A new puzzle: Multiplying bread and fish

They cant have, they are 'leaving the cafe with one pound', not 66 pence. they entered with 10 and leave with one. 3x9=27 and the waiter took 2 pound for a tip still the pound is missing.
 
I think its quiet relevant. Its a puzzle to find how they have spent 9 pound each and the waiter receiving two pounds and paying 25 pound for dinner out of 30 they originally had, the sums don't add up (3x10=30) so far so good. pays for dinner 25 pounds (5pound change in one's) each diner take a pound (3), and the remaining 2 pound goes to the waiter. allgone or is it each has payed (10-1=9) pounds (3x9=27) they gave the waiter 2 pounds (27 +2=29). still doesnt account for the missing pound, no change was given in decimals. each recives a full number.
 
rolyateel said:
I think its quiet relevant. Its a puzzle to find how they have spent 9 pound each and the waiter receiving two pounds and paying 25 pound for dinner out of 30 they originally had, the sums don't add up (3x10=30) so far so good. pays for dinner 25 pounds (5pound change in one's) each diner take a pound (3), and the remaining 2 pound goes to the waiter. allgone or is it each has payed (10-1=9) pounds (3x9=27) they gave the waiter 2 pounds (27 +2=29). still doesnt account for the missing pound, no change was given in decimals. each recives a full number.
Click to expand...
1) 10 in, get 1 back: 3 * 10 = 3 * 9 + 3 * 1 = 30 (27+3=30)
2) change: 25 + 5 = 30 with 5 = 2 + 3 * 1

So you take the 27 from one equation and the 2 pound tip from another. Very curious.
 
Uh....

Has anyone taken a single piece of paper in a LARGE format, cut precisely a right
triangle as shown in the first drawing ONLY, cut the sub-blocks within this rt-triangle,
as shown in the first drawing taking into account that the cut itself is not "taken away"
from the cut-perimeters so that once assembled, it matches EXACTLY in a rt-triangle
to ensure it's proper alignment, and then after precise verification, re-organize the blocks
shown in the 2nd rt-triangle and check the results?

What is the outcome then?

I still find the 'hole' as shown in the 2nd diagram.
 
dant said:
I still find the 'hole' as shown in the 2nd diagram.
Click to expand...
Yeah, me too. I just did it with paper.

Ark is probably having a good time with all our guesses :D
 
Of course! We ALL add to his humor,I think. I am seeing this
picture in my mind of Ark really having a good time! Mirth! :)

I am waiting in suspense when he comes back ready to give
us all a lesson in humility? ROTFL!

The suspense is killing me softly! :D
 
I know the answer! The missing square went underground or to some alternate universe...

Let's add some more entertainment for Ark ;)
I have a feeling that the key must be in triangles. They are the "troublemakers", especially the small one.
Don't have geometry skills to have something more then a feeling :D
 
Here are some scribbles to ponder at and then I will begin to figure this out.

triangles1.jpg


In order to solve the puzzle and feed the fishes, we need to make those little ¼ inch increments into parts, as ¼ X ¼ X ¼ X ¼ would be one part, and this is the part we are looking for.


In Pythagorean Theorem speak, we have area B and area G which are the two different sized triangles in the puzzle. We can figure the area of each to help solve it.

B) 5 + 2 = H

H = 5.3851648 [in parts speak]

G) 8 + 3 = H

H = 8.5440037

Not sure if we will need those figures, saved for now. H is the length of the long side hypotenuses.


In figuring the area, we know, ½ times the base times the [height] equals the area.

B) ½(5)(2)= 5
G) ½(8)(3)= 12
TOTAL= 17 parts


The extra parts included with the first large triangle equals 15 parts. The extra parts included with the second large triangle equal 16 parts. In order to get the extra part, an illusion has occurred. There is not an extra part. It just looks like there is. Maybe that is what Jesus actually did, was to make them think they were eating extra fish.

Both triangles have a total of 17 plus 15 parts, or 32 total parts, but one looks like it has 33 parts.

To ask a question, just because the angle is different does not explain why it looks like a whole part has been added. In reference to the small shaded triangle in my drawing, is not that the difference between both area A’s and equaling one part? Let’s see?

½(2)(1)= 1 part

Yup!

You can see the areas of A look different sizes when one is turned on its side. Well, didn’t need Pythagorean after all.
 
It isn't a straight line , the two triangles do not have the same proportions. So the first image may appear as a triangle, but is in fact a quadrilateral.
 
That's slightly wrong [what I said]. The last part is divided among 4 parts, each part taking a little out, as in .25 parts in difference between one triangle and the other in each 4 sections.

Is that correct?

[added]

Yes, they are two 4 sided figures, quads.
 
The angles on the red triangle are not the same as the angles on the green triangle (except for the rightangles). I seem to remember that you can get the angles using sine/cosine tables with regards to the lengths of the sides, but it's that long since I did it at school I've forget. So the hypotenuse of the whole thing isn't straight.
 
dant said:
What is the outcome then?

I still find the 'hole' as shown in the 2nd diagram.
Click to expand...
I tried it and it appears to be just a trick (and if you look close you do see the hypotnuse bent):



I found a grid background on google images and used Windows Paint to count the boxes the same as in the originals and draw the lines. This is of course a rather crude tool to use, but it's good enough to show what's wrong. See how in the bottom triangle the horizontal green line in the middle of the triangle never reaches the grid line but is slightly below it? That's where the space went, and that's the illusion created in the original by bending the hypotnuse to make it look like the line reaches the grid line perfectly, making the area of the triangle artificially bigger, allowing that little square to exist. It really is impossible to arrange the figures as they are arranged in the bottom triangle without either making the triangle bigger or the figures smaller. The 2 smaller triangles fit, but the other 2 shapes do not, if the triangle is the same size as above. In fact, it's possible that those figures also do not fit in the triangle above as well, the triangle is probably too big, though Paint is just too crude to really tell.

Speaking of Paint though, a good analogy here is taking a bucket of paint and spreading it over a large surface. If the surface is large enough, you won't even see that it has a large volume of paint added to it. But if you take all that paint and put it back into a bucket, kinda like vacuuming a floor that looks clean and then seeing a huge amount of dust in the vacuum cleaner that you never saw on the floor, you realise how much there really is. But what they did was similar, they started out with a clean floor, then they added a very very thin layer of volume over the entire surface (bending the hypotnuse outwards a little in the bottom triangle, and inwards in the top), but you can't really tell. But combined all of that added volume was equal to one small square box, like one bucket of paint, and they could therefore make a hole in the floor but the volume of the floor stays the same cuz they made up for it with a very thin coat of paint.

Also, I'm not sure if you can mathematically solve this because that would require to have correct data to make your mathematical/geometric calculation with - data that we're not given, and measuring data with your eye is always tricky especially if the shapes are designed deceptively from the start. Any measurements will be crude, so you'd need software and to replicate this issue in software that can do this in high resolution and make measurements of higher precision than the eye, automatically measuring the area of these shapes etc in high precision. Otherwise you'd be missing critical data that measurements of angles might not tell you, like if the line is not straight or if the triangle is slightly "fudged" and is not aligned to the grid. The grid is fat and blurry, and so are the lines of the triangle, and I think the problem is not math but the fudging/disguising of data, counting on the assumptions of the human eye.

But how does 4th density make UFO's small on the outside but huge on the inside, that one is a bit more tricky! Here's part of that fascinating session:

May 27, 1995 Frank, Laura, JR and TR, Carla and Roger Santilli
A: Rouswo.
Q: (L) I guess that is the name. Hello, how are you this evening? (T) Who do we have with us?
A: Name given.
Q: (T) Where are you from?
A: Cassiopaea.
Q: (T) How are you tonight?
A: Okay.
Q: (T) We are doing okay too. We have some company this evening.
A: Good.
Q: (T) Roger and Carla Santilli are with us. Roger has some questions to ask later.
A: Hello, Roger.
Q: (RS) Hello. (T) Do you have anything you want to talk about to start off with?
A: Channel open.
Q: (L) Alright, let's start off with the question that S*** and I have wondered about: Is it alright if I give the Reiki
initiation during the Mercury retrograde?
A: Of course!
Q: (L) That's what I thought. Go ahead, Roger, ask what ever you like. (RS) I want to know whether we can have any clues on the propulsion systems of UFOs?
A: Sure!
Q: (RS) Is the gravity experienced by an anti-particle in the field of matter attractive or repulsive?
A: Repulsive when thought of in the way that is parallel to your studies, but, as we alluded to in the previous answer, there are more realms involved besides the one with which you are most familiar.
Q: (RS) The next question is: particles move, matter moves, in our direction of time, do anti-particles, anti-matter, flow backward in time?
A: Think of it as merely one seventh of the equation, Roger!
Q: (L) Can we get an answer on whether this is the case strictly on the third level of density?
A: Backward.
Q: (RS) Yes! I am interested in the propulsion systems of UFOs, the only way that I can perceive travelling the long distances involved in interstellar space is to have what is called a "space/time" machine. We cannot move the enormous distances unless you can fold, somehow, time and space. You cannot fold space unless you join it and fold time. You cannot have interstellar travel unless you have a space/time machine. But, a space/time machine means to also have the ability to move forward and backward in time, to manipulate time. (L) Yes, you would have to cross distances and simultaneously move backward through time so that you would end up arriving wherever you are going essentially at the same moment that you left. (RS) That is why I asked whether we can use anti-matter as a propulsion, because it would be repulsive in the right direction. The second question whether, when we use anti- matter, we would move backward in time. Because, some of those objects, you see them moving, and they can be moving in space but not in time, or they could be moving in time but not in space. If you see a UFO, it does not mean that it is in our time. It could be in a completely different time. (L) And, they disappear sometimes right before the eyes of the observer, and the question is: where do they go? (RS) They could be standing still in space, but moving time. Or moving backwards. (L) And, there are a lot of abductions reported where there is seemingly no time lost at all. They come in, haul the victim out, do whatever they do, and then they slide them back in a fraction of a second away, if not at the identical second they took the victim out! (RS) Yes! This article I presented is exactly about this point! If, indeed, anti- particles have lift, then necessarily they have to go backward in time. Then they manipulate this: you can have an abduction any length of time
inside the craft, but in our time, in our level three, it is zero time! (L) Yes, exactly! And not only that, there is the
phenomenon of the craft that looks small from the outside, but inside is huge! (RS) That is all tied up in it! This is very exciting. I am learning the language. In our third level, the motion in space and time occurs via the change of the unit of time and space, therefore, can we change the unit?
A: Yes, this is precisely what we mean when we speak of "transiting from 4th to 3rd."
Q: (RS) So, when they travel from 4th to 3rd, they change the units. That is precisely what is in the article in the journal! [Holds up book.] This is published in the Ukraine, [turns to page and displays diagrams and equations] this is the experiment to test anti-gravity. There is a two mile long tunnel which is a vacuum inside. They suck the air out. The first measure is to shoot photons to identify, at the end of the two mile tunnel, the no gravity point. The second measure is to shoot a neutron, and we know that a neutron is attractive. So, after two miles, the energy is very, very low. So, there is no gravitational effect when the neutron hits the point. Then, the third step is to shoot an anti-neutron at the same time and see what happens. This experiment will resolve this issue that this board has answered very scientifically. We call it the gravity of anti-particles because we don't know. It can be down... Einstein predicts this as attractive as a neutron, anti-matter and matter have the same gravitational attraction. That's what Einstein says. But, when Einstein's theory was proposed, in 1915, anti-matter wasn't discovered until 50 years later. If now, theoretically, the only way a particle, in our theory, can go up, can have lift, is if time is reversed. There is no other possibility. So, if this experiment is correct, then the space/time machine is absolutely a consequence and can be tested in a laboratory. You can have a particle moving backward and forward in time. [Displays new diagram.] This is the other experiment which is, in this case, is done by
putting a particle which is neutral and subjecting it to... since we don't have a bunch of anti-matter - ideally we would have a pellet of matter and replace it with anti-matter - we don't have a pellet of anti-matter, at this point, there are ways to do it though and it can be measured as to which way it moves, up or down. Now, the question of the units, it is very important, a fundamental question, because, say, you are outside a UFO, and you see the UFO as big as a car, say, and people go inside and report this enormous interior. There is no other way to do this than by changing the unit. What is for us one inch, that unit is completely different inside. For us the unit is the same along the three directions. Now, if you are inside, they can have different units in different directions. This means that if you are outside a cube, and you go inside, the shape, not only the dimensions but the shape even, can be different.
A: Density borderline cross awareness. Does Roger have familiarity with density definitions?
Click to expand...
 
The smallest angles of the red and green triangles are:
red: 20.556 degrees
green: 21.801 degrees

Calculator I used: http://www.carbidedepot.com/formulas-trigright.asp
 
Ok, here is my feeble attempt at simple geometry:

B=Base
H=Height
A=Area, 1/2(b * h)

See: _http://www.1728.com/pythgorn.htm
h=hypotenuse, a=adjacent height angle, b=adjacent base angle

[FIRST DIAGRAM]
=================================================
>>> Dark green block:
B1=2, H1=5
A1= 5
h1=5.3852, a1=21.801, b1=68.199

>>> Red block:
B2=3, H2=8
A2= 12
h2=8.544, a2=20.556, b2=69.444

>>> Orange block
A3= 7 "units" (add the unit blocks)

>>> light green block
A4= 8 "units" (add the unit blocks)

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
>>> Area of entire rt-triangle:
B5=5, H5=13
A5=32.5
h5=13.928, a5=21.038, b5=68.962
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
>>> Verifying sub-blocks against entire rt-triangle:
h=h1 + h2 = 5.3852 + 8.544 = 13.9292 <> h5=13.928 ERROR
AT= A1 + A2 + A3 + A4 = 5 + 12 + 7 + 8 = 32 units <> A5=32.5 ERROR
a1=21.801 <> a2=20.556 : b1=68.199 <> b2=69.444 ERROR

Edit:
a2=20.556 <> a5=21.038 ERROR!

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

So, to conclude:

1) The hypotenuse of two sub-rt-triangles to not add up to the entire rt-triangle. (rounding errors?)
2) I attempted to calculate the areas by adding up the sub-blocks but was unable to match up with
that of the entire rt-triangle. Not sure where the error in my calculation for area is as there is
a 1/2 unit missing somewhere...
3) The two sub-rt-triangles have different angles. Switching them causes misalignment.

OSIT :D
 
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