X).

winkyincanada posted:

Don Atkinson posted:

x² - x² = x² - x²

x(x - x) = (x + x)(x - x)

x(x - x) = (x + x)(x - x)

x = (x + x)

x = 2x

x = 2x

1 = 2       QED

Not sure if this belongs here, or in the Best Jokes thread ?

Logical fallacy is the factoring, multiplication and then division by zero (X-X).

Exactly. Even the first line is just a convoluted way of writing 0 = 0      .................

I did put this one in the "Jokes" thread as well

winkyincanada posted:

4 times. 3 times to roll over the circumference plus 1 time to orbit the big circle.

Well spotted. It's surprising (well, in my experience) how many people don't recognise the additional turn made when rotating around a circle. Roll a couple of Toonies around each other and that extra turn becomes obvious !

The Areas of the large quarter segment and the 2 semicircles are the same πR∧2 so that if the semicircles didn't overlap their total area would exactly match that of the large quarter circle. Therefore the area of the overlap must be equal to the uncovered area of the quarter circle.

Nice explanation sj. Neat.

Some people might find it easier to follow if they draw the full large circle and the four overlapping small circles.

Area of large circle = πR² and

Area of 4 smaller circles = 4π(R/2)² which of course reduces to πR²

We have an 8-litre jug of beer. We also have a 5-litre jug and a 3-litre jug, both of which are empty.

None of the jugs has a measure on it.

Fill one of the jugs with exactly 4 litres of beer.

Fill the 5L jug from the 8L jug and then fill the 3L jug from the 5L jug.

You then have a split of 3L, 2L and 3L.

Empty the 3L jug contents back into the 8L jug and the 2L contents of the 5L jug into the empty 3L jug. Refill the 5L jug from the 8L jug and pour out 1L out of the 5L jug to fill the 3L jug leaving 4L in the 5L jug.

Then go for a lie down after drinking the beer.