can someone explain this?

[DocM]

http://img160.imagevenue.com/img.php?image=93350_trigrid_122_492lo.JPG

 

[warot]

I think we figured it out. If you look at the two triangles (green and red), they have different slopes. One is 2/5 and the other is 3/8. So if you switch them around, the slope is changing constantly.

You can also tell that the spacing on the diagonal like (the white in each square) is not the same as the top one.

that's my guess

 

[donau]

I think we figured it out. If you look at the two triangles (green and red), they have different slopes. One is 2/5 and the other is 3/8. So if you switch them around, the slope is changing constantly.

You can also tell that the spacing on the diagonal like (the white in each square) is not the same as the top one.

that's my guess

I'd agree with the first part of your solution. The gradients are different and therefore the two composite figures that look like triangles are not actually triangles at all. The hypotenuses are not straight lines but each have an angle to them, one inwards and the other outwards. Therefore there is a difference in the space used. And in this puzzle seems to be exactly 1 square.

 

[Osnabrueck]

Warot is right. I put the images on top of each other and got this result:

 

[siko]

Optical illusion is all... your eye cannot really differentiate the slope. Otherwise, it is mathematically impossible.

 

[warot]

Optical illusion is all... your eye cannot really differentiate the slope. Otherwise, it is mathematically impossible.

If you pay attention, you can definitely see that it is not a straight line. But being an engineer, I did rise / run

 

[Michael]

Very impressive guys. I couldn't figure it out myself. Now that I see that the slopes are different, it makes a world of a difference.

 

[SDNR]

I'd agree with the first part of your solution. The gradients are different and therefore the two composite figures that look like triangles are not actually triangles at all. The hypotenuses are not straight lines but each have an angle to them, one inwards and the other outwards. Therefore there is a difference in the space used. And in this puzzle seems to be exactly 1 square.

After obsessing over this puzzle for about an hour, making my own diagrams etc. ...I totally agree with you donau.

I really like these types of mathematical puzzles. An interesting thread DocM

 

[donau]

...I really like these types of mathematical puzzles...

I also like this kind of brain gymnastics. I've always been kinda fond of it but lately I like them even more because I think doing little things like this might help keep bigger problems at bay (referring to memory loss etc.).

 

[DocM]

nice to see that we have some people who like to use their heads on this forum,
I didn't waste a lot of time on this puzzle , but thought that some of you might like to give it a shot...I'll try to find more of this kind of stuff if you guys are into puzzles of that sort

 

[donau]

The solution to this puzzle is not quite as obvious as it might appear.

In actual fact, the angle or gradient of the hypotenuses is identical. The thing that confuses us here is the use of the grid.

Because of the way the hypotenuse is divided, when the triangles are moved to their new position, they no longer meet at the same points. Because the points have shifted off the grid (they now, in effect, exist on a secondary grid and if this altered hypotenuse was extended to the proper points, it would be longer than the original hypotenuse), the triangles are no longer working on the same grid as the other two pieces of the puzzle.

I think you have a point about the position of the point where the 2 triangles meet. It has changed. Osnabruck's diagram where he had overlaid the images shows it nicely. However, the ratios to calculate the gradients prove that they are not equal as Warot had pointed out. The two arrangements will end up in the same end spot because the sum of the hypotenuses is still the same, only the path has been inverted.

Anyhow, this was fun and looking forward to working with you again on the next puzzle...

Thanks DocM

 

[donau]

I was already going to bed but this puzzle would not leave my mind so I had to come back. You are right: the grid is a confusing factor in this puzzle and perhaps is there only to make the puzzle what it is. It would be easier if the grid was not there at all, only space that had been delimited, because the shifting of the components could be "adjusted" so that the "missing" space would not be noticed, since there is no missing (or freed up) space. The grid provides a framework for the illusion that space is "freed" by the re-arranging of the components. I don't know if I make any sense anymore but what I am attempting to say is that you were right about the grid. But the gradients are different as can be easily calculated.

 

[donau]

Here I have superimposed the two diagrams to show the gradient has not changed. What has changed is the points where the hypotenuse ends.

You are correct that the pieces could be arranged like this (having the gradient the same), but then there will be another problem because the sides of the triangles would not be parallel with one another (since the gradients were different) and the rest of the pieces could not be fitted in. Again, I think the grid is the confusing issue here.

 

[donau]

Olli, the gradients of the long sides of the red and dark green pieces are precisely the same. The one important thing which must be taken into consideration is the dark green piece of the puzzle is not, in actual fact, a triangle -- it is, in reality, a quadrilateral. It is not mathematically possible for the dark green piece to be a true triangle with the pieces all fitting onto the grid.

I agree that the dark green "object" could be a quadrilateral instead of a triangle. But even with measurements of the quadrilateral I get the ratios of the sides to be slightly different. Hence the slightly different angle. I don't have any precision equipment for measuring unfortunately, but even with my simple ruler I feel comfortable that the ratios are different. I would be interested in seeing what kind of numbers you are getting. My rough measurements produce ratios of 0.39 for the red triangle and 0.375 for the green triangle/quadrilateral.

 

[Choleric]

In the original picture the green triangle is in fact a triangle not a quad, only the newly drawn was it a quadrilateral

 

[donau]

... It is mathematically impossible for the two sides of the right angle to equal 5X13 square units (as depicted in the diagram) and produce that green triangle if the pieces stay true to the grid.....

Please remember that the solution we came up with showed that the 5x13 composite figure is not a triangle at all. What appears to be the hypotenuse is really two lines forming an angle, and the extra space "freed up" at the bottom is the equivalent of the space taken by inversing of that angle. Our solution also suggested that rather than forcing the angles of the two triangular looking objects to be the same by making one of them a quadrilateral, there appears to be some slight changes in the spacing of the grid to complete the optical deception.

 

[SDNR]

Yes I do remember that donau -- and I do believe warot was right and found the correct solution to the problem. I was kind of hoping it wasn't just an optical illusion -- but could be made to work somehow.

I will finally concede ....it is the only logical answer to the puzzle -- but I hate the deception, I really wanted it to be a real triangle


I will delete my previous posts to prevent further confusion.

 

[donau]

Here's some simple arithmetic to show that the shift of the angle actually equates to 1 square:

Red triangle: 3 x 8 / 2 = 12
Green Triangle: 2 x 5 /2 = 5
yellowish L: 5 + 2 = 7
Green L: 5 + 3 = 8
-----------------------------
Total: 32

If the composite object would be a triangle, the size of it would be:

5 x 13 / 2 = 32,5

So, there is a difference of a space that equals half the size of one grid.
When the red and the green triangles atre swopped, the angel inverts itself and you have the half a space on the other side of the imaginary hypotenuse of the composite picture. Adding the 2 halves together shows us the freed space for one grid block.

I apologize for this overly simple explanation, but since we are beating this thing to death I thought it might not be a bad idea to do this to show how that 1 grid appears to be freed up.

 

[donau]

Yes I do remember that donau -- and I do believe warot was right and found the correct solution to the problem. I was kind of hoping it wasn't just an optical illusion -- but could be made to work somehow.

I will finally concede ....it is the only logical answer to the puzzle -- but I hate the deception, I really wanted it to be a real triangle


I will delete my previous posts to prevent further confusion.

Sorry Rob, I was busy with my previous post and did not see this until after that post. No need to concede anything, I think that was probably the intent of whoever created this puzzle to get people thinking and that's what's important. Hope you don't delete your posts, they just show some cognitive thinking which in itself is impressive. And after all, this kind of things are always just for fun. And that's what this was. Great fun