Re: [HM] 4 triangles in a rectangle

Paul Yiu (Yiu@fau.edu)
Tue, 23 Mar 1999 08:12:16 -0500

Dear Professor Kutler,

I have seen this in a short article by T.M.Apostol in one of the early
volumes of Crux Mathematicorum:

Tom Apostol:
"A property of triangles inscribed in rectangles", Crux Math. 3 (1977)
pp. 242-244.

Related to this is the problem of constructing an equilateral triangle
APQ with P and Q on the sides BC and CD of a given rectangle ABCD.
To do this, one first constructs equilateral triangles BCX and CDY with
X and Y in the interior of the rectangle (*). Extend AX to intersect CD
at Q and AY to intersect at P. Then APQ is equilateral.

(*) This is possible if and only if the ratio k of the sides of the
rectangle lies between (sqrt 3)/2 and 2/(sqrt 3).

Sincerely,

Paul Yiu
Department of Mathematics
Florida Atlantic University
Boca Raton, FL 33431-0991

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From: Samuel S. Kutler[SMTP:s-kutler@sjca.edu]
Sent: Monday, March 22, 1999 8:13 PM
Subject: [HM] 4 triangles in a rectangle

Friends:

I once wrote an article called

Brilliancies involving Equilateral Triangles,

and now I have read a theorem that, had I known it, I would have included
it in the article:

Start with an equilateral triangle and circumscribe a rectangle about it
so that there is one vertex of the triangle in a corner of the rectangle
and the other two vertices are, of course on the other two sides of the
rectangle.

Thus, you have the rectangle tiled with 4 triangles, the equilateral
triangle with which you started and 3 right triangles.

I say that of the three right triangles, the area of the largest is
equal to the sum of the two smaller ones.

I found it on page 19-21 of

MATHEMATICAL GEMS III

by Ross Honsberger. I don't know whether or not he discovered it or
whether he was just passing along a known theorem. Does anyone know its
origin? If not does Ross Honsberger have an e-mail address?

Best wishews,

Sam Kutler