bonanova

For the 2nd 3-3 solution here is a hint: What if there were 5 points to be placed? Here are the solutions:

Very nice! There are two steps to get to 36pi. [1] deducing it doesn't depend on R. [2] computing the R=3 case. Step [1] can be done logically or mathematically. Wordblind absolutely wins the prize for the slickest math.

Verifying Martini's result ... The order on the hill is always [D W J] The positions, and Jenkins' distance, at 7 critical points are as follows: 0. Start at the bottom [0 0 0 => 0] then ...

Plausible, reasonable and simple. Compelling, even. Oh wait. The problem states: "This bridge capacity is up to 80 kgs only, if you exceed this limit the bridge will fall". "up to 80 kg only" [an open interval] does not include 80. Therefore "a perfect 80kg" "exceed this limit," and so "the bridge will fall."

That part matters most.

Correct

If two speak the same language, have access to similar dictionaries and understand rules of grammar and syntax, then we say they can communicate in a literate manner. The meaning of words is a mixture of the "dictionary meaning" and the personal experience of the speaker and listener. The latter component of meaning leads at times to misunderstanding until the context is communicated and understood. Culture and environment can alter generic meanings of words. How can we prove a word has a meaning? One approach might be to agree on a common dictionary as the authority. A more pragmatic approach might be to say that if two people are comfortable with the notion that they agree when they discuss something, then the words they used to reach agreement themselves have an agreed meaning. Is this what you had in mind?

Yup. I didn't want to describe the "found" ones yet - it's fun to find them, as well.

Your solution avoids that situation. I think you solved a more difficult problem than was proposed. I opted for my approach because [1] It takes one fewer crossing [2] I believe the conditions provide for a momentary situation where a woman is without her husband but with another man, so long as she does not remain that way when the boat leaves them. Maybe the author can resolve whether my interpretation is permitted.

You can't do it taking one at a time, and taking them two at a time is trivial. So I'll answer the problem of how they do it themselves. [edit - one of my letters was wrong] xy cross, x returns --- .. .y .. || Ax B. Cz Ax cross, A returns -- .x .y .. || A. B. Cz Cz cross, C returns -- .x .y .z || A. B. C. AB cross, z returns - Ax By .. || .. .. Cz Cz cross ------------- Ax By Cz || .. .. ..

Please clarify - As you take them across the river, there can be no more than one of them with you in the boat?

You're exactly right, I got it backward. Thanks for clarifying.

Who are the three cousins in that case?

I think Melchang wants us to think inside of a logical box. That would lead to an answer like "Do you tell lies?" Melchang, does that cover it? Parable. A Freshman physics student prized his freedom of approach when he solved problems. So when his final exam asked him to find the height of a building using only a barometer, his answer was to throw the barometer from the roof of the building and time the sound of its hitting the pavement. From the speed of sound and the acceleration due to gravity [ignoring air resistance] he could compute the building's height. When the prof gave him a failing grade, he appealed. Upon being given a 2nd chance, he said he would tie the barometer to a long string and measure the period of the pendulum that made - again, from the building's roof - and compute the length of the pendulum. Anticipating another failing grade, he gave an additional answer: measuring the length of the barometer, the length of the shadow it cast on the ground and the length of the shadow cast on the ground by the building would give the answer using proportions. He got another failing grade. He finally appealed to the Dean of the school. In a meeting among the three principals, the Dean gave the student one more shot at giving the "right" answer: measuring the barometric pressure on the sidewalk and on the roof and converting the difference in pressure to inches of air. A really stupid way to use a barometer and one that would require unimaginable precision. Nevertheless that was the answer the prof wanted. The student would have none of it. He approached the Dean and gave his final solution: "I would take the barometer to the basement of the building and knock on the superintendent's door. When he answered, I would say to him: 'Sir, I have this beautiful barometer, which I will give to you if you will tell me the height of this building.'" The Dean gave the student an A.

I intuitively knew what the answer would be, but I have been struggling to come up with the mathematical proof of it. Looking at your math, though, I am puzzled. How do reason that the area of hole's cross-section (r*r) equals the area of the sphere's cross-section (R*R) less the the square of half the hole's length? R*R = r*r + L*L because you can draw a right triangle where R is the hypotenuse [Pythagorus]. Hint: slice the thing along the hole's axis running vertically. Go from the sphere's center horizontally to the surface of the hole - that's a distance r. Go straight up to the top of the hole - that's a distance L [at right angles to r] Go back to the center of the sphere - that's a distance R and is the hypotenuse. When I answered her question, I hadn't proved 36pi is true for all spheres. I just guessed. [it's really a quite surprising result!] But, for all she knew I did all this math in my head in 15 seconds! We were colleagues, and I would place her IQ somewhere north of 160 - high enough to think it could be done, and ... high enough that she usually left me in the dust in our work. She was impressed. It was an moment to savor, and I thought I'd share the story.

If you connect all pairs of 4 dots in a plane [geometrical, not air] you create 6 line segments. This puzzle asks how many ways 4 points can be arranged such that the lengths of the 6 connecting lines share no more than 2 values. The points must be distinct. None of the lengths can be zero. Example: the corners of a square. The 4 sides and the 2 diagonals share common lengths: a so-called 4-2 solution. A moment's reflection and some equilateral triangles reveal there are at least 2 other 4-2 solutions, a 5-1 solution and a 3-3 solution. The real stumper is to find one more 3-3 solution - two in all. You can describe these in words, or attach a graphic. [This question was posed to Jr-High school students in a national competition.]

Here's another variation. A man took his bride to the same hotel the next night. When he settled the bill in the morning the clerk said that will be $9 apiece. The man dutifully paid $81 and left, mumbling to himself. Why?

There's $25 in the till, $3 went back to the men and $2 went to the clerk. 25+3+2=30. Here's a related puzzle. A man took his bride to the same hotel the next night. When he settled the bill in the morning the clerk said that will be $9 apiece. The man dutifully paid $81 and left, mumbling to himself. Why?

Since "ought" is alt spelling of "aught" carrying the meaning of "naught" or stretching it a bit "zero" then the sequence is initial letters of zero [ought] one two three four five six seven, and proceeds with E N T E T ...

Ants don't need gravity, they stick. They took a circular [ok, triangular] path on the ground that slowly made its way up hill. At the top, they stayed in formation, pointed left, and said "he forgot the picnic!"

Writersblock, Still laughing ... I love it! My only reply is ... ok I have two replies ... [1] she never complained and [2] the [apparently lacking] size spec is an important part of the logical solution.

Doing the math and explaining that downward forces are needed to produce upward acceleration, etc. is absolutely correct; even tho the numbers change depending on how, and how high, he throws them. One might be tempted to suggest he could throw them very gently upward ... this approach fails, but more math is needed to prove it. A general argument states: If the total weight of the juggler and the watermelons is NOT being supported by the bridge [the only supporting structure present] then some or all of them will fall, and juggler will fail to get them across. Thus, the bridge DOES support the entire weight, and juggling is seen not to be a solution. ... unless ... the bridge is so short he can toss them into the air before getting on the bridge -- and catch them after he has finished crossing it.

edit: ha! now I've read Riddari's first spoiler. This post adds nothing to what he said. The way I finally understood the answer is this: That one of the two doors you get by switching has a goat is a red herring. you knew that already. Doesn't change your chances. Riddari's explanation #2 is compelling, also - it's the same situation, magnified, and makes intuition favor the correct answer. It took me an hour to believe the 1 of 3 case.