A Fistful of Brain Teasers

Thanks Don, my brain’s a bit frazzled after one of “those days” at work, I’ll try to have a look tomorrow. 

Eoink posted:

Thanks Don, my brain’s a bit frazzled after one of “those days” at work, I’ll try to have a look tomorrow. 

There’s no rush Eoin, these brain teasers are meant to be enjoyable and a distraction from the pressures of work.

The ref to Teresa May etc was simply topical melodrama 

This is my "integration" solution.

I hope it is readable, but I will post a typed-up version later if necessary.

This satisfies Putin's IB's challenge above to "...provide proof that there is a solution"

I await an alternative solution, hint - the one that Eoin outlined above would do nicely and is dead easy to do.

Don Atkinson posted:

This is my "integration" solution.

I hope it is readable, but I will post a typed-up version later if necessary.

This satisfies Putin's IB's challenge above to "...provide proof that there is a solution"

I await an alternative solution, hint - the one that Eoin outlined above would do nicely and is dead easy to do.

It’s easy when you solve the right way Don, my first pass was 3 sheets of A4.

We have a sphere drilled through. The sphere is thus a bead made up of a sphere minus a cylinder through the sphere and minus two caps at each end. Let us assign the drilled hole an arbitrary radius of r. Let us say the caps have height h. (I’m known for originality in unit naming.)

Looking at Don’s picture above, we can easily see that the sphere has a diameter of 6+2*h, the length of the cylinder plus the height of the two caps. So it has radius 3+h.

The volume of the sphere is 4/3 Pi rsphere^3 = 4/3 pi (3+h) ^ 3 =

4/3*pi*h^3 + 12*pi*h^2 + 36*pi*h + 36*pi

——————————-

Google tells me that rsphere = (r^2+h^2)/2*h = 3+h which resolves to r^2 = 6*h + h^2

_____________

The volume of the cylinder is 6*pi*r^2 = 6*pi*(6*h + h^2) = 36*pi*h + 6*pi*h^2

Again from Google, the volume of a cap is pi/3*(3*r^2*h + h^3), substitute for r^2 and the volume of 2 caps is 6*pi*h^2 + 4*pi/3*h^3

Subtract the volume of the cylinder and the two caps from the volume of the sphere, all the h terms cancel out and you have 36*pi^2.

I can’t help thinking my 16 year old self would have seen this as a light workout before starting the hard bit of A’Level maths, not any more.

Great question, thanks Don.

Well done Eoin.

quite satisfying when you get a result like that 

Don Atkinson posted:

Well done Eoin.

quite satisfying when you get a result like that 

Thanks Don, very satisfying, especially after I realised that resolving r and h for r and not for h made the equations about half the length.  Thanks again, a good mental workout.

My version of the "geometry" solution.

Some of the intermediate maths is omitted eg

......in the line starting "2 x DOMES" the second equation is obtained by substituting h = R - 3 (see diagram) then expanding.

.....And in the line starting "CYLINDER", r² is substituted by R² - 9 (see diagram)

Perhaps the purists would have us derive the formulae for a sphere, a dome and a cylinder, but then the previous "integration" method is probably easier

 Anyway, I think the concept of a 6" long hole through the Earth leaving a bead the volume as a 6" long hole through a 12" globe is somewhat surprising !

Chultenham Tin Cup.

My donkey (and donkey jockey) were running in some fancy race yesterday.

The diagram shows part of the “field of play”. Both the three mile wide turf and two mile wide sand extend well to the left and right of the field of play.

My donkey and rider, in competition with others, simply had to cross the field of play from the “Start” to “Finish” by the quickest route possible. It is important to note that all the donkeys travelled at the same speed as each other and could cross the turf at twice the speed they could travel across sand.

Being (slightly) more intelligent than my donkey and my donkey jockey, I felt my role in this race was to figure out the fastest route that I should encourage my donkey and jockey to take – if we all took the same route it would be a dead heat !

The straight line from start to finish didn’t seem to me to offer the fastest route and that was clearly true because we beat all those donkeys that took the straight line.

We also happened to beat all the other donkeys as well.

What route did I choose ?

PS: this one struck me as being a bit more difficult than some of the other teasers. It might be worth sharing your various thoughts...........rather than having a race to see who's first ! (pun intended !)

Thinking logically, and without any maths, I’d take the shortest route on the sand, ie travel two miles, and then go diagonally across the turf, at twice the speed, which I think is about 7.5 miles. 

Oh bugger, this might involve vectors and I was always rubbish at them.

Starting with HH’s answer:distance across turf = 7.6158, sand 2, so at twice speed on turf the time is proportional to turf/2 + sand = 5.8079.

trial diagonal on sand  ending 0.1 mile in from start end to see is increasing: 7.5240 turf, 2.025 sand, time prop to t/2 +s = 5.7645, so does decrease doing some diagonal on sand.

If position on dividing line where the diagonal from across sand ends = distance a from start end, then distance travelled = √ ( a² + 2² ) + √ ( (7 -a)² + 3² ) and time is proportional to same with 2nd part divided by 2

so time is prop. to    ( √(a² - 14a + 58) )/2 + √( a² +4 )

This needs solving for the value of ‘a’ giving the minimum value for this sum.

Hmmm, I will have to go away and scratch my head: but time limited at mo as I’m about to go away for a couple of weeks, so don’t know whether or when I might come back to this.

Hungryhalibut posted:

Thinking logically, and without any maths, I’d take the shortest route on the sand, ie travel two miles, and then go diagonally across the turf, at twice the speed, which I think is about 7.5 miles. 

Good start HH. This is better than the straight diagonal........

.........but as IB has subsequently noted, it is better to do a "bit more" on the sand than the minimum 2 miles !

It's just "how big is that bit more"

sjbabbey posted:

Oh bugger, this might involve vectors and I was always rubbish at them.

My words exactly when I first saw this one

Innocent Bystander posted:

Starting with HH’s answer:distance across turf = 7.6158, sand 2, so at twice speed on turf the time is proportional to turf/2 + sand = 5.8079.

trial diagonal on sand  ending 0.1 mile in from start end to see is increasing: 7.5240 turf, 2.025 sand, time prop to t/2 +s = 5.7645, so does decrease doing some diagonal on sand.

If position on dividing line where the diagonal from across sand ends = distance a from start end, then distance travelled = √ ( a² + 2² ) + √ ( (7 -a)² + 3² ) and time is proportional to same with 2nd part divided by 2

so time is prop. to    ( √(a² - 14a + 58) )/2 + √( a² +4 )

This needs solving for the value of ‘a’ giving the minimum value for this sum.

Hmmm, I will have to go away and scratch my head: but time limited at mo as I’m about to go away for a couple of weeks, so don’t know whether or when I might come back to this.

Nice continuation from HH's good start......

Don Atkinson posted:

sjbabbey posted:

Oh bugger, this might involve vectors and I was always rubbish at them.

My words exactly when I first saw this one

You could follow HH's and IB's lead, avoid vectors and go for a "numerical solution"

Well, taking the formula I derived and using Ecel to cheat and run multiple numbers for ‘a’ I know the answer, and it is a nice round number. That suggests to me tgere should be a simpler way of looking at this, but time to do so eludes for now

Innocent Bystander posted:

Well, taking the formula I derived and using Ecel to cheat and run multiple numbers for ‘a’ I know the answer, and it is a nice round number. That suggests to me tgere should be a simpler way of looking at this, but time to do so eludes for now

 

That's the Numerical Solution method - nothing wrong or cheating ! Well done, you clearly have the answer - it's a neat number !

We used to do that before Excel came along, by judiciously selecting a few values for "a" then plotting a graph etc

Diferentiation often works, but you need to get the function into a format that is easy to diferentiate.

This is the solution.

IB hinted at the  answer with his "nice round number"

And HH correctly figured that crossing the sand at right angles then the turf on the diagonal was a good bit quicker than taking the "direct" diagonal.

I will post my spreadsheet version and graph in a short while.

This is the Excel spreadsheet output. There are a few hidden columns and rows, but the clear picture is that the relative times are a minimum at D = 1

I'm pretty sure this is the method used by IB.

There are other methods......................

for those who prefer the visualization of graphs............

Trying to find a couple of "easy" ones is proving a bit more difficult than I thought ! Hopefully this one should fit the bill, especially if you read the text carefully. I have even emboldened some of the text to help

A farmer owns a square field exactly 60ft x 60ft adjacent to a road. He also owns the land surrounding the field.

He builds a fence as shown by the broken line in the diagram. The fence runs from a corner of the field, past Tree No1 on the field boundary and terminates at Tree No 2 adjacent to the road. The road is parallel to the field boundary and there is a gate from the road in the "SE" corner of the field.

The length of fence from Tree No 1 to Tree No 2 is exactly 91ft.

The distance from Tree No 1 along the boundary to the gate is an exact number of feet.

How far is it from Tree No1 to the gate ?

35 feet from tree1 to gate?

Hi Steve,

good answer, it is indeed 35’.

did you try creating then solving quadratic equations or did you do as I did first.........look for “whole number” solutions for a right angle triangle with hypotenuse 91’ ?

Cheers, Don

Hi Don,

Probably thanks to one of your recent puzzles, I twigged that hypotenuse 91 is a multiple of 13, and used the factor of 7 on a 5,12,13 right angled triangle. I also got comfort from the fact that the triangle within the field is also a 5,12,13 triangle with factor 5.

Cheers, Steve

Brilliant solution Steve. Nicely spotted and nicely explained.