Hi! Remember that Sangaku puzzle from a while back? Well, I received a solution from Thomas Weibull, a mathematician working at the Chalmers University of Technology in Sweden. I got it a while ago, but I forgot to post it here until now! Grab a pencil and some paper, here we go:

Let the big circle be C1 with center C, horizontal diameter, left end
point A, right end point B and radius R (so |AB| = 2R).

Let C2 be the circle internally tangent to C1 at A, midpoint D (on
AB), other intersection with AB E, radius r < R (so |AE| = 2r). The
base of the isosceles triangle is EB and |EB| = 2R - 2r. Let F be the
midpoint of EB, so |EF| = |FB| = R - r. Let G be the top vertex of the
isosceles triangle. By the theorem of the chords, or by the fact that
ABG is a right triangle, |FG|^2 = |AF|*|FB| = (R + r)*(R - r).
[This is R^2 - r^2, so there's a right triangle with hypotenuse R and
sides r and |FG|; oh yes, |CF| = r and |CG| = R.]

EFG is a right triangle, so the Pythagorean theorem gives
|EG|^2 = |EF|^2 + |FG|^2 = (R - r)^2 + (R + r)*(R - r) = 2R*(R - r).

Let the circle C3 be determined by
(1) it is internally tangent to C1
(2) it is externally tangent to C2
(3) it is tangent to EG.
We want to prove
(4) the line from its center to E is perpendicular to AB.
Now, conditions (1), (2), and (3) determine C3 uniquely.
However, conditions (1), (2), and (4) also determine a circle
uniquely; let’s call it C4 and let its center be H. Condition (4)
seems a lot easier to work with than condition (3), so that’s what I
will do and complete the task by proving that C4 is identical with C3.

We apply the Pythagorean theorem to the two right triangles DEH and
CEH. Let the radius of C4 be s.
Then, since C4 is tangent to C2, |DH| = r + s.
Since C4 is tangent to C1, |CH| = R - s.
|CE| = |2r - R| (absolute value, we don’t know which is bigger, 2r or
R, but it doesn’t matter since we square it).
EH is a side in both right triangles, so we get two expressions for
its square: |EH|^2 = |DH|^2 - |DE|^2 = |CH|^2 - |CE|^2, so
|EH|^2 = (r + s)^2 - r^2 = (R - s)^2 - (2r - R)^2.
The left equality gives |EH|^2 = s*(2r + s), while the right equality
and some algebra gives s = 2r*(R - r)/(R + r).

Now let I be the point on the right half of C4 such that EI is tangent
to C4. Then EIH is a right triangle and if we can show that EIH is
similar to GFE, we are done, since then angle HEI equals angle EGF,
which means that the line EI coincides with the line EG, so C4 is
indeed tangent to EG, that is, coincides with C3.

Well, both EIH and GFE are right triangles, so it is enough to show
that |HI|/|HE| equals |EF|/|EG| (these quotients are actually the
sines of angles HEI and EGF, respectively, but we don’t need the
language of trigonometry). Now,
(|HI|/|HE|)^2 = s^2/(s*(2r + s)) = s/(2r + s) = (R - r)/2R
(after a little algebra), and
(|EF|/|EG|)^2 = (R - r)^2/(2R*(R - r)) = (R - r)/2R
so we are (finally) done.
[Of course, now you might start wondering why this squared ratio is
equal to |FB|/|AB|, but maybe it's wiser not to go there ...]

I hadn’t seen this before, so it was fun, and it was also a nice twist
to switch from C3 to C4, which agrees with one of Polya’s heuristics:
start with the unknown.

—Thomas Weibull

Thank you, Thomas!

Tags: brush pen, eeen, marker, paper

This entry was posted on Thursday, February 5th, 2009 at 09:00 and is filed under Comic. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.