Re: [Phys-L] area of the shaded triangle

[John Denker <jsd@av8n.com>]

On 04/29/2018 02:19 PM, Scott Orshan via Phys-l wrote:

> just looking at it, it seems as if the area will go from 1/2 to 1/4 
> of the area, as the slanted line sweeps from the upper left corner
> to the upper right corner.

Yes indeed.
That illustrates a pair of ideas that Pólya emphasized:
  Sometimes given a special case, you can do better by
  solving the general case.

  Also the reverse:  When asked about the general case,
  sometimes it helps to do a few special cases as
  warm-up exercises.

> That's a good way to check the general answer, as it must work in the
> corners and at the midpoint.

Yes, it's a valuable check ... but wait, there's more.
If you're sufficiently crazy, you can solve the whole
problem from scratch that way:

If all you need is a quick estimate, the two points
mentioned above are all you need.  You know absolutely
that the answer is somewhere between 1/4 and 1/2.

Next, consider the area as a function of x.  It can't
possibly be a very complicated function;  there just
aren't enough moving parts to implement anything
complicated.  So we can try equation-hunting.  We
can easily evaluate the area at two points, namely
x=0 and x=1.  What's more interesting is that the
area goes to infinity when x is -1.  This requires
a bit of "outside the box" thinking -- literally --
but not very much.  As Forman Acton said,
    "no polynomial ever had an asymptote,
    horizontal or vertical"
so we can try looking for a rational function.  The
point at x=-1 tells us the denominator has to be
something like 1+x.  The point at x=0 is consistent
with setting the numerator to ½.  The point at x=1
serves as a check.  At this point it we are pretty
sure we know the answer.  For large x geometry tells
us that the area is 0.5/x plus terms of order (1/x)^2
which provides yet another check.  At this point I'm
convinced.  Consider large negative x if you want
yet another check.

This involves a lot of steps, but each step is
very simple.  The whole job is over in less time
than it takes to tell about it.

If this were a timed test, at this point I would
write down the answer and move on.  Quick and
dirty.  Very quick, and not very dirty.

Moral of the story:  There's nothing wrong with
equation hunting *provided* you check the result
to make sure it is correct.  Some of the most
profound results in physics were equation-hunted,
e.g. the Planck radiation formula.  Also beware
that most real-world situations are more complicated
than this, which makes equation-hunting verrrrry
laborious.

> When I first saw the problem, it said that the outer box was a square

The fact that it doesn't "need" to be a square is
an interesting insight.  Y'all know I just love
scaling arguments.  It turns out that you can
treat the figure as a rectangle, and the answer
i.e. the fractional area is independent of the
height H and the width W.  The area of the rectangle
scales like H^1 and W^1, and the area of the shaded
triangle scales the same.  So the desired ratio
scales like H^0 and W^0.

But wait, there's more:  It doesn't even have to
be a rectangle.  A parallelogram will do.  This
affords even more insight.

Shear the diagram so that the top moves half a
unit relative to the base:

	----====----====
	\       |        \
	 \      |         \
	  \     |          \
	   \    |           \
	    \   |            \
	     \  |             \
	      \ |              \
	       \|               \
	        ----====----====


It's now obvious that the height of the triangle
is the solution to a linear equation.  In fact
h=2/3 H.  Now we know the height and we know the
base.  Game over.

The math/physics lesson here is that areas are
invariant with respect to certain transformations,
including shears.

Invariances of various sorts are a valuable part
of the theoretical physicist's stock in trade.

Another moral is that visualization is important.
By shearing the diagram you can make the answer
easy to visualize, leaving no doubt that it is
correct.

> I had to draw a lot of pictures and go down a few different paths,
> before the solution became apparent.
Yeah.  It's important to be honest about that.
Otherwise students get discouraged.  They need
to know that it's OK to struggle with things.

The "shear" solution is not the first thing
that occurred to me.  More like the twelfth.
OTOH it is the one I will remember.