www.BPODigest.com i-Vortal

It seem like this has to do with the angles. The longest boundary of the bottom triangle is slightly raised, giving it a larger area, and therefore room for another square. (in other words, the angle was slightly raised to trick us!)(For example, look at the point at the height of the bottom green triangle and then look at the same point on the above triangle...one can see a little difference in height!)

Well, that little difference that is there is b'cos of the black lines that are intersecting in the bottom green one and there are no intersecting black lines in the top red triangle at the same place. All the blocks are squares and I don't see any reason why the angle should be different. Or I am really dumb to understand your reasoning ;-)

I can see that the point chosen above is probably not the most clear example. I think that there is a better (bigger) example when we look two squares above the "hole". Just look at the area of the square covered by a red color in the bottom triangle as opposed to the top triangle: the top triangle covers a little less red area! (When looking at all of these differences throughout the triangle, all of these "add up" to account for the area of the hole!)

I think all of this is showing that the uppermost (longest) boundary of the two triangles are not straight lines. One dips a little upwards and the other dips a little downwards. Since they are not straight lines, the are actually not true triangles in the first place, but only seem that way from a distance!

Hmm, Yes, now I can see, they are tringles indivdually, but when put together the figure that is formed is not a true triangle. A visual trick.

Here's one for the mathematically inclined:

The green triangle has base=2, height=5 => slope = 2/5 = 0.4

The red triangle has base=3, height=8 => slope = 3/8 = 0.375

Close but no cigar! putting them together edge to edge doesn't make a real triangle.. just an optical illusion here.

Mr. Manan, Mr. Kotamaraja and Mr Gurjeet.

It is very simple. There is no optical illusion, no angle (inclination problem) or boundary problem. Just check your theory of area in quads

Regards
Ravi
Ravi_and_me@yahoo.com

Hi Ravi,

Well, My brain is completely fresh, never used, hehehe. So, can you please explain us what you mean by theory of quads.

Forget about the extra square in the second picture. Let us consider the first picture alone -

1)Calculate the total area of the whole triangle (32.5 sq units) - using the formula = 1/2 x base x height.
2)Sum up the areas of individual components (32 square units)

You would see a difference in 0.5 square units.

The difference in slope as pointed out earlier could be the reason for this 0.5 square units of area difference. So, it could be that the 'whole' is NOT a 'perfect' right angled triangle !!!

A CRY FOR HELP(SOS...SOS...SOS)

this is the most amasing thing i ever saw.
let it be the slope as pointed by Gurjeet
or it be the quad area as pointed out by ravi.
my mind is boggling with ques and i will go crazy if i dont find a solution to it very soon i havent slept proprely for three nights and i am desperate for an explanation so please help me out !!!...

A CRY FOR HELP(SOS...SOS...SOS)

this is the most amasing thing i ever saw.
let it be the slope as pointed by Gurjeet
or it be the quad area as pointed out by ravi.
my mind is boggling with ques and i will go crazy if i dont find a solution to it very soon i havent slept proprely for three nights and i am desperate for an explanation so please help me out !!!...

tarun
((universal_tarun@yahoo.com))

Don't Confuse, Old is Gold

Hey you are wrong and i can prove it.
If you are really interested to get answer of this puzzle send me your reply on " vvk_25@rediffmail.com".
If you have any other query, puzzle or problem then also you can send me on
Regards,
Vaibhav Khamkar

Hi All,
First question which should strike our mind is that, WHY THIS IS A PUZZLE WHEN IT SIMPLY LOOKS A REARRANGEMENT OF DRAWINGS?
Guys analyse it and you will find that just due to this small rearrangement THE SIMILAR ANGLE THEOREM is violated.That implies this rearrangement has something to do with Orientation.Simply speaking FOR A CONSTANT AREA OF THE TRIANGLE THE ORIENTATION SHOULD BE TAKEN INTO ACCOUNT.I know this definition is yet incomplete but what puzzle is all about is told here.
Prateek

Firstly club both the seeming triangles to form a rectangle.

The area of this rectangle L = 13 & B = 5 is 65.

Calculate the area of the first "triangle" by summing up only the parts which make it up. The area is arrived as
follows:

Red triangle - L = 8 & B = 3 & so area is (0.5*L*B) 12
Dark green triangle - likewise is 0.5*5*2 = 5
Yellow portion is 7
Light green portion is 8
Total area of the first "triangle" is therefore 32

On rearrangement, the area occupied by the parts will not change and therefore the total sum of the area of the
original "triangle" and that of the rearranged "triangle" will only be 64.

But you will see that the entire rectangle area is actually 65.

Now you know where the extra "hole" with area of 1 comes from.

Definitely, the big "triangle" is not triangle at all. It is a polygon. In other words, the hypotenuses of the red triangle
and green triangle donot make a single straight line but is bent. But the bend is not very pronounced enough to
make it obvious.

You can calculate the hypotenuses of the red triangle & green triangle and add them up and find that it doesnot equal the hypotenuse as calculated for the big right triangle.

This mail has arrived to me in email :

I am no math genius well my geometry is also flawed ...but i have seen a lot of spam and designs to collect email id'z ...well but reasoning and a lot of Search engine Study tells me a lot .

This is the Link to MIT http://web.mit.edu/ besides this I have done a boolean search in altavista and Google . Suprisingly there is no PROF.John Mentriffe at MIT who has made such a revolutionary theory .....

MIT doesn't mean one can do MAGIC .

Mathematical approach:

Area of the Whole Triangle=(1/2*13*5)=65/2=32.5 square units.--(1)

And We Know That Area of This Triangle = Sum of areas of the composite figures.
That is, Area = areaof (tri1+tri2+rectangle)
Area = (1/2*8*3)+(1/2*5*2)+(5*3)=12+5+15=32 --(2)

Comparing (1) and(2)
We probably have ahint for the solution.
With this are we any closer to the solution??????

I plotted both the geometries on AUTOCAD and found that there is a difference in area given top and bottom. the difference arises because of the two triangles. In the latter figure a void has been created at the expense of extending dimension of the two triangles in secong figure. Anyone who wants to see a closer look of these geometries on CAD can contact me at anees_bst@yahoo.com

the first triangle can be reconstructed to the second by using operations of rotations (R),translation (T),reflections..etc which are common tools of transformation geometry.it is very simple to show that R+T not equals T+R(ofcourse sometimes they do but that's special case).similarly the constructiom of the second from the first can be seen as complex operations of transformations (involving R,T etc) only.so they should differ by structure (i.e misfit of blocks)to keep their area constant.

Oye ... wats the confusion??? They are not triangles. The slopes of the red and green triangles are different ... so putting them together u dont get a triangle ... u det two Quadrilaterals ... altho our eyes arent really good enough to detect a difference of 0.025 in slope.
No problem with pythagoras ... his genius still hold.
Isnt it simple???

The 1st and the 2nd figures are not triangles,because the angle of the red triangle is slightly less than that of the green triangle. Now the area of the 1st figure is=ar(red triangle)+ar(green triangle)+ar(yellow part)+ar(light green part)=12+5+7+8=32
Now ar(2nd figure's shaded portion without the blank space)= ar( 1st figure 's shaded area).
Now ,
ar(blank space) =ar(2nd figure with the blank space)-ar(1st figure).
Now,
ar(2nd figure with blank space)=ar(rectangle formed by joining the yellow part ,light green part and the blank space)+ar (red triangle) +ar(green triangle)=16+12+5=33
Now the ar(blank space)=ar(2nd figure including the blank space)-ar(1st figure)
=33-32=1

to continue with my previous proof I would just like to add that the two figures taken together with other figure as shown above are not at all triangular but form quadilateral where both vary because the two angles of the red and the green triangles are different.
angle of the red triangle

Much ado about nothing... Now we know for sure how frauds are built.

Red trianlgle: 8*3/2 = 12 sq units
Green triangle: 5*2/2 = 5 sq units
Orange piece: 7 sq units
Light green piece = 8 sq units

Total area of triangle with sides 13 & 5: 13*5/2 = 32.5 units (Big triangle)

Slope of Red triangle: 3/8 = 0.375
Slope of Green triangle: 2/5 = 0.4
Slope of Big triangle sides 5 & 13 is: 5/13 = 0.3846 (which is in between)

Area of Top Polygon: Area of Big Triangle - Area of triangle made by slope 0.375, slope 0.4 and slope 0.3846 = sum of all pieces = 32 units.

Area of bottom polygon : Area of Big triangle + Area of triangle made by slope 0.4, slope 0.375 and slope 0.3846 = sum of all pieces + 1 = 33 units.

This gives that area of tringle formed by 3 slopes 0.375 slope 0.4 and slope 0.3846 as = 0.5 sq units.

I think the red triangle is the culprit

You have been misguided. As with many riddles you are bieng decieved through your brilliantly simple reasoning. Clear your mind and you will find the answer. Your preconcieved notions and previous interaction with reality have clouded your mind with irreconsilable assumption.

To solve this, think of how your intuition would lead you in proving that the areas ARE equal. Then knowing that your intuition is always right, percieve what is the underlying conflict in saying that the areas are at the same time equal and also unequal.

Using this method and without looking at the picture at all you find that THE INDIVIDUAL SQUARES ARE NOT ALL THE SAME SIZE AND ARE NOT ALL TRUE SQUARES, ALSO, THE GRID LINES ARE NOT ACTUALLY PARALLEL AND PERPENDICULAR. Were you using the grid squares for measurement? Because any rational visual analysis of alignment, size, and angles requires some comparison to an orthaganal gridspace using the one provided invalidates any and all theroms of clasic geometry.

This shows that your true reasoning was to say that if the sum of the individual areas are equal then the overall areas are equal. The extra white area gap in the second overall area shows that the overall areas are NOT equal. Visual analysis then shows that the relative porportions of the colored areas to thier containing grid squares are equal (where the angled lines of the colored poloygon edges cut through the grid squares...where a gridsquare is divided by a line it is obvious if any nonsimular deviation exists which it clearly appears does not).

With this information all we can say is that within the confidince interval of one or two squares the lines are straight and the corresponding polygons made when these straight lines bisect the corresponding grid squares are simular. This says nothing about the geometry on a scale larger than one square. As optical illusions prove, we are not capable of encompasing analysis on a larger scale unless assumptions are made as our eye can only look in one small area at a time. This is where the tools of euclidian geometry (assumptions) fair us well, as long as we do not violate thier basis in thier use. To do this with the shown image would FIRST require proving that the gridlines are parallel, perpendicular and evenly spaced (all gridsquares are the same size). Measurement of the grid shows that the grid lines do not meet this criteria and the following geometric relations show that this deviation accounts for the extra white square.

Analysis using CAD would be incomplete as it applies to this EXACT image because it would use a truely orthagonal grid space with equal sized squares, which the image was never stated to (and clearly does not) have.

Aren't you guys bored with this yet? To be precise, the green and red triangles must be SIMILAR figures, meaning their corresponding angles are congruent and their corresponding sides are proportional. Since the leg ratio of the red triangle is 3/8 and the leg ratio of the red triangle is 2/5, the triangles are obviously not SIMILAR, so the corresponding angles are not congruent and the rearrangement cannot take place without changing the figure!

Gurjeet is completely right. The slopes of the 2 triangles are not equal, thus the triangle is not a triangle; it is a 4 sided polygon. Thus, the Area formula (1/2 x base x height) does not apply at all. It is an optical ollusion.

This is not an optical illusion. It's simply the way the pieces are rearranged.

Just print out the triangles, cut one of them into it's individual shapes and rearrange them. You'll see that the problem does exist.

dear mr. kotamaraja, the base and the heights are supposed to be switched.

bob,
9th grade

I think my idea is right....

If you look at the overall (big triangle) in both the pictures....
The hypotenuse is not a straight line...

or in other words,
both red and blue traingles do not have the same angle of inclination!

can this be true?

I think my idea is right....

If you look at the overall (big triangle) in both the pictures....
The hypotenuse is not a straight line...

or in other words,
both red and cyan traingles do not have the same angle of inclination!

can this be true?

I think my idea is right....

If you look at the overall (big triangle) in both the pictures....
The hypotenuse is not a straight line...

or in other words,
both red and cyan traingles do not have the same angle of inclination!

can this be true?

I tried it on paper ..it works….BUT

But…

Please check carefully with the hypo area uncovered

Compare the two images…manually you would be able to find some difference…

But if you are image processing psycho .you might be aware of segmentation technique used to diffentiate object from background

Here is the procedure to do this

1.Do segmentation for both images

2. mix both images

3.capture the pixels surrounded by hypo

4.convert into your required format

5.Compare with the area you calculate from previous pic.

Result: you will get the exact amount of area of vacant box(white in color)

Hope this will help to solve the puzzle..

Its very simple. Look carefully at the hypotenuse of the top triangle - its slightly convex in shape. whereas that of the lower triangle is slightly concave in shape. Hence the area of the lower is one block greater than that of the upper one. Neither of them are actually triangles. Who is the idiot who is putting up these sort of cheap puzzles and wasting everyone's time?

Its very simple... The slope of the red and green triangles do not match. Hence, It is a polygon, and not a triangle. Even if it was a triangle (for argument's sake), there would not be any problem, as the answer is hidden in the rectangles, and not the triangles. If you rearrange blocks of a rectangle, you will be able to form various shapes other than a complete rectangle. The second picture shows a 8-sided design, as 3 blocks are forced in a place which can accomodate 2 blocks. Though the resultant is not a rectangle, it accomodates 2 triangles by altering their bases.

Best way is to make these pieces out from a card board and try to arrange them in both ways.
If area in both ways is same then it is really a challenge to the existing theory.

Note: Compare areas by superimposition method!

Best way is to make these pieces out from a card board and try to arrange them in both ways.
If area in both ways is same then it is really a challenge to the existing theory.

Note: Compare areas by superimposition method!

Best way is to make these pieces out from a card board and try to arrange them in both ways.
If area in both ways is same then it is really a challenge to the existing theory.

Note: Compare areas by superimposition method!

This is the most stupid maths challenge i've heard of... Its a sham and total waste of brain power and time for those who took this seriously. The answer is clearly evident : The diagram is erroneous. Now for the Sham part of it there is no "Prof. John Mentriffe" at MIT for that matter. Whoever started this thread please think twice about posting crap about an MIT professor or about a theory that is going to change the understanding of this universe. Please!!!

oh it aint a triangle at all.