Here is a link to a geometry puzzle "http://www.boingboing.net/2009/03/05/tough-geometric-brai.html" Here is a link to my explanation "http://zach.d.cox.googlepages.com/How_Can_This_Be_True_New.pdf". I've seen this puzzle before and even figured it out, but I think writing down a solution in, what I hope is, a simple manner is a good thing to do.
Enjoy!
Hi Zach,
ReplyDeleteI'm afraid that solution is not correct.
The puzzle still works if you make all the shapes the exact size they appear to be: the yellow shape (blue in your re-coloring) is exactly 5 units long, not 5-and-a-bit; the dark-green triangle is exactly 5 units long, not 5.2 as you calculated; the red triangle is exactly 8 units long, not 7.8 as you calculated.
By re-drawing the shape with your new dimensions and overlapping some of them, you are ignoring the evidence right there in the original puzzle. It is clear in the original puzzle that no pieces are overlapped. If you doubt this, use your favorite image drawing program to cut up the pieces and move them around. Or do it with paper.
The problem stems from the fact that you are assuming the big shape is a perfect triangle. In fact, you state this as an assumption in your solution: "...and the hypotenuse of this one is a straight line through the origin."
Once you start from this assumption, then you find that the shape cannot work, and therefore you have to have all these sizes that are slightly off, and overlapping shapes. In effect, you've proven that the shape is impossible.
But when you come across an impossibility, the best thing is to question your assumptions. In this case the hypotenuse is not a straight line.
If you calculate the angle of the top corner of the dark green triangle, you will see that it is not the same as the angle of the top corner of the red triangle. This means that the "hypotenuse" cannot be a straight line. Instead, in the first picture, the hypotenuse "dents in" slightly. In the second picture it "dents out" slightly. (The shapes are therefore not triangles at all: the first has four sides, the second has seven sides, if you count the ones in the missing piece.)
That is where the extra space comes from.
Hi Sam;
ReplyDeleteYou are correct ... in my solution I assume the large figure is a triangle, and given that assumption then all the rest follows. The solution is not only correct but the only solution possible (given the assumption.)
Assuming otherwise (that the large figure is NOT a triangle), as you suggest, leads to a different solution.
And given your assumption the white square is, in your solution, still a square instead of being reduced to a rectangle as in the solution I give.
I invite you to write up your solution and I'll link to it.
Here is another problem like the one above --- this time that author goes further and draws parallels between the puzzle and the Fibonacci numbers.
ReplyDeleteI did prove the Fibonacci formula that [f(n)]^2 = (-1)^n+1 + f(n-1)xf(n+1)
Here is the link to the new puzzle and I'll post my proof once I get it into TeX:
http://mathlesstraveled.com/2011/05/02/an-area-paradox/