Subject: Euclidean Math? Margaret Matthews; "Puzzle level", (Check out this web-site: Simeon's Triangle Puzzle ) I have tried to figure out how this could be, because everything I know about it tells me it can't be. However, I can't seem to make it NOT work. Two right angle triangles. They are each cut up into four identical pieces. In the first, all the pieces fit together so that there are NO empty spaces; in the second, presumed to be identical in size to the first, the pieces are slightly rearranged, and now, there IS a space in the triangle. Thanks again for your help. Margaret Hi Margaret, I have two replies to your question: First from Patrick As with a lot of "Math Magic" puzzles, the trick lies in the unstated assumptions of the puzzle. Anybody who first sees the two shapes assumes that they are triangles, when they are in fact not. To convince yourself of that, look at the slope of the would-be "hypotenuse". The larger triangle has a slope of rise/run = 8/3 ( = 2.66666....). The smaller triangle has a slope of 5/2 = 2.5. A small difference, not enough to notice it at first sight, but enough for the triangle to "bend" inward. When the two triangles are interchanged, the shape "bends" outward, making just enough room in the inside for the hole to appear. (Try putting a flexible plastic ruler of your computer screen along the hypotenuse.) To fully understand how the puzzle works, calculate the area of the the four pieces seperately (you should get 32 square units) and the area of the "assumed" triangle (i.e. base 13, height 5, which yields an area of 32.5 square units). There is half a square missing. Switching from "bending inward" to "bending outward", we gain two of these half squares, which exactly equals the area of the newly created hole. And then from Claude. The small green triangle has sides 2 and 5 for a base angle of about arctan(2/5) = 21.801 degrees. The small red triangle has sides 3 and 8 for a base angle of about arctan(3/8) = 20.556 degrees. These measures are pretty close, and the base angle of a triangle with sides 5 and 13 is in between: arctan(5/13) = 21.038 degrees. That is where the illusion is: The big mosaic looks like a triangle, but it is not a triangle, because the "hypotenuse" is not a straight line. In the first drawing, the "hypotenuse" goes up at 20.556 degrees and then breaks to go steeper at 21.801 degrees, and in the second, it first goes up at 21.802 degrees and then caves in at 20.556 degrees. If you look closely at the drawings, you see the notch in the first one and the bump in the second. This accounts for the extra square: The first mosaic has an area of (3*8)/2 + (2*5)/2 + 3*5 = 32 while a triangle with base 13 and height 5 would have an area of (13*5)/2 = 32.5. The missing half unit is the space between a straight line and the caved in "hypotenuse" that you see. In the second mosaic, there is an extra half unit above the hypotenuse. Together they amount to an extra unit (the white square). Here is a similar puzzle that we were sent a few years ago. Cheers, Harley Go to Math Central