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

No need for the ? Nigel, spot-on ! Well done.

I used a "tree" diagram for this one. It revealed 8 possible outcomes, three of which were the required two heads and a tail.

Nice one Ian.

Got me thinking.

I have a solution (I think) starting with two straight lines that progressively get longer but retain their proportionality......................

.................but i'll wait until others have time to give it some thought !

Cheers, Don

My solution............

Draw straight lines AA' and BB' where A' and B' are on the shoreline and AA' and BB' are perpendicular to the shore.

Temporarily move A' and B' along the shoreline towards each other, maintaining the initial ratio AA':BB' until A' and B' meet at H (harbour).

Triangles AA'H and BB'H are similar triangles.

AH + BH look   as if they minimise the total railway length, but I haven't checked it mathematically.

Don Atkinson posted:

My solution............

Draw straight lines AA' and BB' where A' and B' are on the shoreline and AA' and BB' are perpendicular to the shore.

Temporarily move A' and B' along the shoreline towards each other, maintaining the initial ratio AA':BB' until A' and B' meet at H (harbour).

Triangles AA'H and BB'H are similar triangles.

AH + BH look   as if they minimise the total railway length, but I haven't checked it mathematically.

 

Well done Don, if I understood your answer correctly I think it is correct. There is however a simpler construction which gives the same answer and makes it clear that the railway length is the minimum one. A hint might be to consider using a point in the water as part of your construction.....

I have a parallelogram in mind at the moment..........but only because it "looks" about right.....

..........unless somebody has a better idea,,,,,,,,?

HI Don, 

The construct I have in mind is draw a line from town A perpendicular to the shoreline but extended  into the sea by the same distance as A is from the shore, call this point A'. Now the shortest distance from A' to B is the straight line as drawn and AH and A'H are the same length so AHB  must be the shortest railway to build. As in your solution the the angles between AH and the shoreline and BH and the shoreline are the same.  

Ian 

The other solution (starting with B and B') will also work and gives the same answer. 

Don Atkinson posted:

Ian G. posted:

Don Atkinson posted:

....mealsothinks you guys are correct !!  Really well done !!

Does either of you wish to reveal his working ? Mine is based on 1- (prob of not hitting 3x20)

1 - (0.6)ⁿ where n = number of throws.

Great minds think alike....

....let's not complete that saying..........

for the benefit of anybody who might be wondering.....this is how I finished it (*)

1 - (0.6)ⁿ > 0.9

(0.6)ⁿ < 0.1

n log (0.6) < log (0.1)

n > log (0.1)/log(0.6)

n > 4.5

hence 5 throws are necessary

(*) but as Nick (and Ian ?) probably did, just keep multiplying 0.6 by itself until it gets smaller than 0.1 ......it takes 5 goes (0.6 x 0.6 x 0.6 x 0.6 x 0.6) to get below 0.1

Just wanted to correct a typo (benefit of the doubt) in case of any students in the audience...

Regarding the shoreline, make a running loop of string. Place thumbtacks at A and B. Loop the string around them.Put a third tack in the loop. Keeping the loop taut, run the third tack up and down the shoreline until the non-loop part of the string is the longest. 

Neat solution Ian. Neat !

Here's the picture of the "parallelograms" that I mentioned earlier. Having drawn the "almost straight" shoreline a bit too squiggly, I was struggling to convince myself that my solution was adequate.

Likewise I wasn't too sure about my "similar" triangles or the "proportionally" elongating rods...

Matthew T posted:

Don Atkinson posted:

Ian G. posted:

Don Atkinson posted:

....mealsothinks you guys are correct !!  Really well done !!

Does either of you wish to reveal his working ? Mine is based on 1- (prob of not hitting 3x20)

1 - (0.6)ⁿ where n = number of throws.

Great minds think alike....

....let's not complete that saying..........

for the benefit of anybody who might be wondering.....this is how I finished it (*)

1 - (0.6)ⁿ > 0.9

(0.6)ⁿ < 0.1

n log (0.6) < log (0.1)

n > log (0.1)/log(0.6)

n > 4.5

hence 5 throws are necessary

(*) but as Nick (and Ian ?) probably did, just keep multiplying 0.6 by itself until it gets smaller than 0.1 ......it takes 5 goes (0.6 x 0.6 x 0.6 x 0.6 x 0.6) to get below 0.1

Just wanted to correct a typo (benefit of the doubt) in case of any students in the audience...

Whoops !! Thank you Matthew. Glad somebody is auditing this thread !

JRHardee posted:

Regarding the shoreline, make a running loop of string. Place thumbtacks at A and B. Loop the string around them.Put a third tack in the loop. Keeping the loop taut, run the third tack up and down the shoreline until the non-loop part of the string is the longest. 

Nice, JR.

Took me a moment or two to figure out what you were describing, but once the penny dropped, I thought "Nice !" and "it avoids getting the old feet wet !!"

Ian's shoreline solution, with its triangles, got me thinking back to some of those Triangle rules such as The Sine Rule, The Cosine Rule and.....................memories of maths............

So, I have small plot of grass in the shape of a triangle with sides 8m x 12m x 16m. The area of the grass is ?

Diagrams W, X, Y and Z illustrate triangles with Lines that are concurrent at D, E, F and G

The points D, E, F and G are respectively called...................... ?

Don Atkinson posted:

Ian's shoreline solution, with its triangles, got me thinking back to some of those Triangle rules such as The Sine Rule, The Cosine Rule and.....................memories of maths............

So, I have small plot of grass in the shape of a triangle with sides 8m x 12m x 16m. The area of the grass is ?

46.47 sq m. and just to be contrary solved without using any trigonometry at all. And no I didn't draw it out on squared graph paper or draw it, cut it out and weigh it. ( In fact in ye olden pre computer days this was how scientists sometimes did extract the area under a measured curve - literally they would cut it out and weigh it compared to the weight of a  known (square) shape - honest )   

Good answer Ian.  (ie sq.rt 2160)

Curious how you did it without trigonometry. BTW, nice to learn about the weight/area ratio cf a known weight/area as well.

I used the "semi-perimeter" formula which, for some reason or other, has always stuck in my mind, a bit like the Sine Rule with the "extra" "2R" stuck on the end.

Cheers

Don

Don Atkinson posted:

Good answer Ian.  (ie sq.rt 2160)

Curious how you did it without trigonometry. BTW, nice to learn about the weight/area ratio cf a known weight/area as well.

I used the "semi-perimeter" formula which, for some reason or other, has always stuck in my mind, a bit like the Sine Rule with the "extra" "2R" stuck on the end.

Cheers

Don

Never heard of the semi-perimeter formula - nor as it turns out lots of things about triangles! Googled it and found this interesting (to the likes of us) Wikipedia page.  https://en.wikipedia.org/wiki/Triangle

Here's my working for the answer. 

Nice Wikipedia page Ian. I had to refresh my geometry from my old school book before I drew those triangles and their various "centres". I should have tried Wikipedia first !!!

The semi-perimeter rule shows that for any triangle in which all three sides are defined, the area is

√[s(s-a)(s-b)(s-c)]

where a, b, c are the three sides and s = ½(a+b+c) = 18 in the above case

so Area = √[18.10.6.2] = √[2160] = 46.4758

Don Atkinson posted:

Nice Wikipedia page Ian. I had to refresh my geometry from my old school book before I drew those triangles and their various "centres". I should have tried Wikipedia first !!!

The semi-perimeter rule shows that for any triangle in which all three sides are defined, the area is

√[s(s-a)(s-b)(s-c)]

where a, b, c are the three sides and s = ½(a+b+c) = 18 in the above case

so Area = √[18.10.6.2] = √[2160] = 46.4758

Turns out this formula is also known as Heron's formula and can be derived using exactly the algabraic approach I used to solve the original problem

see https://en.wikipedia.org/wiki/Heron's_formula for this and some other interesting stuff ( to me anyhow ) 

Ian 

I'm not sure that "my" term "Semi-perimeter Rule" is an official name. It might just be a name I gave it to help me remember how it works. I don't recall it ever being called Heron's Formula - you learn something new every day !!

Cheers, Don

The next chapter in my old school maths book covered Combinations and Permutations...............it started of quite easy.............

How many three digit numbers can be made from the integers 2, 3, 4, 5, 6 if :-

a) each integer is used only once.

b) there is no restriction on the number of times each integer can be used.

a) 60

b) 125

Made the initial error of assuming there were 6 numbers to choose from! Always read the question!

Yes Nick, funny how many people either "see" a "1" where none exists, or "assume" that when the last number is 6, there are in fact six numbers to choose from (*)

But well done, and well done again for reading the question a second time !

(*) sorry HH, I guess that should read   "....six numbers from which to choose" ??

Staying with the permutation starters......

How many even numbers greater than 2,000 can be made from the integers 1, 2, 3, and 4 if each time each integer can only be used once ?