## Sunday, May 07, 2006

### Geometry Puzzle

Does anyone out there know how to solve this puzzle? My friends and I are completely stumped by it.

Both the triangles are exactly the same size. The four portions are exactly the same size. How is it then, that when they are moved around, there is an extra square "hole" that appears?

Macht said...

It happens because the hypotenuse (in both pictures) isn't a straight line. Shape B goes horizontally 5 squares and vertically 2 squares while shape A goes horizontally 8 squares and vertically 3 squares. In order for the hypotenuse to be straight, they would have to have the same ratios. So in the top picture, the hypotenuse caves inwards while in the bottom picture, the hypotenuse caves outwards.

1:47 PM
Keith said...

macht,

You are my new hero.

5:22 PM
Anonymous said...

Another way to explain it is that when you move C down so that its top row is now in line with the top row of D, it allows for one box to be unfilled. It does allow the triangle A and B to switch places, but via the previous post, that does not mean much.

6:12 AM
AmericanPascal said...

This is a good example of self deception. On the surface we are inclined to make assumptions that we don’t even realize are being made – the picture (graph in this case) says a thousand words. But we are conditioned to read certain words or things into any picture we look at. However, when we start to analyze and to question our position, we often find the errors in our views. This is such a great example since the “obvious” missing square was in fact non-existent; and the explanation revealed the true error in our perception. This explains many things… if we could only see beyond the obvious.

7:37 PM
Anonymous said...

aNOTHER way to discover the problem is to realize that triangle "A" and triangle "B" can't possibly be similar triangles: one is 8x3, the other is 5x2. Since the 8/3 is not the same as 5/2, the acute angles must be different.

Basically, the "big triangle" made up of the four shapes combined is NOT a real triangle in either configuration.

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Anonymous said...

Does anyone know the answer to this equation?

Three straight lines joined together so that they are rotated symmetric, four straight lines of three different lengths joined together so that they are vertically symmetric and then repeated later on, a semi-circle repeated later on, three straight lines joined together so that they are vertically symmetric, two straight lines joined
together to form a right-angle and three straight lines joined together so that they are laterally symmetric.

10:30 PM
Midnight Essays said...

Another method to clarify it is that when you move C down so that its upper row is now in streak with the highest row of D, it permits for one rectangle to be empty. It does permit the threesome A and B to change spaces, but via the preceding post, that does not nasty much.paying someone to write a paper

10:41 PM
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Three straight lines joined together with the goal that they are turned symmetric, four straight lines of three distinct lengths joined together so they are vertically symmetric and after that rehashed later on, a semi-circle rehashed later on, three straight lines joined together so they are vertically symmetric, two straight lines joined together to shape a right-point and three straight lines joined together so they are along the side symmetric.
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