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plasmid

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  1. plasmid

    I'm afraid that a VERY similar riddle has already been posted here at Hotel Bill. I can't blame you for not finding the earlier post by searching the archive, though -- the old post was about people staying at a hotel instead of getting drinks at a bar. Good riddle, though!
  2. Aww, no one else is going to post a wiseguy answer? Ok... [spoiler='Here's how I would do it ']First, since there's a 20 L jug and a 5 L jug, hopefully that means my measurements won't need to be any more precise than 5 L to get the machine to work safely. Also, the scale said 83 kg and not 83.0 kg, so I only know that the weight is somewhere between 82.5 and 83.5: if the solution requires more accuracy than that, then I'm out of luck no matter what I do. But I'd still try to play it as safe as I could. I'd first fill the counterweight with 80 L. Then I'd fill the 5 L jug and take advantage of the fact that 1 L of water is about as much as I can drink in one sitting without really pushing myself overboard. After drinking about 1 L (aiming toward the high end with my estimate – better to go over rather than under as you'll see in a moment) I would pour the rest in the counterweight to bring it to 84 L (= 84 kg). Since I just drank a liter of water, the weight of myself and the goods is now 84 kg instead of 83. Now I load up the basket with myself and my goods, and prepare to release the break, but first... ...I unzip my pants. As scsw pointed out, if the pulley system is very low resistance, then I might end up plummeting down in virtual free-fall and not have the time and coordination to make adjustments. However, if I were quite startled by going into free-fall, and if I just drank a liter of water, then my natural reflexes would take care of releasing the extra weight over the side without any coordinated action on my part. As long as my pants are unzipped, that is.
  3. Holy cow, Grayven, too bad I can't quote your signature in this response but CONGRATULATIONS! There have been some good ideas so far, I think I'll hold off on posting what I had in mind and leave some time for more to come in. After all, any problem that depends on trickery is bound to have multiple good answers. But for those of you who wanted to send the goods up alone and uncontrolled to be collected later at the top, your stuff would run a big larceny risk.
  4. Clarification: Remember, the path took you halfway up the cliff face, which is where all this machinery is. The brakes are currently on, allowing you to fill the counterweight and then board the basket before releasing the brakes.
  5. In your travels, you come to a crossroads with two men who only answer yes/no questions, one of whom always tells the truth and another always lies. You are about to make short work of the puzzle, but then you notice that there are actually dozens of possible paths out from the crossroads. Mildly flustered, you turn to one of the men and ask, “If I were to ask the other fellow what your answer to this question is, would he say 'yes'?” After thinking a moment and stuttering a bit, the man's head explodes. The remaining one is so scared witless that he gives you directions, and you know that even if he were the liar he would at this point not dare to cross you. (He volunteers something along the lines of, “If you were to ask me which way you wanted to go, I'd say that way.”) You then come across a river, a fox, a chicken, a bag of grain, and a rowboat that can only carry you and one piece of cargo. Since crossing the river in a rowboat is a lot of work, you take off a shoelace and use it to muzzle the chicken, and then get everything across in five crossings instead of seven. You recover your shoelace with a bit of chicken spit (not misspelled) and continue onward. You bring your fox, chicken, and grain to a sheer cliff face. The path takes you to a spot halfway up, and at the top is the market where you want to sell them. The cliff has a pulley mechanism with a basket to bring up you and your goods, and a counterweight container to be filled with water, along with a warning: "Fill the counterweight with too little water and you'll go crashing down. Fill it with too much water and you'll be hurled up out of control. Fill it just right, and a small push will take you safely to the top. Be sure you're balanced before releasing the brakes! The empty counterweight container weighs just as much as the empty basket." You weigh yourself and your goods on an electric scale and it reads 83 kg, and runs out of power and shuts down immediately afterward. There is a 20 L jug and a 5 L jug and a waterfall to fill them with. How do you get up the cliff? (Remember from the previous two encounters that you're a bit of a wise-a**.)
  6. An isosceles triangle is a single triangle where two of its sides have the same length. The single triangle formed by the two S2 lines and the dotted line would be an isosceles triangle (regardless of what the angle between S1 and S2 is -- S1 is not part of the isosceles triangle).
  7. First some background on geometry for those who have either never taken it or have forgotten it. Two triangles are congruent if they have the same shape and size. They may be reflected or rotated or translated relative to each other, but by applying these transformations it is possible to make the two triangles superimposable. Given that there are two triangles, it is well known that if one triangle has sides of length {S1, S2, S3} and the other also has sides of length {S1, S2, S3} then the two triangles are congruent (this is sometimes called the SSS or side-side-side theorem in geometry). Similarly, if one triangle has two sides of length {S1, S2} and the angle between those two sides is {A3}, then any other triangle that also has two sides of length {S1, S2} and an angle in between them of {A3} must be congruent with it (this is called the SAS or side-angle-side theorem). There is a similar ASA theorem if you know the length of one of the sides and the angles at each end of that side. In grade school, I was quite distraught that there was no theorem stating that if you knew that two of the sides and one of the angles that does NOT fall between the two sides are equal then the triangles are congruent (that's because it's not true!), which would naturally be called the A$$ theorem. So I made up a proof for such a theorem on my own. The proof is below, see if you can spot the flaw in it. Proof: Take two triangles where you know the lengths of sides S1 and S2 and you know the angle A1 that is opposite from side S1. Position the two triangles so that their sides with length S1 are overlapping, the sides of length S2 for each triangle are both emerging from the same end of the common S1, and the triangles are pointing in opposite directions, as shown in the attached figure. Now imagine a line passing from the A1 of one triangle to the A1 of the other triangle. Consider the triangle formed by this line and the two lines of length S2. We next take advantage of a theorem on isosceles triangles: if a triangle has two sides of the same length, then the two angles opposite of those sides must have the same angle. We therefore know that the two angles labeled A1a are the same. Since the two angles A1 are the same, and the two angles A1a are the same, then the remaining angles labeled A1b in the figure (which span from the imaginary line to the side of unknown length for each triangle) must also be the same: just A1 minus A1a. Considering the triangle formed by the imaginary line and the two sides of unknown length, we can again use a theorem on isosceles triangles: if two angles of a triangle have the same angle, then the sides opposite those angles must have the same length. That proves that the two sides that were originally of unknown length must actually have the same length. Then the two triangles must be congruent, by the SSS theorem.
  8. plasmid

    Here's my explanation of the answer
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