Hey ! hang on a minute, something's not quite right here !

Are my eyes deceiving me ?

The hypotenuse on the upper triangle looks ever so slightly concave and the lower one looks convex.

dd

You must have good eyes then

You must have good eyes then

Maybe I should have more faith in what my eyes are telling me then

dd

Maybe I should have more faith in what my eyes are telling me then

dd

That's a good idea, also in trading

its a weird one!

the orange block = 7 squares
light green block = 8 squares
green triangle = 5 squares ( (5 x 2) / 2)
red triangle = 12 squares ( (8 x 3) / 2)

total surface area = 7 + 8 + 5 + 12 = 32 squares

big triangle = 32.5!!! ( (13 x 5) / 2)
I dont understand how the big triangle surface area is 0.5 square bigger than the smaller pieces added together.

Am I making false assumptions about the surface area of the triangles? (Length x Height / 2) ??

Exactly. The illusion of a large triangle is made by the two smaller triangles. But, if you draw a new hypotenuse from the two larger sides there is indeed a difference in the area. God bless AutoCAD

1) weigh 6 against 6 - heavier ball is on either left or right - discard all balls on lighter side.

2) weigh remaining 6 - 3 against 3 - heavier ball is either on left or right - discard lighter balls

3) weigh 2 of the 3 balls against each other; if both are equal, heavier ball is the one remaining ball not on the scales... if heavier ball IS being weighed, it will tip the balance in it's favour.

" ... whether it's heavier or lighter"

arent you assuming its heavier?

" ... whether it's heavier or lighter"

arent you assuming its heavier?

where do I say that?

where do I say that?

I dont know how you account for possibility that the offending ball is lighter.

If I weigh 6 against 6, I dont know if its (tipping to one side) caused by the heavier ball being on one side of the scale, or caused by the lighter ball being on the other side of the scale.

I dont know how you account for possibility that the offending ball is lighter.

If I weigh 6 against 6, I dont know if its (tipping to one side) caused by the heavier ball being on one side of the scale, or caused by the lighter ball being on the other side of the scale.

That's right Trendie.

Sorry MrGecko, it ain't that easy

You say that after you weigh 6 against 6 you are discarding all balls on the lighter side, thus you are assuming that among the 6 balls you take to weigh 3 against 3, one is heavier. This can't be assumed since there might be one lighter ball in one of the 6 you choose to ignore.

I dont know how you account for possibility that the offending ball is lighter.

If I weigh 6 against 6, I dont know if its (tipping to one side) caused by the heavier ball being on one side of the scale, or caused by the lighter ball being on the other side of the scale.

Aah.. ermmm.... yes, well....

I had mis-read the question, and excluded the possibility that the ball could be lighter...

[insert general abuse here]

OK Folks, my competitiveness really came out here trying to be the 1st to get it. Must say the volatility today has helped, I'm much less inclided to take on positions and play with the puzzle instead...

this was DIFFICLULt... anyway, here is my second, rather less c0cky attempt at solving it. I havent covered all the possible permutations, but the methodology is there (i.e. if 1a 1b 1c 1d < 2a 2b 2c 2d, do the same thing except the other way around. Or just re-label the balls).

Throughout, > means "is heavier than" and < means "is lighter than"... e.g. Car > Dog, duck < feather

OK here goes.....

split the balls into 3 groups of 4;

Group 1 = [1a, 1b, 1c, 1d]
Group 2 = [2a, 2b, 2c, 2d]
Group 3 = [3a, 3b, 3c, 3d]

then, weigh Group 1 vs. Group 2:

If they are EQUAL, then group 3 must contain the odd ball (If we keep an equal number of balls on each side, for every "go", if they balance the ODD ball must NOT be present, and if they DONT balance then the ODD ball MUST be present).

Then collect group 1 (which we know are all EVEN) and compare them with group 3 (which we know one of which is ODD)...

remove one ball from each group, and weigh:

1a 1b 1c vs. 3a 3b 3c

Scenario 1:
If they Balance, we know ball 3d must be the ODD one, so compare it with 1d (which we know is EVEN) to determine whether is it heavy or light.

Scenario 2:

1a 1b 1c > 3a 3b 3c

therefore we know 3d must be EVEN, and one of 3a 3b 3c is LIGHT...

weigh 3a vs 3b:

if 3a > 3b, then 3b is LIGHT
if 3a < 3b, then 3a is LIGHT
if 3a = 3b, then 3c is LIGHT

reverse the process if 1a 1b 1c < 3a 3b 3c, except the odd ball must be HEAVY

______________________________________________________________________

Now it gets harder...

back to weighing go 1, where 1a 1b 1c 1d vs. 2a 2b 2c 2d

say 1a 1b 1c 1d > 2a 2b 2c 2d

COMMENT:

We can do 3 changes to the arrangement of the balls. we can...

perform [X] = remove an equal number of balls from either side

perform [Y] = replace an unknown ball with a known ball

perform [Z] = switch any number of balls from one side to the other

** note: re-name all group 3 balls to E (e.g. 3a = 3b = 3c = 3d = E) as we know they are all EVEN, because the odd ball MUST be present for the scales NOT to balance **

now, after our 1st weighing of balls, we have

1a 1b 1c 1d > 2a 2b 2c 2d

then, perform [X] and remove ONE ball from either side, 1d and 2d, leaves us with

1a 1b 1c vs. 2a 2b 2c

then perform [Y], replace ONE UNKNOWN ball with a KNOWN ball E from the 3rd group, say 2c, leaves us with

1a 1b 1c vs. 2a 2b E

and lastly perform [Z]. swap one ball on each side, say swap 1a with 2a, leaves us with

2a 1b 1c vs. 1a 2b E

now weight them, our 2nd "go"......

Scenario 1 part 1

2a 1b 1c > 1a 2b E

then either 1b or 1c is HEAVY, as they have remained on the heavy side

OR

2b is LIGHT, as it has remained on the LIGHT side

in which case.....

Scenario 1 part 2

remove 2b, which is either LIGHT or EVEN, and weigh 1b vs 1c....

if 1b > 1c then 1b is HEAVY
if 1b < 1c then 1c is HEAVY
if 1b = 1c then 2b is LIGHT

Scenario 2 part 1

2a 1b 1c = 1a 2b E

then we have removed the ODD ball, so either

1d is HEAVY (ball removed from heavier side)

OR

2c or 2d is LIGHT (the balls removed from the LIGHT side)

then....

Scenario 2 part 2

remove 1d, which must be HEAVY or EVEN...

weigh 2c vs 2d

if 2c > 2d then 2d is LIGHT
if 2c < 2d then 2c is LIGHT
if 2c = 2d then 1d is HEAVY

Scenario 3

2a 1b 1c < 1a 2b E (i.e. the heavier and lighter side have swapped over)

then we have removed 2 balls much must be EVEN, as the ODD ball MUST be present.

then either 1a is HEAVY (swapped over, keeping on the heavy side)

OR

2a is LIGHT (swapped over, keeping on the light side)

then

Scenario 3 part 2

weigh 1a vs E

if 1a > E then 1a is HEAVY
if 1a = E then 2a is LIGHT

note 1a < E is impossible

______________________________________________________________

thats it... if the scales go the other way at any stage, just flip the notation around.

I would scan the workings out, but I did them all on A3 paper and only got an A4 scanner

OK Folks, my competitiveness really came out here trying to be the 1st to get it. Must say the volatility today has helped, I'm much less inclided to take on positions and play with the puzzle instead...

:clap: Well done Mr Gecko

well done.
I now realise my 6 against 6 is a wasted "go", as no informaton can be gleaned from it.
weighing by 4s was the way to go, as you can deduce something from the first go.

well done again.

OK Folks, my competitiveness really came out here trying to be the 1st to get it. Must say the volatility today has helped, I'm much less inclided to take on positions and play with the puzzle instead...

Excellent stuff MrGecko :clap:. I can see you put a lot of work in that. My "eureka" moment came when I realized I had to switch balls from sides. It's not really the first thing that comes to mind. I can't rep you anymore right now, but you definitely deserve one for being the first to solve it, and writing it all out.

I bet you want to challenge me with a similar difficult puzzle now

This is my answer to this puzzle:

The itinerary is:

A -> B -> C -> I > C -> D -> E -> F -> C -> D -> E -> F -> G -> H -> I -> A

Yes, that one will do it FW (there are others)

These are rather simple problems to solve, trial and error will do it - however, it is possible to find the length of the result WITHOUT doing any trial and error... you can deduce the length is 4800 without taking a step out of the door, as it were....

Also, I added an extension - Each letter MUST be visited, but not each path... what is the length of the shortest route??

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