Math problem, last people I can think of

[dvd_master]

My psycho teacher wants us to write an ESSAY on this one problem, and it's just worth so much of our grade I have to make sure the answer is right.

Please? I know this isn't very on-topic, but I can't think of where else to ask and you can delete this thread as soon as it's done.


http://www.hamsterhouse.com/dvd_master/sa.jpg

Thanks!

[Doobie]

You don't know the Pythagorean theorem? You just find the hypotenuses of the four triangles that make up the shrouds.

What's with units in feet? I thought everyone in school was having the metric system shoved down their throats?

[Kedirekin]

You know, I started to right up the answer, but I decided that would be helping too much.

Here are some hints:
- right triangles
- pythagorean formula
- 12 and 3
- 9 minus 6 divided by two and 24

Wow! Even that feels like giving you the answer.

[dvd_master]

That's evil! I still don't get it. I only see 2 triangles, the ones at the top, or else I'd be able to get it.

As far as I know, I got the top triangle's needed length, but the bottom I'm getting messed up with.

A squared + B squared = C squared... I just don't see what goes into what for the top large part.

Just tell me if this is right:
Top part, 23.2 (rounded)
Bottom, 47.1 ? (also rounded)

[r6d2]

Whatever numbers you get, remember to add some wire for the knots.

[sysKin]

Originally posted by dvd_master
Just tell me if this is right:
Top part, 23.2 (rounded)Hm, no. 2*sqrt(12^2 + 1.5^2) =~ 24.18677
Bottom, 47.1 ? (also rounded)That is less than double 24ft alone.... if they were vertical it would be 48.000 right? Can't be less.

it's 2*sqrt(24^2 + 3^2) =~ 48.37355

[edit]Ok I just noticed Kedirekin's post and he's probably right. It's all about a question - is 3ft only one half (on the top) or two halves? 9ft is two halves at the bottom, that's obvius - but if only one on top, Kedirekin is right.

[edit2] Kedirekin lol, I had written "2.25" but then I realized that "9" for lower part might be confusing. - so I changed to 1.25^2 lol. This is exactly how I make xvid bugs, too.

[edit3]You are most likely right about upper half being 6 total. So, you're right.

Radek

[Kedirekin]

Sorry, that's not the answer I get.

Wish I could draw you a picture.

Draw a line from the bottom of the upper shroud straight down. Now how long is the little piece from that line straight across to the lower shroud? Hint: 9/2 - 3. Another hint: the line you just drew forms a new right triangle. Once you get that, you have the base and height of the lower right triangle and you can use the pythagorean formula to calculate the hypotenuse.

Here's the answers I get:
One upper shroud: ~12.369 ft (root of 153))
One lower shroud: ~24.047 ft (root of 578.25)

Total of all shrouds: 2 upper shrouds + 2 lower shrouds.


I had one too many zeros for the lower shroud. Sorry about that - always check your work.

[Kedirekin]

Oops! We're giving you conflicting answers.

Is the top bar one 3 foot section or two 3 foot sections? I thought it was two 3 foot sections.

@sysKin
One half of 3 is 1.5. I think you used 1.25.

[dvd_master]

It means the entire rod is 3 feet, not half. I'm searching through the equations and trying to see where I made my mistake.

You guys are helping a lot, thanks a billion!

[mpucoder]

If the upper rod is three feet total, then the drawing is not accurate, the shrouds would not bend over the upper rods (hint: 3:12 :: 9:36)
Are you supposed to write an essay, or solve the problem?

[bit-wise]

Having lived on a sailboat for a while, the drawing is accurate in the fact that shrouds "bend" over the stays. So, there are two stays in the drawing, (off each side of the mast), each 3 feet in length. (Instead of a 3 foot total length, each 1.5 feet in length).

dvd_master, be thankful you are not in advanced calculus, because the exact formula takes into account the number of strands in the wire, the total sail area, and some serious equations to factor in the strechability of the shroud and the amount the mast can bend (yes, it does bend quite a lot). On a 24 foot mast, you will get about 3/4 a foot of stretch in the wire when subjecting the sail to 2000 pounds of pressure (the equivalent of around 15 knot winds across the sail).

[r6d2]

Originally posted by bit-wise
the drawing is accurate in the fact that shrouds "bend" over the stays.Regarding accuracy, isn't the keel too short?

[bit-wise]

Heh. You would be surprised at some of the keel designs from the mid-seventies. The total keel area dictates a big portion of its effectiveness, so they made the keel very shallow, but had it run the length of the boat. The best way I can describe the performance of these boats is a bathtub with a sail. However, they didn't ground as much since they only took about 3 feet of water.

I was thinking more that the boat is a bit low in the water - guess there is a leak somewhere. Maybe that's the next problem, calculate the volume of water the boat has taken on based on how low it is sitting, factoring in the buoyancy (or lack thereof) of the interior items such as beds, cusions, galley, engine...

[r6d2]

Originally posted by bit-wise
guess there is a leak somewhere.Maybe the keel is not actually there and water is entering through the slot. :)

Well, this is quite off topic but isn't this whole thread too? ;)

A small divertimento:
A house has a row of square tiles in the garden put one after the other in this fashion: <> (diagonally connected). The tiles go from the house door and exactly to the border of the garden.

The lady of the house got tired of looking always at the same arrangement so she asks the gardener to put them side by side, with no space between them, and she strongly remarks she expects to have it filled using only whole tiles.

The gardener begins his work, but when he is just on the middle of the path, he notices that each time he moves a tile he has to go further for the next one. In a moment of inspiration under the burning sun, he also concludes that there will be not enough tiles to finish the path.

He goes to the lady and tells her. She asks:

-How many more tiles do you need to finish the job?So, how many more?

[mpucoder]

about 40% more. more exactly square root of 2 times as many.

[r6d2]

Close, but no cigar... :cool:

[avih]

Originally posted by r6d2
Whatever numbers you get, remember to add some wire for the knots. LOL

[Kedirekin]

You mean the tiles are layed out roughly like this (white is tile - sorry, I can't make them look square):

********************
**** ******** ****
*** ****** ***
** **** **
* ** *
** **** **
*** ****** ***
**** ******** ****
********************
**** ******** ****
*** ****** ***
** **** **
* ** *
** **** **
*** ****** ***
**** ******** ****
********************



Based on surface area alone, the path either needs to be half as wide or will require twice as many tiles.

Is that correct? Do I win a prize? ;)

[r6d2]

Originally posted by Kedirekin
You mean the tiles are layed out roughly like this (white is tile - sorry, I can't make them look square)Nope, it is only one row of tiles. You drew two. I'll post the right answer tonight if nobody gets it before then.

The prize is the position of moderator for the upcoming Math Forum. :)

[Soulhunter]

50% maybe... Damn, I hate math... :devil:

Bye

[bit-wise]

Hmmm, I would have thought .414 more tiles would be needed. I just tried it with 3 scraps of square paper 1" by 1", and I needed exactly 1.2 pieces of paper more to make the path if they are to be laid side by side.

I guess my answer is incorrect because that .2 does not satisfy the fact that she "she strongly remarks she expects to have it filled using only whole tiles".

Am I allowed to move the house?

[Kedirekin]

It requires 1.4142135623730950488016887242097... more tiles, and the path is 1/1.4142135623730950488016887242097... as wide as before.

However, in a mathematical sense, it is not possible to fill the length of the path exactly using only whole tiles. Root 2 is an irrational number. No matter how many tiles are in the original path, there is no ratio of tiles that will exactly fill it using the new layout.

[r6d2]

@Kedirekin, :goodpost:

You got it right, except that you need only sqrt(2) - 1 more tiles, just like bit-wise posted, i.e. 41.4...% more than what you have.

But the catch it precisely that there is no number of tiles that can satisfy the restriction. You are hereby named as the Official Mod of the Math Forum (coming soon to a Forum near you).

@bit-wise, :goodpost: too. However, even though you posted first, you failed to arrive to the correct conclusion. You also need to have your scissors replaced, 1.2 is only 40%. :)

Anyway, you were close to another solution (albeit impossible in real world). There was no restriction on moving the house or the street, so that would be a good theoretical solution to the problem.

But there is another practical issue to this: they usually don't sell tiles by the unit, so the lady would probably have to round up at the store. :)

Also, if you use an infinite number of tiles, and move the house to infinity, you could speculate that's a solution too.

[r6d2]

The previous one was easy, right? Let’s see this one.
Suppose the Earth is a perfect sphere with a flat surface, and you tie a rope around its longest path (let's say the Equator), so it is perfectly tense and has no slack. Now release the knot, loose the rope and enlarge it by adding 1 meter. Tie again the knot at the new position.

Then, spread the extra meter evenly along the globe by retightening the rope simultaneously from all 360 degrees.

(Use Lanczos resizer to keep the most quality, and also to not being accused of being off-topic.) :)

Now you stand by the rope at any point in the globe.

The question is: Can your hand pass under the rope, without pulling it?Think about it. The extra meter slack is no longer there at your feet. It is evenly spread around the globe.

[bit-wise]

Actually, rack the 1.2 up to pure laziness in the fact that I didn't feel like typing lots 'o decimal places...

Now this sphere thing. Question. The rope, in its first position, (before it is resized) - is it at the largest circumference of the sphere, or does it not matter? Because if you start the rope horizontally at the equator and move the entire rope north it no longer touches the globe at any point, your hand slides right under. But I don't think that is in the spirit of the question.

[r6d2]

Originally posted by bit-wise
The rope, in its first position, (before it is resized) - is it at the largest circumference of the sphere?You are right. That's the spirit of the question. Both before and after resize the rope is around the Equator. (I edited it already).

[Kedirekin]

Is this a pure math problem?

The additional meter of rope will create an infinitesimal increase in the radius of the (perfect?) circle formed by the rope. Assuming the rope isn't simply allowed to fall to the ground after retightening, there will be an infinitesimal gap between the surface of the earth and the rope.

With a sphere the size of the earth and a hand of typical thickness, one additional meter wouldn't create a gap nearly large enough for your hand to pass under the rope, assuming we're not allowed to pass our hand within the confines of the earth's surface (no digging or tunneling), and that the definition of 'under' is between the rope and the earth's surface (not simple south of the rope).

I have no idea if I've covered all the bases. Don't keep us in suspense too long.

[r6d2]

Originally posted by Kedirekin
Is this a pure math problem?Yes, it is. There is no reading between lines or infinity stuff.

Your reasoning is correct. It comes from your intuition. You did not do the actual math, but described what you think would happen.

Well, you should do the math. ;)
Don't keep us in suspense too long.I could answer the question with a simple yes or no, but let's be patient. Perhaps we can get another mod for the Math Forum showing up...

[mpucoder]

The relationship between circumference and radius is linear, circumference = 2*pi*r. Regardless of the circumference of the Earth, the radius will increase by 1/(2*pi), about 6". So yes, your hand will slide underneath easily.

[r6d2]

We have a winner! :goodpost: This time you made it, @mpucoder, and quite precisely. The result presented above contradicts our intuition. Let’s try to prove it.

Let R be the Earth radius and L the original length of the rope. Thus we have that,
(1) L = 2*Pi*R
Now we increase the length of the rope by 1 meter, getting L’. Our intuition tells us that the radius of the second circle will be somewhat bigger, right? Let R’ be the new radius. Thus we have that,
(2) L’ = L + 1m
and also,
(3) L’ = 2*Pi*R’
Let h be the radius increase. Thus we have that,
(4) R’ = R + h
Replacing (2) and (4) in (3) yields,
(5) L + 1m = 2*Pi*(R + h)
Now we replace (1) into (5) which yields,
(6) 2*Pi*R + 1m = 2*Pi*(R + h)
Distributing the right side,
(7) 2*Pi*R + 1m = 2*Pi*R + 2*Pi*h
Subtracting the term 2*Pi*R from both sides yields,
(8) 1m = 2*Pi*h
Then we get that h, the increase in radius is,
1m
(9) h = ---- ~= 16cm ~= 6”
2*Pi
As you can see, that infinitesimal increase is high enough not only a hand could pass underneath, but also a whole rabbit, a cat, even a small dog.

The most surprising part of this result IMHO, is that it does not depend on the radius of the sphere in question. You could do the same at home with a tennis ball and the result would be exactly the same. If you don’t believe it, you are welcome to try.

Now we have two mods for the Math Forum. :D

[Kedirekin]

I just did the math and it makes me question my sanity.

Why is intuition so far off here? I can see the numbers and I still don't believe it.

This is truly a wierd feeling. It doesn't matter if I wrap the rope around a bowling ball or around the circumference of the galaxy - add one meter of rope and the radius increases by the same amount.

GAH! Where's the number for Bellevue?

[jel]

GAH! Where's the number for Bellevue?
in case of further maths questions i've taken the precaution of putting the number on 'speed dial'...

[r6d2]

Originally posted by Kedirekin
Why is intuition so far off here? I can see the numbers and I still don't believe it.Ah... The Lord works in mysterious ways... :)

Intuition is an important part of our survival skills, and in spite of that it plays those tricks on us quite often. Want another ride? :)
Suppose you have a huge, and I mean huge sheet of paper, 0.1mm of thickness (10 sheets add up to 1mm).

Also suppose you can fold it in two without any problem, regardless of the number of times you do it (in practice, there is a limit of about 6 to 8 times you can fold a sheet by its half every time).

Then begin to fold it, repeatedly, in two by its half (remember the sheet is huge). With each turn you double the thickness, so the folded sheet begins to grow.

How many times do you have to fold the sheet to get to the Moon?
This time say a number, from your guts, then do the math.

[Kedirekin]

Gut, 25 times.

Now I'll do the math.

[KpeX]

I googled quickly to find the distance from the earth to the moon is 384,400 km.

d=384,400 km =384,400 * 10 ^3 m = 384,400 * 10^3 * 10^3 mm=384,400,000,000 mm


thickness =.1 mm (per sheet)*2^n (thickness doubles every fold, n=number of folds)

384,400,000,000=.1 * 2^n

n = log(384,400,000,000/.1)/log(2)=41.8, so 42 folds would be required.

My intuition didn't really have an answer, but it probably would have been more than 42.

[r6d2]

Originally posted by KpeX
My intuition didn't really have an answerFirst you cheat, then you spoil the problem. :)

Well, I wonder why NASA spends billions of dollars in R&D instead of building a sheet of paper big enough.
Derived question: How big does the sheet need to be to hold a person standing on top?Again guts, then math.

BTW, @KpeX, since you are an audio mod (and also to keep you busy so you don't spoil this one), ;) would you pay a visit to this (http://forum.doom9.org/showthread.php?s=&postid=441940#post441940) thread and clear up if sound can travel in the vaccum through a space-time shockwave?

[KpeX]

@r6d2

About the audio thing, I'm pretty sure you're correct and audio will need a medium to transmit. You can't vibrate vacuum ;)

Regarding the math - Gut instinct, assuming that we need, say, 1 square foot to stand on, i'd guess maybe a few thousand square miles. I'll do some math now.

I doubt my guess is very close however. That's why I like math, cuz my intuition isn't worth much :D


Edit:

area to stand=1 square foot

# of folds=42

area = (original area) * (1/2)^(number of folds) (i.e., each time you fold the paper the area is halved, since one side is half as long)

Original area= 1/( (1/2) ^ 42 )=2^42=4398046511104 square feet

Length of one side= 4398046511104^.5 = 2097152 feet = 397.2 miles

So using a square piece of paper one side would have to be about 400 miles long.

Fun stuff guys :D

[BoNz1]

You guys are absolutely correct in order to hear sound you need medium for it to travel in. Sound is simply the compression and rarefraction of molecules in the air. So all those big explosions that you see in movies, ie Armageddon would never happen. BTW, a couple of months ago I had a math question of the week in my sig usually they were quite hard. Possibly I should resurrect this, no? :)
EDIT: There we go, done enjoy 8D

[r6d2]

Originally posted by BoNz1
You guys are absolutely correct in order to hear sound you need medium for it to travel in.Please, BoNz1, let's keep on topic, will you? :D:D:D

Interesting question that of yours, but did they really ask you to use differential calculus to solve it? My bet is simple geometry is quite enough.

[bit-wise]

There was a recent publication that sound, does, and is, traveling through space. It appears that the Chandra X-Ray laboratory detected sound from a distant black hole as a nice B-flat 57 octaves below middle C.

From my skimming the article, it appears that bubbles in the escaping plasma jets at the edge of the black hole produce enough force to create this sound. It requires something with as much energy as a black hole to produce enough energy for the sound waves to travel to Earth. Since space is only 99.999(to some decimal place) a pure vacuum, there is enough force to produce a sound wave that can travel 250 million light years back to Earth. I leave the mathematical proof as an exercise for the reader. But no, Hollywood explosions would not be heard.

But your reasoning is correct, sound can not transfer through a pure vacuum.

Oh, on another note. r6d2: On behalf of my client, they all would like to express thanks in causing them to collectively scratch their heads and ponder the rope question. The answer caused quite a stir.

So, did we ever answer the sailboat question?

[sysKin]

Originally posted by r6d2
(...)if sound can travel in the vaccum through a space-time shockwave? Now *this* got me interested (but I haven't seen the thread yet). A sound is a wave. A wave means that the particlaes (of the medium) are accelerating. An accelerating particle (with a mass) creates gravitational wave that is 90 degrees phase shifted compared to the sound wave, and has the same direction and frequency and everything.

Such wave would have to be emmited by something thin, much thinner than the (sound) wavelength - otherwise different particles, which have different phase of the same wave, would emmit together and interfere one with another. This is because of differet speeds of sound and light.

If such wave would hit something thin again, it would make it vibrate and in the end, you would recieve the sound again (90 degrees phase shifted again, doesn't matter).

:D
Radek

[sh0dan]

You should change your description to:

"sysKin: inventing new XviD bugs and solving deep space communication problems!"

:D

[SeeMoreDigital]

Is the answer to the question 42?

In HitchHiker's Guide to the Galaxy... that's the answer for everything!

[r6d2]

Originally posted by SeeMoreDigital
Is the answer to the question 42?Yes, it is.
Originally posted by bit-wise
So, did we ever answer the sailboat question? I don’t know. I think it was left as an exercise to the reader. :)

About the spa question, the absolute error in volume is,
Pi*(r^2 + 2*r*h)*eWhere r is the radius, h the height and e the error of the instrument. In this case, the error in volume amounts to about 5 cubic feet, which is about 5% of the volume of the spa. As with any error, it is a +/- value.

You see intuition again is misleading. It’s a huge error you get, eh? Considering the instrument error is only about 1.7% of the magnitudes in question.

[r6d2]

Let’s have another stroll at a intuition divertimento:
Suppose you are among a group of people attending a class. There is a final exam approaching and there’s a motion from the students to postpone the test for another week.

(Nobody has studied enough because you all have been encoding too many movies using the university computers and the last version of AutoGK. OK, the reason is actually irrelevant, it’s just to be on topic.) :)

The teacher says,

-OK, I’m willing to postpone the test, but that makes me uncomfortable and would not look good if I just decide it on my own based on your reasons. Let’s leave it to chance. If any two of you have a birthday on the same date, I’ll do the test on schedule. If none of you has the same birthday as any other member of the class, I’ll postpone it. So take a sheet of paper and write down your birthday. I’ll collect them in one minute.

The students get very happy. Having the year as much a 365 days, and being a group of only 50 people, it is very unlikely two of them were born on the same day.

The question: What is the probability of the exam to be postponed?Now this one is hard. As usual, use your guts, then do the math.

[duartix]

Ok, 3 guys (A,B & C) check into a hotel. The owner says the room is 30$. They gather 10$ each and give them to guy A.
A goes down the stairs to pay the hotel owner.
Hotel owner tells A he got it wrong because the room is 25$.
Hotel owner gives back 5$ to A.
A thinks:
- "How am I going to distribute 5$ between 3 people?"
- "I know, I'll tell them it's 27 (9$ each), split 3$ between the 3 of us and I'll keep the other 2$."
So he does.

Now you people tell me:
They payed 27$ (9 each) right?
27$ + 2$ (kept by A) makes 29$. The room was 30$.
There is 1$ missing. Where is it?

[downimp]

[QUOTE]Originally posted by duartix
"split 3$ between the 3 of us and I'll keep the other 2$."

That means the A guy has 3$ (3/3 +2).
Then 9*3 + 3 = 30

Corect?

[duartix]

Yes, A has kept 2$ for himself + 1 (from the split).
Your alternative explanation is correct, but what I want is someone to explain why is 1$ missing from my reasoning. That is the fun part. :D

Oh, by the way, downimp, welcome to the forum !!! :D :D :D

[Wilbert]

It seems that many of you are bored :) Well, here is another nice puzzle:


Consider n number of poeple standing in a circle. All of them are numbered from 1 to n. Now, Man no. 1 is given a sword. He kills the person next to him (i.e Man No. 2) and passes the sword on to Man no.3. Man 3 in turn kills the man next to him (No. 4) and passes the sword on to the next man (i.e Man No. 5). This sequence continues until only one man is alive. Find which number is it.

For eg, take n=5

then:
|1| 2 3 4 5 (the number in | | has the sword)
1 |3| 4 5
1 3 |5|
|3| 5
|3|

Thus in the end man no. 3 remains.

Suppose we have 3141 man, which person stays alive after doing this?

[duartix]

which person stays alive after doing this? This is definitely the kind of problem where it's easier to write a computer program than to think about it. ;)