Triangle paradox

Moo Of Doom

Certainly the word RandallB is thinking of is parallelogram?

T@P

this is probably the most "popular" puzzle ive seen on these forums... really its kind of ridiculous now that im seeing it for the 87th time.

RandallB

Your correct - You'd be surprized how many smart ones "Know" this Puzzle - so they snooze through it without completely recognizing this source area for the more easily seen square area. Yet even the most novice can find it with a bit of diligence.

Thanks for keeping it in white for those new to the forum wanting to work it on their own.

pack_rat2

The smaller triangles aren't similar, but they're close enough so that you think they are, and you think they're similar to the big triangles. If you place the edge of a piece of paper along the hypotenuses of the the large triangles, you notice the upper one sags, and the lower one bulges.

pack_rat2

"Below the four parts are moved around. The partitions are exactly the same as those used above."

Indeed they are...no fibbing here. We tend to assume the triangles are similar because on first glance, they certainly seem to be.

niranarch

There is no optical illusion. Its just a careful manipulation with line thickness checkout this drawing i made in CAD ( you can try for yourself)

http://sphotos.ak.fbcdn.net/hphotos-...3_549897_n.jpg

davee123

Although yes, you could do that, it's not what's going on in this problem. In the problem, there are 4 shapes:

A) Teal right triangle measuring 2 units by 5 units
B) Red right triangle measuring 3 units by 8 units
C) Orange irregular hexagon with area of 7 square units
D) Green irregular hexagon with area of 8 square units

These shapes don't change, and neither do they overlap. In your example, you'll notice that you've drawn shape (A) with a height slightly less than 2, and shape (B) with a height of slightly more than 3, so the shapes are misrepresented in your version. The OP has nothing to do with line thickness whatsoever-- you can see this for yourself if you'd like to make a similar CAD drawing with the shapes as defined above, with a miniscule line thickness.

DaveE

Nick89

Haven't read the whole thread but the answer is really simple:

niranarch

@dave123- have you tried drawing it? its impossible to draw as you see in the original one.
If you are trying do try it in a CAD software so that thickens of lines dont effect you. In a small drawing line thicknesses seem to be irrelevant but can easily hide such differences

davee123

No, it's not impossible. Heck, it's right there in the very first image. I think the problem you may be having is that you're attempting to draw it by STARTING with a triangle, and then subdividing it into 4 sub-parts, which is correctly impossible. Don't do that. Start with the 4 sub-parts, and arrange them as shown. Arrangement #1 doesn't have a square-shaped gap, and Arrangement #2 DOES have a square-shaped gap.



DaveE

aidanapc1359

OK Guys,

First of all the problem states that the partitions in both large triangles are all equal.
So please just accept this. No measurements are necesary or relevant.
Also the the two large triangles are equal.
The problem arises due to topology.
or simply "there is no conservation of area or volume".
With a 2D shape you can maintain area while changing perimeter.
For example, two rectangles a 1m x 4m, area = 4 m2,
For a 2m X 2m rectangle, area also equals 4 m2
However the two rectangles have different perimeters!, The first = 10m, the second = 8m
So by changing shape you can maintain area and change perimeter.
This is similar to the 3D version of volume and surface area.
The reason why your small intestine has so many villi and microvilli in it, is because it greatly increass the surface area for absorption.
Also why a piece of sodium reacts much slower than powded sodium (crushing into a powder greatly increases surface area without changing volume.
Getting to the point:
Basically because the partitions no longer slot together in the bottom diagram, this has changed the shape but maintained the area (however the effective area for covering has decreased and created a hole)
Thus the perimeters of the two figures are different!
Another way of answering this is because the overall shape is now irregular it can no longer cover all of the area of the bottom diagram, even though there has been no loss of area in the partitions!
Forget about measuring squares, this problem is not about deception but mathematics.
Hope this makes sense.
By the way, this is my first posting, very pleased to meet you all

Aidan.

nashsur

I like to keep things simple and maybe that is my mistake. To me it is just the difference in length of the orange and green thinner sections.
Please note the overall length of the red triangle is 8 blocks. There is only one way to combine the length of the orange and green blocks to make 8. However, the thinner sections of the two blocks are of different lengths. One is 3 and one is two. No matter what you do to combine them to an overall length of 8, you will have the resulting hole. That is the long and the short of it. It seems to me that if you are trying to create a base for the larger red triangle it would be to your advantage to create a base that is the same length as the red triangle. To that end you would combine the orange and green blocks in a configuration that would total an equal length of the red triangle. I am sure there is a mathematical expression to describe this but it is not needed here. Other than to say 4 + 4 = 8 and 5 + 2 = 7. That leaves 1 not accounted for. Why add all these other complications to it?

To wrap up; overall length is red triangle base plus green triangle base. Green triangle, orange block and green block are of equal base length. Red triangle base length = 8 . All other shapes base length = 4. Red triangle is 3 high. All other units are 2 high on one end. The two block shapes have one end that is 1 high. The 1 high end on the orange block is 3 longer than the 2 high end and the 1 high end on the green block is 2 longer than the 2 high end. The only way to combine the two blocks to equal the correct length of the red triangle and the height of the green triangle is to put the two 1 high ends of the orange and green blocks on top of each other so as to cause the overall height to be 2 and the overall length of the combined orange and green blocks to equal 8. This will automatically leave a space between the orange 2 high level and the green 1 high level because 4 + 2 does not = 8.

Again, if I miss the point of the exercise please let me know.