Re: Quite complex problem from the euclidean geometry

On Fri, 28 Sep 2007 12:37:30 +0200, "Philippe 92"
<nospam@xxxxxxxxxxxx> wrote:

(supersedes <mn.e2db7d79f70b10a4.22155@xxxxxxx>)
superseded because of typo error.

As Angus noted, let R the perpendicular projection of X on BC.
Then a well known theorem : Simson line, see for instance
http://mathworld.wolfram.com/SimsonLine.html

says that X on circumcircle if and only if (iff) PQR in line.
(for a given X anywhere on the plane).
Here X is by definition on circumcircle, hence PQR are in line,
for any triangle.
The classic proof of this uses _oriented line angles_ to deal
with "different locations of points" and answer to Angus' fears.

(Some day I'd like to learn Euclidean geometry properly.
Is Hilbert's presentation in the public domain, now? I
did once order a printed copy, but it was out of stock.)

[...]
Hence for any triangle, intersection points of PQ with circle (DX)
are at least this point R.

(Because XRD is a right angle.)

Angus method is to find angle OR with PQR, O midpoint of DX.
I won't write down the full proof, just a guide line.
To answer Angus' "how to deal with ASCII", figure at
http://i22.tinypic.com/35cla4z.gif

(Neat. What software did you use to generate this?
I've been making do with pencil, paper and a rather
inflexible pair of compasses - because my best pair
of compasses seems to be permanently out on loan.)

Let expand a few properties of Simson line PQR.
Let (d) the parallel to BC from the circumcenter I.
The symetric Y of X from (d) is on the circumcircle.
XY being perpendicular to BC, R is on XY, and the Simson line
PQR is parallel to AY (classical).

Here is a proof of this (I think), but it's rather messy
and unenlightening:

Let angle YAQ = d (for Greek delta).

(I'm sorry about the clash with your "(d)", but I'll
get fatally confused if I try to change notation now.)

It is enough to prove that AQP = d (by alternate angle
criterion for parallel lines).

Once again we use the fact that APXQ is cyclic (two
of its opposite angles are right angles). From this
it follows that the angles AQP and AXP are equal.

So it is enough to prove that angle AXP = d.

Because APX is a right angle, it is enough to prove
that angle PAX = 90 - d.

We do this by looking at the triangle ABX:

(As well as joining B to X by a straight line segment,
also join B to Y, and P to Y.)

(Unfortunately once again I'm working from a diagram
which differs from my two previous diagrams and from
yours! I'm sorry, but I will have to sort out this
difficulty later, because I must type out the proof
before I've forgotten it - at the moment it's just a
diagram, with no text.)

We calculate BAX = 180 - ABX - AXB. The calculation
is fairly ugly, so I hope that someone comes up with
a better proof. (You say the result is "classical",
so a reference would do.)

Of course, BAX and PAX are the same angle.

Let angle ABY = e (for epsilon). Angles subtended by
the same arc are equal, so also angle AXY = e.

Let CAX = g (for gamma). Then (angles subtended by
same arc again - I won't keep mentioning this) also
angle CBX = g.

Since BRX is a right angle, angle BXR = 90 - g. Hence
angle AXB = 90 - g - e.

To calculate angle BAX from the triangle ABX, we now
need to know the angle ABX.

By the usual argument (definition of d = angle YAQ,
which is the same as angle YAC; and two angles both
subtended by arc YC), angle YBC = d.

So we have:

angle ABX = angle ABY + angle YBC + angle CBX
= e + d + g

therefore:

angle BAX = 180 - angle ABX - angle AXB
= 180 - (e + d + g) - (90 - g - e)
= 90 - d

as required.

End of proof. (Yes, it's ugly. I apologise. I only hope
it isn't wrong as well as ugly.)

Let A' be the symetric of A from (d), A' on the circumcircle.
Hence AX and A'Y intersect at F on (d).
YFX is then an isosceles triangle similar to ROX, hence AY is

superseded : should be A'Y of course !

parallel to OR and angle QRO = angle AYA' = constant.

This angle will be a right angle iff AA' is a diameter.
That is iff altitude of A goes through circumcenter I, that is
iff altitude is perpendicular bisector of BC, that is iff AB=AC.

Beautiful (except for my contributions!). The problem
makes sense now. However, the introduction of Y and A'
seems rather magical, i.e., unmotivated. I suppose it
requires expertise in (what they used to call) "modern
geometry"?
--
Angus Rodgers
(twirlip@ eats spam; reply to angusrod@)
Contains mild peril
.

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