bonanova
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Posts posted by bonanova
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Yup.I assume that is the easy one, and you want us to find another.I didn't want to describe the "found" ones yet - it's fun to find them, as well.
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p = pi
D = the diameter of the sphere
d = the diameter of the hole
R = the radius of the sphere (.5D)
r = the radius of the hole (.5d)
H = the height of the hole (end to end)
h = the height of the hole (centre to end = .5H)
After drilling, the remaining ring shape is equivalent to a barrel of unknown width (D), 6" high (H) and with the central cylinder portion removed.
The formula for a barrel with sides bent to the arc of a circle
= pH(2DD + dd) / 12
= (1/12)pH(8RR + 4rr)
cpotting,
kudos for finding all these neat [cap and barrel] formulas.
That's does all of the calculus work.
I found it useful to put everything in terms of R and h.
You can do this by noting that RR = rr + hh [Pythagorus]
V[barrel] = ph [8RR + 4rr] /12 = ph [8RR + 4RR - 4hh]/6 = 2ph [RR - hh/3]
V[cylinder] = height x area = Hprr = 2ph [RR - hh]
V[barrel] - V[cylinder] = 2ph [RR - hh/3 - RR + hh] = 2ph [2hh/3] = 4phhh/3.
Recalling that h=3,
V[barrel] - V[cylinder] = 36p [a constant].
Looking at your derivation, everything is correct.
If you add like terms in your expression for V,
you'll see that all the rr terms add up to zero.
Your derivative expression is correct, also,
except that you should include the constants (1/12)pH8, etc...
[if y(x) = Cxx, then dy/dx = 2Cx, not just 2x]
You'll see again that the r-dependent terms add up to zero,
and the derivative is zero.
OK?
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Your solution avoids that situation.That will work bonanova, except that at some points a wife is in presence of a man that isn't her husband.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.
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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 || .. .. ..
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1. You need to take all 6 to the other side ...
Please clarify - As you take them across the river, there can be no more than one of them with you in the boat?
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I think bonanova was close, but he made one mistake; "A if only B" should be A->B, not B->A.
You're exactly right, I got it backward. Thanks for clarifying.
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Note: It is also possible ...
... to have D be the daughter of A, if E becomes the sister of C.
Who are the three cousins in that case?
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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.
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Here's the mathematical solution:
Radius of sphere = R, radius of hole = r, length of hole = 2L [so L=3], height of cap = h.
Since r*r = R*R - L*L and h = R-L, we can eliminate r and h and do everything in terms of R and L...
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.
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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.]
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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?
you shouldn't have to peek for this answer ...
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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?
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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 ...
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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!"
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Writersblock,The moral of the story is: your girlfriend lied. Size DOES matter.
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.
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The volume of the spherical caps is given by:
where
[*] h = the height of the cap (difference between r and the distance from the centre of the[/*:m:1cc31]
sphere to the centre of the circular end of the hole)
Kudos to cpotting for the cap formula.
Radius of sphere = R, radius of hole = r, length of hole = 2L [so L=3], height of cap = h.
Since r*r = R*R - L*L and h = R-L, we can eliminate r and h and do everything in terms of R and L.
Swizzling cpotting's cap formula, V[cap] = pi/3 (2R*R*R - 3R*R*L + L*L*L)
Cylinders are ho-hum, V[cyl] = 2pi*L*r*r = (2pi/3) (3R*R*L - 3L*L*L)
V[removed by drilling] = V[cylinder] + 2V[cap]
doing the math,
V[removed] = (4pi/3)R*R*R - (4pi/3)L*L*L
Pretty amazing: the volume removed by a hole of length 2L is the difference of the volumes of two spheres: one of radius R, the other of radius L.
So the remaining volume is simply the volume of a shpere with radius L. [hint-hint at the logical solution]
V[remainder] = (4pi/3)L*L*L = 36pi.
My friend wouldn't have posed a math problem [boring] and she wouldn't have left out critical information. Therefore the answer couldn't depend on the radius of the sphere. I chose a sphere size [radius=3] that would make 0 volume removed [a hole of length 6 and diameter 0]. With nothing removed, the remaining volume is the original volume: (4pi/3) 3*3*3.
It took me about 15 seconds to say 36pi, and the moment [if not the girl!] was mine.
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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.
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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:
if you stick with your first selection your chances of winning are 1 out of 3. If you switch, you cover the other two possibilities and your chances have to be 2 out of 3.
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.
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Well,I must not be a genius because the easy way is not jumping in my head. The difficult way would be to determine the volume of the cut out section and deduct it from the volume of the sphere. I am too lazy to do all the work actually involved with doing that though.
I did the calculus afterward, and however you do the calculation, it's difficult. The remaining volume is a volume of revolution with requires finding the cross sectional area and spinning it thru 360 degrees. It's no easier to compute the cylindrical volume removed, cuz there are spherical "caps" on the cylinder which don't have formulas that I could find.
The logical way is easier, and I'll post it in a day or so if you don't get it... have fun.
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To clarify, I believe bonanova is speaking of a hole 6 inches long of an unspecified diameter. Am I correct bonanova?
Yes.
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Clarifications:
[1] the hole is a circular cylinder of empty space whose axis passes through the center of the sphere - just as a drill would make if you aimed the center of the drill at the center of the sphere and made sure you drilled all the way through.
[2] the length of the hole [6 inches] is the height of the cylinder that forms the inside surface once the hole is drilled. picture the inside surface as viewed from inside the hole and measure the length of that surface in the direction of the axis of the drill.
in this sense, you could for example drill a 6-inch hole through the earth. the diameter of the hole would be huge, and you'd just have a tiny remnant of the earth left. but if you could set it on a table [a big table] it would be 6 inches high.
you of course could not drill a 6-inch hole through a sphere whose diameter was less than 6 inches. it was actually this fact that led me to the logical answer and made me a genius for a couple of minutes.
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He says, "I am lying." Which doesn't have a truth value.
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Maybe this has already been posted. A friend asked me this a while back, and I answered her in less than a minute.
She said I was a genius. But I said there were two ways to arrive at the answer, and I simply chose the easier way.
A 6-inch [long] hole is drilled through [the center of] a sphere.
What is the volume of the remaining portion of the sphere?
The hard way involves calculus. The easy way uses logic.
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Riddari,
You're correct on both comments [physical attributes and spoiler for mistake.]
Thanks ... !
Literarily thinking
in New Logic/Math Puzzles
Posted
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?