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Posts posted by bonanova


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  .. .. ..

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?

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.

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?

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.

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 = RL, 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 crosssection (r*r) equals the area of the sphere's crosssection (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 socalled 42 solution.
A moment's reflection and some equilateral triangles reveal there are
at least 2 other 42 solutions, a 51 solution and a 33 solution.
The real stumper is to find one more 33 solution  two in all.
You can describe these in words, or attach a graphic.
[This question was posed to JrHigh 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?
you shouldn't have to peek for this answer ...

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,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.

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 = RL, 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 hohum, 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. [hinthint 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.

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:
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.

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.

To clarify, I believe bonanova is speaking of a hole 6 inches long of an unspecified diameter. Am I correct bonanova?
Yes.

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 6inch 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 6inch 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.

He says, "I am lying." Which doesn't have a truth value.

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 6inch [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.

Riddari,
You're correct on both comments [physical attributes and spoiler for mistake.]
Thanks ... !

within the red circle I found no possible ways to get this answer following the riddles rules.
I'm missing something. What rule is being broken?

A variation on the Lone Ranger theme:
A man left home one day and began to run. After he'd gone a ways, he turned left and kept running. A bit later he turned left again and ran a little faster. Later on, he turned left a final time and ran even faster than before.
In the distance he saw two masked men.
Who were they?
One of them was not the Lone Ranger.

I first heard it posed as ... if a hen and a half can lay an egg and a half in a day and a half, how many hens would it take to lay 18 eggs in 3 days?
It's tempting to figure a hen can lay an egg a day so it would take 6 hens. But it takes a hen and a half to lay an egg a day, so it takes 9 .
Cross the 3 Couples
in New Logic/Math Puzzles
Posted
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.