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

# A Fistful of Brain Teasers

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 exactlywhen 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.

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.

*Now this one should be straightforward and easy…………!*

**Bruce's Bashers**

In the Yorkshire Dales inter-village cricket Cup Final, Arthur’s Eleven batted first and were over the moon with their run score.

Bruce’s Bashers responded with a somewhat mediocre three-quarters of Arthur’s Eleven score.

In their second innings, Arthur’s Eleven, only managed three quarters of Bruce’s Bashers first innings score, and Bruce’s Bashers responded with a disappointing three quarters of those runs in their second innings !

So the run score for each innings was three quarters of the previous innings.

Arthur’s Eleven were nevertheless delighted to have won by 50 runs.

**How many runs did each team score in each innings ?**

Don Atkinson posted:

Now this one should be straightforward and easy…………!

Bruce's BashersIn the Yorkshire Dales inter-village cricket Cup Final, Arthur’s Eleven batted first and were over the moon with their run score.

Bruce’s Bashers responded with a somewhat mediocre three-quarters of Arthur’s Eleven score.

In their second innings, Arthur’s Eleven, only managed three quarters of Bruce’s Bashers first innings score, and Bruce’s Bashers responded with a disappointing three quarters of those runs in their second innings !

So the run score for each innings was three quarters of the previous innings.

Arthur’s Eleven were nevertheless delighted to have won by 50 runs.

How many runs did each team score in each innings ?

128, 96, 72, 54

Nicely scored Eoin !

**Cycle Journey !** *(but of course it could be a car, or a motor bike, or simply a walk.......)*

Two cyclists, on the road at Aldermaston wish to go to Burfield which for the sake of brevity are marked as A and B on the map above.

Boris the Biker decided to go via the crossroad at Downton (D) which is 6 miles, then take the straight road to Burfield which is another 15 miles.

(All roads are dead straight and the junction at Downton is a 90 deg crossroad)

Charlie the Challenger said he would take the upper road via Compton (C)

Strangely enough, they found the distance each way to be exactly the same.

Can you answer the simple question……**” How far is it from Aldermaston to Compton ?”**

Picked this up on holiday, and seemed easy enough to not take any time, and indeed it was: 3 1/3rd miles.

pythagorus

bc^2=(ac+6)^2+15^2

Ac+bc=21

Ac=21-bc

Bc^2=(27-bc)^2+225

Bc^2= bc^2-54bc+729+225

954-54bc=0

954=54bc

Bc=17.667

Ac=21-17.667=3.333

Innocent Bystander posted:Picked this up on holiday, and

seemed easy enough to not take any time, and indeed it was: 3 1/3rd miles.pythagorus

bc^2=(ac+6)^2+15^2

Ac+bc=21

Ac=21-bc

Bc^2=(27-bc)^2+225

Bc^2= bc^2-54bc+729+225

954-54bc=0

954=54bc

Bc=17.667

Ac=21-17.667=3.333

yep ! I noticed that I posted a couple of "moderate" ones and nobody seemed willing to tackle them, so I searched high and low for a couple of "easy" ones. Well spotted

......and nice explanation of your solution.

Another Easy one......

Talking to two of my students the other day, I asked them to state their ages. They did so.

Then to test their arithmetical powers I asked them to add the two ages together.

One gave me 44 as his answer and the other gave me 1280.

I immediately realised that the first had actually subtracted their ages, whilst the other had multiplied them together.

What are their ages ?

Don Atkinson posted:Another Easy one......

Talking to two of my students the other day, I asked them to state their ages. They did so.

Then to test their arithmetical powers I asked them to add the two ages together.

One gave me 44 as his answer and the other gave me 1280.

I immediately realised that the first had actually subtracted their ages, whilst the other had multiplied them together.

What are their ages ?

Never mind that, when do their lessons in English comprehension start?!

The ages should be 64 and 20, but Christopher has made a very valid point

Don Atkinson posted:Another Easy one......

Talking to two of my students the other day, I asked them to state their ages. They did so.

Then to test their arithmetical powers I asked them to add the two ages together.

One gave me 44 as his answer and the other gave me 1280.

I immediately realised that the first had actually subtracted their ages, whilst the other had multiplied them together.

What are their ages ?

64 and 20, by trying a couple of pairs of factors of 1280. Formal method would be have x-y=44, x*y=1280, get the quadratic and solve.

Christopher_M posted:Don Atkinson posted:Another Easy one......

Talking to two of my students the other day, I asked them to state their ages. They did so.

Then to test their arithmetical powers I asked them to add the two ages together.

One gave me 44 as his answer and the other gave me 1280.

What are their ages ?

Never mind that, when do their lessons in English comprehension start?!

If the question was “what is the sum of your ages”, I could understand, if neither’s education had included the verb to sum.

Mind you, with those ages, one probably has had no contact with maths since leaving school best part of 50 years ago, and the other probably dossed through it ... and both must have been desperate otherwise they wouldn’t have chosen Don as a teacher!

Innocent Bystander posted:Christopher_M posted:Don Atkinson posted:

If the question was “what is the sum of your ages”, I could understand, if neither’s education had included the verb to sum.

My inner pedant forces me to point out that “sum” in the question is using the word as a noun.

Chris has a valid point

Mulberry has the right answer

Eoin also has the right answer and a neat outline of the “formal” methodology

IB raises an interesting aside. Fortunately I didn’t use the word “sum”........whether it’s a noun or a verb

makes me realise just how carefully I need to be when drafting these teasers ........i’ll recheck my next one very, very carefully - I might be some time.....

Eoink posted:Innocent Bystander posted:Christopher_M posted:Don Atkinson posted:

If the question was “what is the sum of your ages”, I could understand, if neither’s education had included the verb to sum.

My inner pedant forces me to point out that “sum” in the question is using the word as a noun.

Touché!

Well, after the little diversion over the word "sum". I thought "words" and "riddles" might be more up your street.

*He went to the wood and caught it*

*He sat himself down and sought it*

*Because he could not find it*

*Home with him he brought it*

*What was it ?*

Of course, it could be a number of things I guess, but the thing about riddles is that * only my answer is correct* no matter how reasonable, logical or obvious all the other answers might be

A flea or tick?

Eoink posted:A flea or tick?

Not sure that fleas and ticks are particularly associated with wood or woods.

Not that I’m any further! Many things you could catch in a wood, but why **sit down** to seek it? And the implications are both that it was difficult to find (so, likely quite small) and that he would not have brought it home if he had found it when he sought it (maybe undesirable, or just better left in the wood). From the latter two points I get the flea/tick idea, but I don’t feel that fits the whole.

I struggled with the first 2 lines, tick was my attempt to think of something you might get in the woods. (I spend most of my life in a city.) I rejected teddy bears' picnic for not fitting the other 3 lines. Sitting down to find something felt like something in your shoe, but that got me no further really.

Well, the concept of ticks is reasonable. But as I said in the post, reasonable per se, isn’t good enough ! Nice try !

bear in mind that most of us don’t know we’ve picked up a tick, so wouldn’t be seeking it whilst still in the woods.

But your connection with “wood” and “felt like something in my.....” are worth developing.

Is it a splinter in a toe? So the wood is a piece of wood which you catch with a bare foot, get a splinter in a toe, sit down to try to remove it, then come home still with the splinter under your skin.

Eoink posted:Is it a splinter in a toe? So the wood is a piece of wood which you catch with a bare foot, get a splinter in a toe, sit down to try to remove it, then come home still with the splinter under your skin.

Sounds good!

I

Eoink posted:Is it a splinter in a toe? So the wood is a piece of wood which you catch with a bare foot, get a splinter in a toe, sit down to try to remove it, then come home still with the splinter under your skin.

I think that will do nicely.................although the official object is a "thorn"

It just goes to show how irritating these Riddles can be !

Here's another to fascinate or irritate.........

*Before my birth I had a name,*

*But soon as born I chang'd the same;*

*And when i'm laid within the tomb,*

*I shall my father's name assume.*

*I change my name three days together,*

*Yet live but one, in any weather.*

Don Atkinson posted:I

Eoink posted:I think that will do nicely.................although the official object is a "thorn"

It just goes to show how irritating these Riddles can be !

My challenge, if challenges were allowed, is that you don’ catch a thorn (or splinter), rather it catches you...