Where'd the Square Go...
Check the slope of the lines.
There is an angle point in the hypotenuse after rearrangement.
OZ?
Yep, he cheats by not having the pieces lay flat on the grid. In the first picture the slope of the left triangle is 3/8 = 0.375 but the top triangle is 2/5 = 0.4, and it the second picture they are reversed so the kink goes the other way.
Was that a Sam Loyd puzzle? I recall Martin Gardner writing about him decades ago. I don't recall that he cheated that way, though, in the puzzles I saw.
(What a memory I have! But then I forgot to eat breakfast this morning.)
Before AND after, in opposite directions.
@bill93:?ÿ I have a really good memory too.?ÿ It's just short.
I am famous for both sides of my memory.?ÿ On the one side I remember minute details from decades ago.?ÿ On the other I can't remember your name though we were classmates in school for six years.?ÿ It will come to me, but embarrassingly late.
My Grandmother could remember the slightest detail of something that happened many many years ago but could not remember how many time she had told that story to the same person and that same day.
I'm getting closer to that every time I drive by some place that I surveyed. "Yeah, we surveyed that place back in 1987.?ÿ Mid-March as I recall.?ÿ It rained all day but we stuck with it as the job had to be delivered two days later."
I can remember details about surveys I did decades ago.?ÿ I could not remember my wedding anniversary.?ÿ I'm happily divorced and happily still surveying.
Looks like a teaching aid to demonstrate small differences in angles.
The four figures do not form a triangle in either arrangement. Because they are close you believe him when he says they are triangles, but he is a liar. They first is a quadrilateral, the second is an eight sided figure.
Optical delusion.
Paul in PA
It actually didn't "go" anywhere. The presenter would have you believe that the smaller triangle's base = 5 units and its height = 2 units. But that's not true. While the base does equal 5, the height is 1.923076923 units, to 10 digits. The large triangle has base = 8 units, but its height is actually 3.076923077 units, again to 10 significant digits.
The two rectangular areas, where the purportedly missing piece is, actually have identical areas when the heights of the triangles are computed correctly. 1.923076923 * 8 = 3.076923077 * 5 within the limits of precision of a 12-digit calculator.
So, the "missing piece" is explained by a misrepresentation of the heights of the two triangles. The presenter should be made to cut his triangles from inch-thick, six foot by ten foot steel plate so that the rounding difference would show up.
?ÿ
That's a different interpretation from mine, but equally valid since we can't put a micrometer to his cutouts.
Indeed. The differing slopes reveal the fact that the heights of the triangles can't be 2 and 3. And the slopes can't be different if the two pieces are part of the same line.
But, 3.076923077/8 = 1.923076923/5 = 5/13 = 0.3846153846.
?ÿ
I don't see how you're able to call the height of the triangles off.?ÿ You could just as easily make the argument that the height is spot on and the base is the dimension being misrepresented, and defeat the puzzle that way.
The way Bill93 did it exposes the trick quickly and easily, imo.
That's true. You could also say that both the heights and the bases are spot on, but that would make the sum of the two smaller hypotenuses greater than the hypotenuse of the big triangle. Which error the presenter made is not certain. I chose the height because I would have cut out the figures by cutting vertically first, but it could certainly be done by cutting horizontally first.?ÿ
I'm not refuting Bill's analysis nor Dave's analysis. Rather, I'm adding to them. But remember, the question posed was, what happened to the missing square. To answer the question, I think you have to show that both rectangular areas are equal, and, in order to do that, you have to find the dimensions of the two smaller triangles.
There are indeed three possibilities for those dimensions. @paul-in-pa added to the differing slope situation, so that leaves one to be evaluated.
When I taught Geometry and drew freehand diagrams on the whiteboard, I sometimes missed a point with a line that was supposed to go through it. I could just thicken the line and the problem was solved.?ÿ
In this case, line thickness creates a problem that doesn't exist.
?ÿ