HYACINTHOS 6538

In message #6469, I wrote:

>> 1. (Floor van Lamoen in message #4547)
>>
>> Let A'B'C' be the reflections of ABC through a
>> point D, then the circumcircles of AB'C', A'BC'
>> and A'B'C concur in a point.
>>
>> 3. (Reduced to elementary version of 2.)
>>
>> The point of concurrence lies on the circumcircle
>> of triangle ABC.

And I asked if there are any synthetic proofs. Now I have found a
proof easier than I could ever imagine:

PROOF.

Let T be the intersection of circles AB'C' and A'BC' different from
C'. Then, we calculate angles:

ATB = ATC' + BTC'
= AB'C' + BA'C' (concyclic)
= AB'A' - A'B'C' + BA'B' - C'A'B'
= (AB'A' + BA'B') - (A'B'C' + C'A'B')
= 180° - (A'B'C' + C'A'B')
(since AB'A'B parallelogram)
= 180° - (ABC + CAB)
(since triangles ABC and A'B'C' congruent)
= ACB.

Therefore, T lies on the circumcircle of ABC. In other words: The
intersection of circles AB'C' and ABC different from A lies on the
circle A'BC'. Analogously, we can prove that the intersection of
circles AB'C' and ABC different from A lies on the circle A'B'C. This
proves 1. and 3..

A-----
/\ -----
/ \ -----
/ \ ------B'
/ \ /
C' / \ /
/ \ /
/ \ /
/ D \ / T
/ \
/ / \
/ / \
/ / \
B/-------------------------/----------\C
----- /
----- /
----- /
------A'

>> 4. (J. R. Musselman, see message #1026)
>>
>> If D is the nine-point center of triangle ABC,
>> then the point of concurrence is the Feuerbach
>> point of the tangential triangle of ABC.

This is not so easy, of course. I see that A', B', C' are the
reflections of the circumcenter O of ABC in BC, CA, AB. But I see
that this is somewhere in Lev and Tatiana Emelyanov's "mysterious
points", but in a quite different form.

What is more interesting: a variation of 4..

5. If D is the symmedian point of triangle ABC, then the point of
concurrence is the Feuerbach point of the tangential triangle of ABC,
too.

Here is a list of the points of concurrence:

|-------|------------------------------------------------|
| D | point of concurrence of the circles AB'C', |
| | A'BC', A'B'C, ABC |
|-------|------------------------------------------------|
| X(1) | X(100) = anticomplement of Feuerbach point = |
| | Feuerbach point of anticomplemetary triangle |
| X(2) | X(98) = Tarry point |
| X(3) | undefined (point jumping on circumcircle) |
| X(4) | X(107) |
| X(5) | X(110) = focus of Kiepert parabola = Feuerbach |
| | point of tangential triangle |
| X(6) | X(110) = focus of Kiepert parabola = Feuerbach |
| | point of tangential triangle |
| X(7) | not in ETC |
| X(8) | not in ETC |
| X(9) | X(100) = anticomplement of Feuerbach point = |
| | Feuerbach point of anticomplemetary triangle |
| X(10) | not in ETC |
| X(11) | X(104) = antipode of X(100); lies on ON, where |
| | O circumcenter and N Nagel point; lies on HF, |
| | where H orthocenter and F Feuerbach point. |
| X(12) | not in ETC |
| X(20) | not in ETC |
| X(21) | not in ETC |
|-------|------------------------------------------------|

Well, it's time to find a general formula!

Darij Grinberg