Guess where is the missing square

[The Lone Gunman]

Bob explain it to me.

What is the area of the geometrically perfect triangle you propose?
What is the area of the small sub triangle?
What is the area of the bigger sub triangle?
The area of the rectangle containing the empty square?

Now subtract the 3 sub figures and tell me you get a nice round zero.

Do that for both examples.

[xoggoth]

Total area of components = 32. Total area of complete triangle assuming it it had a straight hypotenuese = 32.5. Er um therefore um

QED

[AtW]

I hope you guys will keep argueing until this thread reaches 1000+ posts. I won't leave CUK until after I bean shaunbhoy's record.

[Lucifer Box]

I hope you guys will keep argueing until this thread reaches 1000+ posts. I won't leave CUK until after I bean shaunbhoy's record.

Alexei, you have unleashed a beast.

[BobTheCrate]

AtW,
Here is my next contribution toward your attempt for a new thread record.

Lone Gunman,

Example true right angle triangle ABC with right angle at ACB
Adjacent line BC at 13 units
Opposite line AC at 5 units

Area of triangle = 32.5 sq units.
Angle ABC = angle ADF = 21.0375 degrees by tangent rule.

Angle ABC = angle ADF because any diagonal line bisecting two parallel lines creates identical angles at both intersects.

Now draw in the sub-shapes. Excuse the apalling diagram below - it is only meant to act as clarity for where additional points approximately meet when creating the sub-shapes and does not represent anything more. Finite line lengths are supplied below it.

Stick to the lengths supplied below it and you should have all straight lines that fit properly within the true triangle ABC. Pretty confident I've relayed the measurements correctly. If you suspect a mistake let me know and I can sort it.

.................................................. ..............A
.................................................. .. /..........|
............................................./..................|
...................................../..........................|
................................D___________________F
........................... /...|...............................|
....................../.........|...............i_________H
................/...............|...............|...............|
........../.....................G_________J...............|
...../...........................|...................... ........|
B___________________E___________________C

Lengths to use

Red triangle DBE base length BE = 8 units
Lime shape base length EC = 5 units
HC = 2 units
GE = 1 unit
GJ = 2 units (important length)
iJ = 1 unit
iH = 3 units (important length)
Green triangle ADF base length DF = 5 units

Area calcs for each shape you wanted
(I'll prove these calcs for you seperately if you like)

Red triangle = 12.3076 sq units
Green triangle = 4.8078 sq units.
Orange shape = 7.3845 sq units
Lime shape = 8.0000 sq units

Add 'em up and you'll get 32.4999 sq units which you may (or may not) accept as what we also got above for the parent triangle the shapes produced in AtW's top diagram.

Re-arrange the shapes as per AtW's bottom diagram.

Sum of shapes at respective base is Green 5 + orange 2 + lime 5 = 12 units

1st down is green triangle. Base length DF = 5 units
Next down is orange shape. Base length GJ = 2 units
Next down is lime shape arranged so that iH of 3 units sits on top of orange shape line GJ of just 2 units.
3 - 2 = 1 unit gap produced at base.
(Total length of whole base is 12 + 1 unit gap = 13.)

Or another way.
Green base = 5 units, met by
Orange top line DF = 5 units, met by
Lime shape top line iH = 3 units
Total units = 13
Base units still = 12
Difference = 13 - 12 = a gap of 1 unit.

To finish off, the 8 unit base line (BE) of red triangle goes atop the 5 unit top line (DF) of orange and the 3 unit top line (iH) of lime.

The area of the shapes remain unchanged because we haven't changed the size of any of them. The collective area of the sub-shapes being constant of course.

The bottom parent triangle is no longer a triangle of course because its got a bloody great gap of 1 unit in its base making it a 6 sided shape. But it was produced using a geometrically true right angle triangle and its 2 geometrically true right angle sub triangles within. And the two 5 sided shapes of course.

(Image below posted to widen display area for crap diagram)

[AtW]

AtW,
Here is my next contribution toward your attempt for a new thread record.

Keep 'em coming -- so many on topic posts that don't talk about sex, threaded, hostages or politics are unknown in history of CUK.

[The Lone Gunman]

Just for you ATW.

Bob, I disagree with your maths but that does not matter.
You have given a set of areas covered by the individual components which you say adds up to the total area of the overall triangle in which they all sit.

I dont see how you can then justify that all those pieces and one extra square unit can fit in the bottom diagram if the overall "triangle" has the same area.

I cant actualy see the original drawing any more so give me a break here.

You agree the area of a triangle is 1/2 base times height?
So main = 13 x 5 which is 6.5 x 5 = 32.5
other 2 are 8 x 3 and 5 x 2 which gives us 12 and 5 respectively.
The other 2 shapes form a rectangle in the first which is 5 x 3 which is 15. That all ads up to 32.
The triangles are invarient, but the rectangle in the second figure (including the space) is 8 x 2 which is 16 which adds up to 33.

The differences in measurement can be accounted for in the bowing of the hypotenuse.

[oliver]

Red triangle DBE base length BE = 8 units
Lime shape base length EC = 5 units
HC = 2 units
GE = 1 unit
GJ = 2 units (important length)
iJ = 1 unit
iH = 3 units (important length)
Green triangle ADF base length DF = 5 units

Area calcs for each shape you wanted
(I'll prove these calcs for you seperately if you like)

Red triangle = 12.3076 sq units
Green triangle = 4.8078 sq units.
Orange shape = 7.3845 sq units
Lime shape = 8.0000 sq units

Your figures are obviously incorrect! LOL

For instance, please explain how you got the figure of 7.3845(?) for the area of the orange shape (other than subtracting your other calculations from 32.5 in an effort to make things add up)?

Using your (incorrect) measurements you have in fact shown that this shape has an area of 7. Doh!

But this is not even the main issue.

Add 'em up and you'll get 32.4999 sq units which you may (or may not) accept

LOL hahahahahahahaha - when 'proving' your maths you could at least PRETEND it adds up! And this STILL isn't the main issue...

The bottom parent triangle is no longer a triangle of course because its got a bloody great gap of 1 unit in its base making it a 6 sided shape.

Wrong! It is NOT a 6 sided shape! You will NEVER accept this will you? Of course not because it would force you to understand how wrong you are.

Maybe you should have said "fill in the bloody great gap of one 'unit' and then squint at the resulting shape - possibly from a distance. It looks like a triangle and hopefully no-one is smart enough to see that it isn't"

10 out of 10 for providing me (and possibly others) with remarkable value for money in the entertainment department.

[xoggoth]

Where did those decimal places come from? Have I entered some strange alternative reality or have we got onto discussing a different problem somewhere?

32.4999 square units???? I calculated it in primary school mode as 32.5, as did LG, but I now realise I did not take proper account of line thickness. How stupid of us, that 0.0001 make all the difference, I can see it now.

[xoggoth]

Ah!! I see the really big mistake here. We have been thinking of this as a purely two dimensional problem when, if we were capable of a bit of lateral thinking, it would be plain that the third dimension is important.

What we need to do is consider the thickness of each of the individual segments. In the third dimension the first triangle has a profile whcih may be represeneted as a Cissoid of Diocles, in the second, spatial reallocation causes it to have one near to (to within 0.01% at the point of greatest deviation) a Conchgoid of Nicomedes. If we apply non-linear integration to the difference between these, using spatial transposition to take full account of the interating subsets, we arrive at the correct solution. BC is right.

PS Don't think this is half as puzzling as the Monty Hall problem

http://www.letsmakeadeal.com/problem.htm