21 replies to this topic

[TimeSpaceLightForce]

I totally agree--Badhri's principle is much more practical in terms of masses.

..too many moves and uncertainties

[famous1304]

If using the LCD screen, I would put all 4 on the scale, and remove one until one took off more weight than the others.  only using the scale once.  Or you could go the opposite route, and you would have to put them all on the scale one at a time as to not spill.  So just watch the weights as you put a new cup on and you would know which one had the poison.

[CaptainEd]

Spoiler for Badhri's principle is the winner, details need cleanup

Badhri's principle is to construct a situation with differing fractions of the four glasses, so that we can compute which glass it was that had added mass.

However, the OP says that the glasses are "full", which means there's no room to add from one into another.

So, here's an alternative:

(1) I gather that we are given:
* A, the weight of one glass full of water
* X, the weight of added poison.

(2) remove all glasses from the scale, without peeking at the scale.
(3) Using a volume measuring device, such as a pipet, measure the volume in G1, and empty it, call that volume V.
(4) measure V/6 from G2, transfer to G1
(5) measure V/3 from G3, transfer to G1
(6) measure V/2 from G4, transfer to G1
(7) At this point, G1 is full
(8) place G1 on the scale.
(9) If G1 weighs A, original G1 had the poison
--if G1 weighs A + X/6, original G2 had the poison
--if G1 weighs A + X/3, original G3 had the poison
--if G1 weighs A + X/2, original G4 had the poison

[TimeSpaceLightForce]

If using the LCD screen, I would put all 4 on the scale, and remove one until one took off more weight than the others.  only using the scale once.  Or you could go the opposite route, and you would have to put them all on the scale one at a time as to not spill.  So just watch the weights as you put a new cup on and you would know which one had the poison.

Your solution is right unless making the machine do its job means using it
(in that case reading weight trice). It is resting if nothing is on it. If there is

a reading already using once means having a new reading..

Use the scale tray if you would like take them  off  or place them on the scale

Edited by TimeSpaceLightForce, 17 December 2012 - 06:45 PM.

[CaptainEd]

Spoiler for OK, a=a solution=solution with=with no=no pouring=pouring at=at all'=all

on the laboratory bench next to the digital scale, place a knife edge as a fulcrum, and balance a plank of length 2 meters on the knife edge, with one end touching the digital scale.
At this point, the plank is adding exactly no weight to the scale.

Now,
* place G1 on the plank directly over the scale,
* place G2 on the plank at a distance of 1/2 from the fulcrum
* place G3 on the plank at a distance of 1/3 from the fulcrum
* place G4 on the plank at a distance of 1/4 from the fulcrum

* read the scale

* Without the poison, the scale would read G1 + G2*1/2 + G3*1/3 + G4*1/4 will be (A+A/2+A/3+A/4) =

(12A + 6A + 4A + 3A)/12 = (25A/12)

However, the poison of weight X will either appear as an addition of X (if in G1), X/2 (if in G2), X/3 (if in G3) or X/4 (if in G4).
(Firefox doesn't seem to play well with this spoiler editor.)

Edited by CaptainEd, 17 December 2012 - 08:02 PM.

[TimeSpaceLightForce]

Spoiler for Badhri's principle is the winner, details need cleanup

Badhri's principle is to construct a situation with differing fractions of the four glasses, so that we can compute which glass it was that had added mass.

However, the OP says that the glasses are "full", which means there's no room to add from one into another.

So, here's an alternative:

(1) I gather that we are given:
* A, the weight of one glass full of water
* X, the weight of added poison.

(2) remove all glasses from the scale, without peeking at the scale.
(3) Using a volume measuring device, such as a pipet, measure the volume in G1, and empty it, call that volume V.
(4) measure V/6 from G2, transfer to G1
(5) measure V/3 from G3, transfer to G1
(6) measure V/2 from G4, transfer to G1
(7) At this point, G1 is full
(8) place G1 on the scale.
(9) If G1 weighs A, original G1 had the poison
--if G1 weighs A + X/6, original G2 had the poison
--if G1 weighs A + X/3, original G3 had the poison
--if G1 weighs A + X/2, original G4 had the poison

if you have estimated half a glass and it  weighs more than half the water weight, the additional weight can be water or poison's..thats why the volume must be precisely measured as done here..still that is troublesome.. the glass weight shall be considered too in that case

[TimeSpaceLightForce]

I'm not sure I would accept this challange. But if I had to , place two glasses on scale. Either these two, or the other two weigh 2A+x. We now know that the poison is one of of the two heavier pairs. Now the hard part : drink one of these glasses. If you don't die, you know the other glass is poison. If you die, you go to your grave knowing you were also correct

This is one solution if you have a goldfish in the bowl

Edited by TimeSpaceLightForce, 18 December 2012 - 10:09 AM.

[TimeSpaceLightForce]

Spoiler for OK, a=a solution=solution with=with no=no pouring=pouring at=at all'=all

on the laboratory bench next to the digital scale, place a knife edge as a fulcrum, and balance a plank of length 2 meters on the knife edge, with one end touching the digital scale.
At this point, the plank is adding exactly no weight to the scale.

Now,
* place G1 on the plank directly over the scale,
* place G2 on the plank at a distance of 1/2 from the fulcrum
* place G3 on the plank at a distance of 1/3 from the fulcrum
* place G4 on the plank at a distance of 1/4 from the fulcrum

* read the scale

* Without the poison, the scale would read G1 + G2*1/2 + G3*1/3 + G4*1/4 will be (A+A/2+A/3+A/4) =

(12A + 6A + 4A + 3A)/12 = (25A/12)

However, the poison of weight X will either appear as an addition of X (if in G1), X/2 (if in G2), X/3 (if in G3) or X/4 (if in G4).
(Firefox doesn't seem to play well with this spoiler editor.)

I can not say no to this one  CaptainEd it is so resourceful and using mechanics is cool..using knife and bench and plank.

Spoiler for

You might as well look at the corner cabinet for the poison scanner

[CaptainEd]

I can not say no to this one  CaptainEd it is so resourceful and using mechanics is cool..using knife and bench and plank.

Spoiler for

You might as well look at the corner cabinet for the poison scanner

Glad you like it, but now, in the absence of a poison scanner dog, what in the world was the solution you were envisioning?

[TimeSpaceLightForce]

Spoiler for Lethal Solution

actually the lever method is the best next to pipet solution ..

or

Take glasses 1 2 3 4 off the scale

Spill some of 2 then refill from 1

Weigh mixture 2 with 3. If reading:

a bit > A+A : glass1 has it

a bit < A+A+x :glass2

=A+A+x :glass3

=A+A :glass4

 

note: taking the 4 glasses at the same time with two hands needs caution

         the purpose of spilling some of glass2 content is to make sufficient

         difference on poison weight (e.i about 2% to 20% water volume).

         The scale can read milligram anyway 2%gram=20mg   98%gram=980mg.

Warning: DO NOT try this at home !