Triangles

Rev Hellfire

Heres one for you to figure out :-

Recently a professor in MIT has put a new theory on right angled triangles that has challenged some accepted norms in Geometry. This theory, Prof John Mentriffe says, will revolutionalize area in mathematics that deals with calculation of motion objects in space and design of the Universe.

What do you think ?




[This message has been edited by Rev Hellfire (edited 15-04-99).]

EgoManiac

This, as you must surely know, is all my doing.
Feel my power over generally accepted principles of geometry and KEEL

Rev Hellfire

Does this mean that I'm like the priest of Ego?

FreaK_BrutheR

****in hell that's so cool!!!!

consider my head wrecked!

nice un rev.

Scarab

The graph is ++4><0r3d.

Rev Hellfire

Or as the boss-man just said :-

Hmm,

Even more interesting, add up the areas of each of the components:

Green triangle: (1/2 base by perp height) = 2.5 *2 = 5
Orange shape: = 7
Green shape: = 8
Red triangle = 4*3 = 12

Total area = 32

Total area of the triangle (using .5 base * perp height) = (13 * .5) * 5 = 32.5

Explain that!

Dave

Cloud

It looks to me that the angles in the red and turquoise triangles are different. Put a piece of paper on your screen and see

J.
--

Cloud

I guess one thing is that the full shape is not a triangle at all - the angle in the red triangle is 21.8, and in the turquoise is 23.2.

J.
--

Cloud

Okay, okay - so the big sloped side curves in in the shape with no hole, and curves out in the shape with a hole.

Whew!

J.
--

LegacyUser

Or as a guy on IRC put it -

<QL-Kalliath> i know why, the c unts have drawn it wrong

Cloud

A guy here in the office went a bit farther than most!

Here's his findings...

J.
--

This is what I said

Well taking it pixel by pixel, it looks like the diagonal line on the
upper one is cut lower, therefore the possibility is that if you add up
the difference this makes across each square it would add up to one
square.

and this is what I got back


> The hypotenuse of the two triangles positioned in either model is NOT a
> straight line because for;
>
> T-red: tan(angle) = 3/8
> T-green: tan(angle) = 2/5
>
> They aren't the same (angle) and this can also be seen by careful
> inspection.
>
> So in the top model, the total area is affected by the hypotenuse of the
> combined red and green triangle hypotenuses sort-off "sucking-in"
> towards the bottom right corner whereas in the bottom model, it sort-of
> "sucks-out" in the opposite direction.
>
> If you were to take the red and green triangles together from the top
> model and paste it directly on the bottom model and matching the corners
> the difference in area exposed by the overlap just happens to be
> precisely one square unit. If I had trigonometric tables on quasar, I'd
> work it out mathematically. Alternatively, print it out then cut out
> each model and see for yourself.

---

DeVore

I cant believe you all fell for that one. IN the top one, look at where the red "triangle" touches the green "triangle".
Its exactly at the intersection of 4 squares.

Now look at the SAME FOUR SQUARES in the bottom triangle (ignoring the convienent colour coding Rev has added to confus the issue) you can clearly see the triangle doesnt go through the same intersection.

That gives it all away... that and the fact there there is no Prof John Mentriffe currently employed by MIT.

DeV (not sitting in a pile of cut out coloured paper triangles no no not at all...)

Rev Hellfire

I'll have you take that back DeVore, why myself and John (to the Profs friends) were playing golf this morning

Rev Hellfire

OK, Stop looking at the graphics, and explain the maths to us. As from Boss-man, its the maths man, explain the area's calculated...

Dupre

>From where comes this "hole"?

Small (nearly) unnoticable volume gains due to the fact that the bottom-left angle is slightly greater in the lower triangle than the upper one. Visible by comparing the empty space of of squares through which the non-axial line goes in both cases.

Quite a good way to fool humans - small gains over a large area. Works well for money laundering and stealing stuff - small bits over a large period of time and nobody notices (so I've been told)


>and explain the maths to us

Fek-off

koloth

The big back lines make it look like a crude kiddies drawing, but zooming into it you see very precise pixel differences (black line v's grid line) between the two sets

Rev Hellfire

OK, the final part of the puzzle is this,

Calculate how much time each of your offices have spent trying to figure it out.

now get back to work you lot.

Dupre

A lot less time than I've spent today trying to figure out how to crack secured pdf files so I can print them

Anybody have any idea how this can be done ? I'm talking about pdf's that have the stupid security thing added so that printing is disabled.
There's supposed to be a documented way through converting to Postscript and back, but I can't seem to get it to work.

Any else know how to get of this REALLY annoying crap security addition ?

Spent 4 hours trying it so far ... ****in' waste of time s@£$%

Jaden

BTW all, I have mocked this up in CAD, and can confirm it is not a trick. I'm not Joking, and have DXFs to prove it.....

The solution is that Half the base by the perpendicular height is NOT the area a triangle. It is merely an approxamation...

Now THATS Scary.....

Hail To The King, Baby.

Hecate

does this mean that the moon will soon collide with the earth?

Scarab

Jesus i cant believe some of you are falling for this, its so simple to make it untrue..

the height of both is the same and the lenght of both is the same, therefore the angle of the hypothenuse should be equal but all u have to do is look at one of the graph lines ( say second from top ), on the top drawing the line intersects the hypothenuse at the edge of a box in the second one it intersects the hypothenuse at the middle of box, therefore either the lines on the graph are of differing size or the hypothenuse is curved.

El_Presidente

The hypothenuse is not in fact a hypothenuse at all, the red and green triangles have different slopes in BOTH pictures. Print it out and it becomes much easier to see.

I think we've all seen how this works, still should give my maths teacher a headache.

Jaden

Gahhhhhhhhh,

I've been up half the night with this, both working out the explanation, and explaining it to others.

Firstly, the drawing shown is not really accurate, it's a whole lot better to draw it again for ureself. I myself, have done it in Cadstar (bloody waste of a 14 grand PCB design package, but anyway.)

IT IS possible, by NO trickery, eg overlapping, curving lines, to re-arrange the shapes and create a hole. The reason this hole "appears", is because the hypotanuse of the 2nd triangle created by the 4 shapes is not real. The two lines that comprise it are not parallel. (The implications of this frighten me). This creates, as was pointed out, a "dip" in the line, that explains where the box came from. This is possible, because laws of geometry are APPROXAMITE, and not exact, as we have all been lead to believe.

All those of you who had makrks in the Inter, junior, or leaving, docked for incorrect answers on trigonometry, may now petition for a recheck.

Hail To The King, Baby.

PS - Hellfire,
Or as the boss-man just said :-
Hmm,

Even more interesting, add up the areas of each of the components:

Green triangle: (1/2 base by perp height) = 2.5 *2 = 5
Orange shape: = 7
Green shape: = 8
Red triangle = 4*3 = 12

Total area = 32

Total area of the triangle (using .5 base * perp height) = (13 * .5) * 5 = 32.5

Explain that!

Dave

The area of the triangles have not changed (overlay them to prove it), so the logical explanation is that the formula to calculate their area is incorrect.

Now go get some Brownie points.....

Dupre

.

[This message has been edited by Dupre (edited 16-04-99).]

Dupre

Don't just dig yer nuclear bunkers yet. Theres nothing wrong with the maths.

You used the word approximation and thats how this whole trick works.
The thick black lines used give it away.

Forget completely about the bottom triangle.
The whole problem with the top triangle is the two black lines that divide out the two sub-triangles.
They are approximate, but not exact. So when you do yer (1/2 base * height) calculation for the full triangle you get the correct volume of 5*(13/2) = 32.5.

But when you break the triangle up and are doing the volume calculation for the two sub-triangles, you are "approximating" their boundaries to be on the grid ... They _nearly_ are on the grid. Hence the thick black lines and the 0.5 difference in the two overall volumes. This approximate, but inaccurate figure feeds into the volume equation, which, understandibly kicks out the wrong answer.
As I said in my earlier post, (nearly) unnoticable volume gains over a widish area.

In effect, the figure Total area = 32 is an approximation. If you could draw the top triangle with an infintely thin pen, you would see that the point of intersection of the two dividing lines doesn't lie on the hypothenuse at all. If you drew it better you would get a volume nearer 32.5, and as you got nearer to drawing it correctly you might get something like 32.521325353456356 or god knows what. But 32.5 is more accurate than 32 and 32.5567573453455 is more accurate than 32.5.
Using thick black lines however, there is no _point_ of intersection between the lines, just an _area_ of intersection, so you can't be accurate about it. Because of this, you can juggle the thing around to give the 'missing square' effect shown above.

There - I don't think I could have explained that any worse

In short, ya's have been comparing two different triangles in the real-world, even though they appear to be the same in graphics land.

Now go forth, and make lots of money from drunk people in one of those pub scam things



[This message has been edited by Dupre (edited 16-04-99).]

Shiminay

This is all pure mad!

Ye are all pure mad!

I am great...
KEEL... over
heeheehee

---

All the best,

Dav


[This message has been edited by Kharn (edited 16-04-99).]

Koopa

NNNNGGHHH WHY USE CRAPPY TRIANGLES TO DO THIS, WHY NOT JUST PUT UP A PICTURE OF A BIG HAIRY ELEPHANT AND SAY "THIS *CIRCLE* IS THE SAME AREA AS THE CIRCLE WITH THE HOLE IN IT UNDERNEATH" WHERE YOU HAVE DRAWN A (CRAPPY) CIRCLE WITH A BIG FAT H0LE IN IT??

Dupre

Coz Einsteins 10th law of Fat Hairy Elephants states quite clearly:

Don't PUT UP A PICTURE OF A BIG HAIRY ELEPHANT AND SAY "THIS *CIRCLE* IS THE SAME AREA AS THE CIRCLE WITH THE HOLE IN IT UNDERNEATH" WHERE YOU HAVE DRAWN A (CRAPPY) CIRCLE WITH A BIG FAT H0LE IN IT or else time-space will collapse in on itself.


Man, how do you not know that - every bleedin primary school kid knows it.

Koopa

BUT WOULD IT STILL COLLAPSE IF THE PROPOSED HAIRY ELEPHANT HAD IN FACT THE SAME AREA AS THE AFOREMENTIONED 2 DIMENSIONAL SPHERE with a hole missing?? can Dr. Dupre verify this? (if that IS his real name..)

Koop-(yes_i_AM_currently_writing_a_thesis_on_the_subject)-a.

Dupre

Oh ffs - Thats going too far.
You mean you haven't read the Appendix to Einsteins 1st Law of Fat Hairy Elephants either?


APPENDIX A. Paragraph 1 Line 15

3. TIME-SPACE WILL STILL COLLAPSE IF THE PROPOSED HAIRY ELEPHANT HAS IN FACT THE SAME AREA AS THE AFOREMENTIONED 2 DIMENSIONAL SPHERE with a hole missing*

*my collegues, in particular Dr. Dupre, can verify these findings.


P.S. I am actually typing up my final year project report as I send in all these bull**** posts. Its amazing the wastes of time that can be found when you have a ton of work to do

[This message has been edited by Dupre (edited 16-04-99).]