socratic method
FoR THiS MoNTH’S Puzzle, we travel back in time 2,401 years.
in the toe of italy’s “boot,” a young seeker names Socrates (no, not that
Socrates) who is seeking admission into the cult of Pythagorus is being interviewed by the head of the school, Philomath. As we join them,
Philomath has just asked Socrates why he wishes entry into the mysteries.
“Well,” says Socrates, “I’ve always been
into triangles, right? And Pythagorus
came up with this theorem, I hear.”
“A squared B, something something,”
Socrates eagerly answers. “You can con-
struct an infinite number of them, even
of a particular area, by subdividing the
continuum into segments of every con-
ceivable length.”
“You forget, Socrates, that we are Py-
thagoreans, and so we cannot conceive
of any irrational number,” Philomath
scolds. “It’s a holdover from the old
man’s musical obsessions. You know—
thirds, fifths, etc. But the square root of
two is right out.”
Socrates sits in thought for a mo-
ment. Then he asks, “So it’s probably
tricky, puzzling even, figuring out how
to construct two right triangles with in-
teger sides but the same area.”
“Right you are, Socrates,” says
Philomath. “But if you manage to find
four right triangles with integer sides
of the same area, we’ll let you in.”
Of course, having more tools at your
disposal than Socrates, you just might find
a shockingly large number of quadruples
of triangles. To keep it simple, I would
like the four smallest such triangles.
Previous Issue’s Puzzle
learning from the Past
One of my favorite aspects of writing this
column is telling some crazy (although
possibly true) story that sets up the puz-
zle. The puzzle itself rarely takes more
than a paragraph, but I have been known
to throw together pages of introductory
material (Editor’s note: This is true). But
since I’m running way behind deadline
this month (Editor’s note: This is also
true), I’m going to go meta on you and
talk about how I come up with puzzles.
When I took on the task of creat-
ing Contingencies puzzles, I wasn’t
sure I’d be able to come up with
that many new ones. But I saw
a presentation by Scott Kim
(one of the great puzzle
and game creators of
Solutions may be e-mailed
to the author at cont.puzzles@
gmail.com. In order to make the
solver list, your solutions must by
received by May 31, 2011.
our age) that showed me how to do it.
One of his best tricks is to start with an
existing puzzle, and vary the parameters
or extend the rules somehow. Sometimes
you create something unsolvable, but
you also have a good chance of creating
something new and cool.
I have a fairly extensive collection
of math puzzle books and every other
month or so—generally, but not always,
just before my deadline—I sit down with
a stack of them and find one that strikes
my fancy. For this month’s puzzle, for example, I found my inspiration in puzzle
No. 81 in H.E. Dudeney’s Amusements in
Mathematics, originally published in 1917.
The observation that starts puzzle No.
81 is that you can take the digits 1 through
9, and arrange them into two multiplication problems with the same product.
The example Dudeney gives is 158* 23 =
79* 46= 3,634. The puzzle that he presents
is to find other arrangement (or arrangements) of the digits 1 through 9 that make
two equal products like this ( 3 digit times
2 digit equals 2 digit times 2 digit) in such
a way as to maximize the product.
It’s not a bad puzzle, but I considered
the problem too easy (to tell the truth, I
nearly always consider the puzzles in my
various books too easy, because none of
them were written for actuaries). So I set
myself the task of making it a bit harder.
I recently learned that
when you count in base 12,
the numbering system goes
1, 2, 3, …, 8, 9, T, E, 10,
11,…. (my source
