## Re: [EMHL] Congratulations for 10th Anniversary

Expand Messages
• Dear friends of Hyacinthos Congratulations for 10 years. Ricardo Barroso [Non-text portions of this message have been removed]
Message 1 of 8 , Dec 22, 2009
Dear friends of Hyacinthos
Congratulations for 10 years.

Ricardo Barroso

[Non-text portions of this message have been removed]
• Dear Friends Thank you for your messages on Hyacinthos 10th Anniversary. Wishing you a Merry Christmas and a Happy and Heamthy New Year Antreas [Non-text
Message 2 of 8 , Dec 22, 2009
Dear Friends

Thank you for your messages on Hyacinthos 10th Anniversary.

Wishing you a Merry Christmas and a Happy and Heamthy New Year

Antreas

[Non-text portions of this message have been removed]
• Dear Hyacinthians, ... I hope that everyone noted that H_A can be replaced by any point on BC. So for any point P on BC the quad formed by O, A and the
Message 3 of 8 , Dec 28, 2009
Dear Hyacinthians,

I wrote:
> Here is a little anniversary-observation:
> H_A be the perpendicular foot of A on BC. Let X_B and X_C be the perpendicular feet of H_A on AC and AB (yes, well known for the Taylor circle). The polygons A X_B O X_C and B X_C O X_B C have equal area (O: circumcenter).

I hope that everyone noted that H_A can be replaced by any point on BC. So for any point P on BC the quad formed by O, A and the perpendicular feet of P on AC and AB has area half of triangle ABC.

Kind regards,
Floor.
• Dear Floor, very good problem. I had not noticed the generalization. Let the projections of P on AB, AC be D, E AF be the altitude of ABC and AF be the
Message 4 of 8 , Dec 28, 2009
Dear Floor,
very good problem.
I had not noticed the generalization.
Let the projections of P on AB, AC be D, E
AF be the altitude of ABC and
AF' be the altitude of AED.
then the line AP is the diameter of
AEPD and is isogonal conjugate with AF'
and AO is isogonal conjugate with AF.
Hence for angles
(AO,AF') = (AP, AF) or
(AO,ED) = (AP,BC) = f
Hence for areas
(ABC) = (1/2).BC.AP.sinf = R.sinA.AP.sinf =
= R.ED.sinf = 2.(1/2).OA.ED.sinf = (AEOD)

Best regards

[FvL]
> I wrote:
> > Here is a little anniversary-observation:
> > H_A be the perpendicular foot of A on BC. Let X_B and
> X_C be the perpendicular feet of H_A on AC and AB (yes, well
> known for the Taylor circle). The polygons A X_B O X_C and B
> X_C O X_B C have equal area (O: circumcenter).
>
> I hope that everyone noted that H_A can be replaced by any
> point on BC. So for any point P on BC the quad formed by O,
> A and the perpendicular feet of P on AC and AB has area half
> of triangle ABC.
>
> Kind regards,
> Floor.
>
>
>
> ------------------------------------
>
>
>
>     Hyacinthos-fullfeatured@yahoogroups.com
>
>
>

___________________________________________________________
Χρησιμοποιείτε Yahoo!;
Βαρεθήκατε τα ενοχλητικά μηνύματα (spam); Το Yahoo! Mail
διαθέτει την καλύτερη δυνατή προστασία κατά των ενοχλητικών
• Sorry for a typo ... the correct is Hence for areas (ABC) = (1/2).BC.AP.sinf = R.sinA.AP.sinf = = R.ED.sinf = 2.(1/2).OA.ED.sinf = 2.(AEOD) and (AEOD) =
Message 5 of 8 , Dec 29, 2009
Sorry for a typo

> Hence for areas
> (ABC) = (1/2).BC.AP.sinf = R.sinA.AP.sinf =
> = R.ED.sinf = 2.(1/2).OA.ED.sinf = (AEOD)

the correct is

Hence for areas
(ABC) = (1/2).BC.AP.sinf = R.sinA.AP.sinf =
= R.ED.sinf = 2.(1/2).OA.ED.sinf = 2.(AEOD)
and
(AEOD) = (1/2)ABC

Best regards

___________________________________________________________
Χρησιμοποιείτε Yahoo!;
Βαρεθήκατε τα ενοχλητικά μηνύματα (spam); Το Yahoo! Mail
διαθέτει την καλύτερη δυνατή προστασία κατά των ενοχλητικών
• And we can replace the point O by any other point O on the line AO and the (signed) area of the polygon A P_B O P_C is still constant. Happy New Year and
Message 6 of 8 , Dec 31, 2009
And we can replace the point O by any other point O' on the line AO and the
(signed) area of the polygon A P_B O' P_C is still constant.
Happy New Year and cheer Antreas.
Francois
PS
Line P_BP_C envelope a parabola tangent to the sides AB and AC and to the
altitudes from B and C, line AO is the asymptotic direction.
It's just a genral property of the tangents to a parabola.

> Sorry for a typo
>
> > Hence for areas
> > (ABC) = (1/2).BC.AP.sinf = R.sinA.AP.sinf =
> > = R.ED.sinf = 2.(1/2).OA.ED.sinf = (AEOD)
>
> the correct is
>
> Hence for areas
> (ABC) = (1/2).BC.AP.sinf = R.sinA.AP.sinf =
> = R.ED.sinf = 2.(1/2).OA.ED.sinf = 2.(AEOD)
> and
> (AEOD) = (1/2)ABC
>
>
> Best regards
>
>
>
>
>
> ___________________________________________________________
> �������������� Yahoo!;
> ���������� �� ���������� �������� (spam); �� Yahoo! Mail
> �������� ��� �������� ������ ��������� ���� ��� �����������
>
>
>
> ------------------------------------
>
>
>
>
>

[Non-text portions of this message have been removed]
• I forget to say that H_A is of course the focus of the parabola. Francois 2009/12/31 Francois Rideau ... [Non-text portions of
Message 7 of 8 , Dec 31, 2009
I forget to say that H_A is of course the focus of the parabola.
Francois

2009/12/31 Francois Rideau <francois.rideau@...>

> And we can replace the point O by any other point O' on the line AO and the
> (signed) area of the polygon A P_B O' P_C is still constant.
> Happy New Year and cheer Antreas.
> Francois
> PS
> Line P_BP_C envelope a parabola tangent to the sides AB and AC and to the
> altitudes from B and C, line AO is the asymptotic direction.
> It's just a genral property of the tangents to a parabola.
>
>
> Sorry for a typo
>>
>> > Hence for areas
>> > (ABC) = (1/2).BC.AP.sinf = R.sinA.AP.sinf =
>> > = R.ED.sinf = 2.(1/2).OA.ED.sinf = (AEOD)
>>
>> the correct is
>>
>> Hence for areas
>> (ABC) = (1/2).BC.AP.sinf = R.sinA.AP.sinf =
>> = R.ED.sinf = 2.(1/2).OA.ED.sinf = 2.(AEOD)
>> and
>> (AEOD) = (1/2)ABC
>>
>>
>> Best regards
>>
>>
>>
>>
>>
>> ___________________________________________________________
>> �������������� Yahoo!;
>> ���������� �� ���������� �������� (spam); �� Yahoo! Mail
>> �������� ��� �������� ������ ��������� ���� ��� �����������
>>
>>
>>
>> ------------------------------------
>>