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Posts posted by bonanova


watery eyes?

Right ... the pentagon is the stumper. The clue, What if there were five points? makes it come to mind, cuz the points of a pentagon create sides and diagonals of only one type, just like as square does. Then you just remove a point to get a 4dot solution. It's seldom you get one solution to be fundamentally different from all the others.

cpotting, amazing story. You might find this amusing: A mathematician, an engineer and an accountant were asked What is 2+2?never underestimate the ability of accountants to ignore facts and plod on anyway ("Numbers be d***ned  the amounts don't matter, just as long as they balance!").Mathematician replied 4
Engineer replied 4.000 plus or minus 0.001
Accountant replied  What to you want it to be?

Can you share your solution?

My answer is correct, because
I was lying.

How about...
"Do you refuse to answer this question"?
If he answers "Yes," he lies.
If he refuses to answer [attempting an affirmative response] he hasn't answered "Yes".
Melchang, are you saying this question can, at some point be answered yes, according to situations and environments?

Answer for a jet:
Yes.
Answer for a propeller plane:
Yes.
Reasoning:
An airplane takes off when its wings generate a lift force greater than its weight. Lift force is determined by air speed. A backwardrunning treadmill will not impede the plane's forward motion, nor limit its air speed, only make its wheels spin twice as fast!
But what if an airplane's engine drove its wheels? Then the answer would be No. It could not achieve forward motion nor generate lift.
That would be an interesting kind of airplane  after it took off it would become a glider!


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 ...
2. Jenkins meets Wife [9.6 14.4 14.4 => 17.6]
3. Jenkins reaches top [10.4 13.6 16 => 19.2]
4. Wife meets the Dog [12 12 14.4 => 20.8]
5. Jenkins meets Wife [11.04 13.44 13.44 => 21.76]
6. Jenkins reaches top [9.75 12.16 16 => 24.32]
7. Jenkins reaches bottom [ 0 0 0 => 40.32]

... the man carry one water melon with him, weighing a perfect 80kg, put down the water melon on the other side, than walk back to the other watermelond, weighing 79kg, and then carry that watermelon back to the other side, weighing a perfect 80kg. This answer is plausible, reasonable, and simple, is it not?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.interprets the word into a concept. 
CorrectAlso, by defining a plane, I assume we are limited to 2 dimensions? 
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 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.

p = piD = 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 rdependent terms add up to zero,
and the derivative is zero.
OK?

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.

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.]
One Girl  One Boy
in New Logic/Math Puzzles
Posted · Report reply
A friend insisted the answer "had to be" 50%. So I asked him this:
Of all the couples in the world with 2 children, statistically speaking,
what fraction have a 1 boy and 1 girl?
He answered: one half.
and what fraction have 2 boys and 0 girls?
He answered: one fourth.
So, just taking these couples,
what fraction have 2 boys and 0 girls?
He answered: One third; and then: Oh.... OK I get it now.