Once upon a time there was a triangle …

There is an infinity of tales on triangles. Or circles. Or conics. Some of them, though well known, I will try to tell once more, changing a word here and there.

- Center of gravity
- Bisector of an angle of a triangle
- ABC - 2250 years old problem
- Tangent and Ellipse
- Square

As far as the center of gravity is concerned

The calculation is:

One can also realize it without really calculating, perhaps just counting
the number of equal segments:

The bisector of an angle between two sides of a triangle divides the third side into segments proportional to the sides that meet them in the vertices of the triangle.

Perhaps you prefer another proof? If you produce a paper model with
two copies of the triangle on the left glued one on top of the other
along the bisector - and make a necessary incision in the lower part
of the bisector than you can make a visual demonstration performing
two consecutive paper foldings.

Finding the midpoint of an interval is *the*
most elementary construction, right?

- Setting the problem
- How to do it if the interval were broken at some point?

- Solving it
- Use the broken line to make a triangle, circumscribe a circle on it.
- Bisect the base, go up to meet the circle.
- Project the meeting point onto the broken line (
*broken chord*). You've found the midpoint.

- Why does it work?
- If the chord were rectified...
- Check the angles: the new vertex and the
ones at the base lie on a circle with the center
**C**. - The projection line passes through the center of the new circle,
so it halves its chord
**PR'**.

*Archimedes' Broken Chord*? Well, where does this name come from?

All this story is told in *one* picture on the
very first page of **Episodes in Nineteenth and Twentieth Centure Euclidean
Geometry** by Ross Honsberger (AMS, NML, Washington 1995) - what is the
point in repeating it?

Well, it is just seven points, not a point. Bytes are cheaper than paper and I can afford to make a story somewhat longer. I get confused when the same sketch transmits many different arguments at the same time.

Given a point on the ellipse, how to trace a tangent line?
Well, perhaps we had better ask first: what does it mean?
No derivatives are permitted; the same goes for coordinates,
we are talking about geometry. And for many centuries before
Descartes and Newton human beings managed to work with lines
tangent to ellipses. Was it terribly difficult?

Perhaps not. Just take a look at four pictures below. The first of them introduces the notions. Two others show how to construct the tangent. The last one should convince you that the constructed line catches the ellipse in only one point.

The exercise appears in our (Waldir Quand & A.S.)
collection (ex.15)
and we thought that we had invented it. Well, what I mean is that
we did not borrow it from any other collection. One has to calculate
the area of the red square:
**a**^{2} = [(1-m)^{2}] _{/} (m^{2}+1)
is ready.

A part of its charm stems from the fact that there are really many ways to approach it. I used to suggest to my students that this solution

asked for a minimum of calculations: as **x=ym**, you use the
Thales theorem to get **a=y(1-m)**, then Pythagorean theorem gives
you **y ^{2}(m^{2}+1)=1** and the final result

One day I realized that I have already seen a nice solution for the specific
case of **m=1/2**; it appeared in one of the first issues of *Revista
do Professor de Matemática* in the early eighties. The solution was
attributed to Raul Neder Porrelli:

Exchanging light-blue triangles for dark-blue ones you see that there are 5 small squares in the big one.

Some more years passes and something led me to apply to myself the advice
that I always offer to my students (even if they do not ask for it): revisit
the old solutions, ask yourself why it worked. So I thought: the formula for
**a** looks like a tangent of a difference of some angles, why not to try
it once more - after all I know the angle if the value of the function
**tan** is 1 … Yes, the obsevation resulted in another, much
shorter, proof: if you name the angle and give some cosmetic touches

the you get something that is easier to visualize

and the new proof is ready to enter:

Obviously, I liked and I was ready to say it good-bye but then unexpectedly I met two articles on the Internet dealing with the very square: this one and one more. What a surprise … perhaps it is too early to say goodbye to my little friend?

My friend writes me:

If you discard the prolongation of half-lines that stick beyond the small square then you get the figure of Bhaskara's proof of Pythagorean theorem. (Sure. Shame on us not to note it.) By the way, the square appears also in nature.It does in fact - and the internal divisions are nicely used to point to different directories of the site:

In order to have an encounter with the most recent offspring of this
family of figures have a look at *Crux Mathematicorum with Mathematical
Mayhem, vol.27 (6, October 2001), p.391*. The Mayhem staff proposes
that the area of the small square is 1/2001 and asks what is **m**.

Moving in the other direction in time one finds the article “A Round-Up
of Square Problem” by Duane Detemple and Sonia Harold in *Mathematics
Magazine, vol.69 (1, February 1996), p.15-27*; in its turn it directs the
reader to the article “Dörrie Tiles and Related Miniatures” by
Edward Kitchen in *Mathematics Magazine, vol.67 (2, April 1994),
p.128-130*; the case of **m=1/2** solved just as I have found it in
*RPM* stands there - but attributed to Heinrich Dörrie; the reference
sends to his book *Mathematische Miniaturen* published in Breslau in
1943.