## Re: [EMHL] Altitudes as diameters of circles

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• ... Dear Paul, A general problem that Floor posted says that the circles with diameters three concurrent cevians have always as radical center the orthocenter
Message 1 of 20 , Feb 1, 2002
On Thursday, January 31, 2002, at 01:59 PM, Paul Yiu wrote:

> Dear Antreas and friends,
>
> [APH]: Let ABC be a triangle, and AHa, BHb, CHc its altitudes.
>
> Draw the circles (Ka), (Kb), (Kc) having as diameters the
> altitudes AHa, BHb, CHc, resp. [Ka = midpoint of AHa, etc].
>
> *** Where is the radical center?
>

Dear Paul,

A general problem that Floor posted says that the circles with
diameters three concurrent cevians have always as radical center
the orthocenter of ABC.

Here are some other rad. center problems on the altitudes
configuration:

1. Which is the locus of the radical center of the circles
(Ka, t*h_a), (Kb, t*h_b), (Kc, t*h_c) as t varies?
[For t = 1/2 we have the case of the altitudes as diameters.]

2. Which is the locus of the radical center of the circles
(Ka, t*R), (Kb, t*R), (Kc, t*R) as t varies?

3. Which is the locus of the radical center of the circles
(Ka, t*a), (Kb, t*b), (Kc, t*c) as t varies?

Antreas
• On Thursday, January 31, 2002, at 10:59 AM, Antreas P. Hatzipolakis ... Let A A be the common chord of (Kb), (Kc) [A near to A] B B be the common chord of
Message 2 of 20 , Feb 1, 2002
On Thursday, January 31, 2002, at 10:59 AM, Antreas P. Hatzipolakis
wrote:

> Let ABC be a triangle, and AHa, BHb, CHc its altitudes.
>
> Draw the circles (Ka), (Kb), (Kc) having as diameters the
> altitudes AHa, BHb, CHc, resp. [Ka = midpoint of AHa, etc].

Let
A'A" be the common chord of (Kb), (Kc) [A' near to A]
B'B" be the common chord of (Kc), (Ka) [B' near to B]
C'C" be the common chord of (Ka), (Kb) [C' near to C]

Are the triangles

1. ABC, A'B'C'

2. ABC, A"B"C"

perspective?

Antreas
• ... In my figure, the triangles seems to be perspective, and the midpoint of the perspectors line segment is the Nine Point Circle center N. Proof ? Antreas
Message 3 of 20 , Feb 1, 2002
On Friday, February 1, 2002, at 02:21 PM, Antreas P. Hatzipolakis wrote:

>
> On Thursday, January 31, 2002, at 10:59 AM, Antreas P. Hatzipolakis
> wrote:
>
>> Let ABC be a triangle, and AHa, BHb, CHc its altitudes.
>>
>> Draw the circles (Ka), (Kb), (Kc) having as diameters the
>> altitudes AHa, BHb, CHc, resp. [Ka = midpoint of AHa, etc].
>
> Let
> A'A" be the common chord of (Kb), (Kc) [A' near to A]
> B'B" be the common chord of (Kc), (Ka) [B' near to B]
> C'C" be the common chord of (Ka), (Kb) [C' near to C]
>
> Are the triangles
>
> 1. ABC, A'B'C'
>
> 2. ABC, A"B"C"
>
> perspective?
>

In my figure, the triangles seems to be perspective, and the midpoint
of the perspectors' line segment is the Nine Point Circle center N.
Proof ?

Antreas
• Dear Antreas and all hyacinthists, ... In both case, there are not perspective (a = 11 , b = 10.50 , c = 6). Best regards, Gilles
Message 4 of 20 , Feb 1, 2002
Dear Antreas and all hyacinthists,

> [APH]
>
> > Let ABC be a triangle, and AHa, BHb, CHc its altitudes.
> >
> > Draw the circles (Ka), (Kb), (Kc) having as diameters the
> > altitudes AHa, BHb, CHc, resp. [Ka = midpoint of AHa, etc].
>
> Let
> A'A" be the common chord of (Kb), (Kc) [A' near to A]
> B'B" be the common chord of (Kc), (Ka) [B' near to B]
> C'C" be the common chord of (Ka), (Kb) [C' near to C]
>
> Are the triangles
>
> 1. ABC, A'B'C'

>
> 2. ABC, A"B"C"
>
> perspective?
>
> Antreas

In both case, there are not perspective (a = 11 , b = 10.50 , c = 6).

Best regards,

Gilles
• Dear Antreas, [APH]: Let ABC be a triangle, and AHa, BHb, CHc its altitudes. Draw the circles (Ka), (Kb), (Kc) having as diameters the altitudes AHa, BHb,
Message 5 of 20 , Feb 1, 2002
Dear Antreas,

[APH]: Let ABC be a triangle, and AHa, BHb, CHc its altitudes.
Draw the circles (Ka), (Kb), (Kc) having as diameters the
altitudes AHa, BHb, CHc, resp. [Ka = midpoint of AHa, etc].

[PY]: Where is the radical center?

[APH]: A general problem that Floor posted says that the circles with
diameters three concurrent cevians have always as radical center
the orthocenter of ABC.

*** Thank you for reminding me of this very interesting result. I remember
it now.

[APH]: Here are some other rad. center problems on the altitudes
configuration:

1. Which is the locus of the radical center of the circles
(Ka, t*h_a), (Kb, t*h_b), (Kc, t*h_c) as t varies?
[For t = 1/2 we have the case of the altitudes as diameters.]

2. Which is the locus of the radical center of the circles
(Ka, t*R), (Kb, t*R), (Kc, t*R) as t varies?

For (2), it is a fixed point P.

For (1), it is a line through the point P.

For (3), another line through the same point P.

3. Which is the locus of the radical center of the circles
(Ka, t*a), (Kb, t*b), (Kc, t*c) as t varies?
best regards
sincerely
Paul
• Dear Antreas, [APH]: Let ABC be a triangle, and AHa, BHb, CHc its altitudes. Draw the circles (Ka), (Kb), (Kc) having as diameters the altitudes AHa, BHb,
Message 6 of 20 , Feb 1, 2002
Dear Antreas,

[APH]: Let ABC be a triangle, and AHa, BHb, CHc its altitudes.
Draw the circles (Ka), (Kb), (Kc) having as diameters the
altitudes AHa, BHb, CHc, resp. [Ka = midpoint of AHa, etc].

Here are some other rad. center problems on the altitudes
configuration:

1. Which is the locus of the radical center of the circles
(Ka, t*h_a), (Kb, t*h_b), (Kc, t*h_c) as t varies?
[For t = 1/2 we have the case of the altitudes as diameters.]

2. Which is the locus of the radical center of the circles
(Ka, t*R), (Kb, t*R), (Kc, t*R) as t varies?

For (2), it is a fixed point P.

*** This is indeed the circumcenter of KaKbKc.

Best regards
Sincerely
• Dear Antreas, [APH]: Let ABC be a triangle, and AHa, BHb, CHc its altitudes. Draw the circles (Ka), (Kb), (Kc) having as diameters the altitudes AHa, BHb,
Message 7 of 20 , Feb 2, 2002
Dear Antreas,

[APH]: Let ABC be a triangle, and AHa, BHb, CHc its altitudes.
Draw the circles (Ka), (Kb), (Kc) having as diameters the
altitudes AHa, BHb, CHc, resp. [Ka = midpoint of AHa, etc].

Here are some other rad. center problems on the altitudes
configuration:

1. Which is the locus of the radical center of the circles
(Ka, t*h_a), (Kb, t*h_b), (Kc, t*h_c) as t varies?
[For t = 1/2 we have the case of the altitudes as diameters.]

2. Which is the locus of the radical center of the circles
(Ka, t*R), (Kb, t*R), (Kc, t*R) as t varies?

For (2), it is a fixed point P.

[PY]: This is indeed the circumcenter of KaKbKc.

*** This is true if the orthocenter is replaced by an arbitrary
point. If circles of equal radii are constructed with centers at the
midpoints of the cevians of P, then the radical axes are precisely
the perpendicular bisectors of the segments joining these midpoints,
and the radical center is the circumcenter of this midpoints of
cevians triangle.

Best regards
Sincerely
Paul
• ... Dear Paul, Here is another problem: Let A B C be the circumcevian triangle of H and (Ka), (Kb), (Kc) the circles with diameters AA , BB , CC
Message 8 of 20 , Feb 4, 2002
On Thursday, January 31, 2002, at 01:59 PM, Paul Yiu wrote:

> Dear Antreas and friends,
>
> [APH]: Let ABC be a triangle, and AHa, BHb, CHc its altitudes.
>
> Draw the circles (Ka), (Kb), (Kc) having as diameters the
> altitudes AHa, BHb, CHc, resp. [Ka = midpoint of AHa, etc].
>
> *** Where is the radical center?
>

Dear Paul,

Here is another problem:
Let A'B'C' be the circumcevian triangle of H and
(Ka), (Kb), (Kc) the circles with diameters AA', BB', CC'
respectively.

The same question in general: A'B'C' = the circumcevian triangle
of a point P.

Antreas
• ... Let me rephrase the problem in this way (in order to compose a variation): Let ABC be a triangle, and AHa, BHb, CHc its altitudes meeting at H. Let A , B ,
Message 9 of 20 , Feb 4, 2002
On Monday, February 4, 2002, at 06:10 AM, Antreas P. Hatzipolakis wrote:

> Let A'B'C' be the circumcevian triangle of H and
> (Ka), (Kb), (Kc) the circles with diameters AA', BB', CC'
> respectively.
>
> Which is the radical center?
>

Let me rephrase the problem in this way (in order to compose
a variation):

Let ABC be a triangle, and AHa, BHb, CHc its altitudes meeting at H.

Let A', B', C' be the reflections of H in Ha, Hb, Hc, resp.
[now AA', BB', CC' are the circumcevians of H]

Let (Ka), (Kb), (Kc) be the circles with diameters AA', BB', CC'
resp..

VARIATION:

Let A', B', C' be the reflections of H not in Ha, Hb, Hc, resp.,
but in A,B,C resp.

Let (Ka), (Kb), (Kc) be the circles with diameters AA', BB', CC'
resp..

Which is now the radical center of the circles?

Antreas
• Dear Antreas, [APH]: Let A , B , C be the reflections of H not in Ha, Hb, Hc, resp., but in A,B,C resp. Let (Ka), (Kb), (Kc) be the circles with diameters
Message 10 of 20 , Feb 4, 2002
Dear Antreas,

[APH]: Let A', B', C' be the reflections of H not in Ha, Hb, Hc, resp.,
but in A,B,C resp.

Let (Ka), (Kb), (Kc) be the circles with diameters AA', BB', CC'
resp..

Which is now the radical center of the circles?

*** These circles all pass through H, which is therefore the radical center.
This applies to an arbitrary point in place of H.

Best regards
Sincerely
Paul
• ... I meant to say: with diameters HaA , HbB , HcC [anyway, and the above one is another variation!] Antreas
Message 11 of 20 , Feb 4, 2002
On Monday, February 4, 2002, at 02:43 PM, Antreas P. Hatzipolakis wrote:

>
> On Monday, February 4, 2002, at 06:10 AM, Antreas P. Hatzipolakis wrote:
>
>> Let A'B'C' be the circumcevian triangle of H and
>> (Ka), (Kb), (Kc) the circles with diameters AA', BB', CC'
>> respectively.
>>
>> Which is the radical center?
>>
>
> Let me rephrase the problem in this way (in order to compose
> a variation):
>
> Let ABC be a triangle, and AHa, BHb, CHc its altitudes meeting at H.
>
> Let A', B', C' be the reflections of H in Ha, Hb, Hc, resp.
> [now AA', BB', CC' are the circumcevians of H]
>
> Let (Ka), (Kb), (Kc) be the circles with diameters AA', BB', CC'
> resp..
>
> Which is the radical center?
>
> VARIATION:
>
> Let A', B', C' be the reflections of H not in Ha, Hb, Hc, resp.,
> but in A,B,C resp.
>
> Let (Ka), (Kb), (Kc) be the circles with diameters AA', BB', CC'
> resp..
>

I meant to say: with diameters HaA', HbB', HcC'
[anyway, and the above one is another variation!]

Antreas

> Which is now the radical center of the circles?
>
> Antreas
>
• ... Dear Paul, Yes. Sorry for not being careful! I sent a correction. Antreas
Message 12 of 20 , Feb 4, 2002
On Monday, February 4, 2002, at 02:52 PM, Paul Yiu wrote:

> Dear Antreas,
>
> [APH]: Let A', B', C' be the reflections of H not in Ha, Hb, Hc, resp.,
> but in A,B,C resp.
>
> Let (Ka), (Kb), (Kc) be the circles with diameters AA', BB', CC'
> resp..
>
> Which is now the radical center of the circles?
>
> *** These circles all pass through H, which is therefore the radical
> center.
> This applies to an arbitrary point in place of H.
>

Dear Paul,

Yes. Sorry for not being careful!
I sent a correction.

Antreas
• Dear Antreas, [APH]: Let A , B , C be the reflections of H not in Ha, Hb, Hc, resp., but in A,B,C resp. Let (Ka), (Kb), (Kc) be the circles with diameters
Message 13 of 20 , Feb 4, 2002
Dear Antreas,

[APH]: Let A', B', C' be the reflections of H not in Ha, Hb, Hc, resp.,
but in A,B,C resp.

Let (Ka), (Kb), (Kc) be the circles with diameters AA', BB', CC'
resp..

I meant to say: with diameters HaA', HbB', HcC'
...
Which is now the radical center of the circles?

*** Still the same orthocenter!

Best regards
Sincerely
paul
• ... Dear Paul, I am afraid that there is no way to go away from H ! :-) Let s try other variations: Let Ea, Eb, Ec be the midpoints of AH,BH,CH, resp. [ie
Message 14 of 20 , Feb 4, 2002
On Monday, February 4, 2002, at 03:08 PM, Paul Yiu wrote:

> Dear Antreas,
>
> [APH]: Let A', B', C' be the reflections of H not in Ha, Hb, Hc, resp.,
> but in A,B,C resp.
>
> Let (Ka), (Kb), (Kc) be the circles with diameters AA', BB', CC'
> resp..
>
>
> I meant to say: with diameters HaA', HbB', HcC'
> ...
> Which is now the radical center of the circles?
>
> *** Still the same orthocenter!
>

Dear Paul,

I am afraid that there is no way to go away from H ! :-)

Let's try other variations:

Let Ea, Eb, Ec be the midpoints of AH,BH,CH, resp.
[ie EaEbEc is the Euler triangle of ABC]

Consider the circles with diameters EaHa, EbHb, ECHc.

Still H !
[EaHa, EbHc, EcHc are chords of the nine point circle of ABC
intersecting at H. So, this is the very first variation

Variation:

Let Ma, Mb, Mc be the midpoints HHa, HHb, HHc, resp.

Consider the circles with diameters AMa, BMb, CMc.

Still H????

Antreas
• ... Another Variation: Consider the circles with diameters EaMa, EbMb, EcMc [AHa = the a-altitude; Ea, Ma = the midpoints of AH, HHa, resp., etc] Which is
Message 15 of 20 , Feb 4, 2002
On Monday, February 4, 2002, at 03:49 PM, Antreas P. Hatzipolakis wrote:

>
> On Monday, February 4, 2002, at 03:08 PM, Paul Yiu wrote:
>
>> Dear Antreas,
>>
>> [APH]: Let A', B', C' be the reflections of H not in Ha, Hb, Hc, resp.,
>> but in A,B,C resp.
>>
>> Let (Ka), (Kb), (Kc) be the circles with diameters AA', BB', CC'
>> resp..
>>
>>
>> I meant to say: with diameters HaA', HbB', HcC'
>> ...
>> Which is now the radical center of the circles?
>>
>> *** Still the same orthocenter!
>>
>
> Dear Paul,
>
> I am afraid that there is no way to go away from H ! :-)
>
> Let's try other variations:
>
> Let Ea, Eb, Ec be the midpoints of AH,BH,CH, resp.
> [ie EaEbEc is the Euler triangle of ABC]
>
> Consider the circles with diameters EaHa, EbHb, ECHc.
> Which is their radical center?
>
> Still H !
> [EaHa, EbHc, EcHc are chords of the nine point circle of ABC
> intersecting at H. So, this is the very first variation
> in this thread of problems]
>
> Variation:
>
> Let Ma, Mb, Mc be the midpoints of HHa, HHb, HHc, resp.
>
> Consider the circles with diameters AMa, BMb, CMc.
>
> Which is their radical center?
>
> Still H????
>

Another Variation:

Consider the circles with diameters EaMa, EbMb, EcMc
[AHa = the a-altitude; Ea, Ma = the midpoints of AH, HHa, resp., etc]

Enough for today!

Good night!

Antreas
• On Thursday, January 31, 2002, at 10:59 AM, Antreas P. Hatzipolakis ... Perspector: (cotA ::) in barycentrics. Variation: (Ka),(Kb),(Kc) = the circles with
Message 16 of 20 , Feb 6, 2002
On Thursday, January 31, 2002, at 10:59 AM, Antreas P. Hatzipolakis
wrote:

> Let ABC be a triangle, and AHa, BHb, CHc its altitudes.
>
> Draw the circles (Ka), (Kb), (Kc) having as diameters the
> altitudes AHa, BHb, CHc, resp. [Ka = midpoint of AHa, etc].
>
> A1 = (second tangent to (Ka) from B) /\ (second tangent to (Ka) from C)
> [the first one is BC]
>
> B1 = (second tangent to (Kb) from C) /\ (second tangent to (Kb) from A)
>
> C1 = (second tangent to (Kc) from A) /\ (second tangent to (Kc) from B)
>
> The triangles ABC, A1B1C1 are perspective.

Perspector: (cotA ::) in barycentrics.

Variation:

(Ka),(Kb),(Kc) = the circles with diameters (HHa), (HHb), (HHc), resp.

(K'a) = the reflection of (Ka) in the perp. bisector of BC.
(K'b) = the reflection of (Kb) in the perp. bisector of CA.
(K'c) = the reflection of (Kc) in the perp. bisector of AB.

A' = (2nd tangent to (K'a) from B) /\ (2nd tangent to (K'a) from C)
[the first one is BC] [the first one is BC]

B' = (2nd tangent to (K'b) from C) /\ (2nd tangent to (K'b) from A)

C' = (2nd tangent to (K'c) from A) /\ (2nd tangent to (K'c) from B)

The triangles ABC,A'B'C' are perspective.

tanA
Perspector: ( ----------- ::) in barycentrics
5 - 4tan^2A

Antreas
• On Thursday, January 31, 2002, at 10:59 AM, Antreas P. Hatzipolakis ... Another variation: AHa, AHb, AHc = altitudes. Hab = Orth. Proj. of Ha on AB, Hac =
Message 17 of 20 , Feb 7, 2002
On Thursday, January 31, 2002, at 10:59 AM, Antreas P. Hatzipolakis
wrote:

> Let ABC be a triangle, and AHa, BHb, CHc its altitudes.
>
> Draw the circles (Ka), (Kb), (Kc) having as diameters the
> altitudes AHa, BHb, CHc, resp. [Ka = midpoint of AHa, etc].
>
> A1 = (second tangent to (Ka) from B) /\ (second tangent to (Ka) from C)
> [the first one is BC]
>
> B1 = (second tangent to (Kb) from C) /\ (second tangent to (Kb) from A)
>
> C1 = (second tangent to (Kc) from A) /\ (second tangent to (Kc) from B)
>
> The triangles ABC, A1B1C1 are perspective.
>

Another variation:

AHa, AHb, AHc = altitudes.

Hab = Orth. Proj. of Ha on AB,
Hac = Orth. Proj. of Ha on AC.

(Kab) = the circle with diameter HaHab
(Kac) = the circle with diameter HaHac

A'= (2nd tangent to (Kab) from B) /\ (2nd tangent to (Kac) from C)
[the 1st tangent is AB] [the 1st tangent is AC]

Similarly B', C'.

A
/\
/ \
/ \
/ \
/ \
/ \
/ \
Hab \
/ Hac
/ Kab \
/ Kac \
B-----------Ha----------C

A'

The triangles ABC, A'B'C' are perspective.

1
Perspector: ( ----------------- ::) = (tanAsin^2A ::) in Barycentrics.
cotA*(1 + cot^2A)

Antreas
• On Thursday, February 7, 2002, at 02:00 PM, Antreas P. Hatzipolakis ... BHb, CHc ... Similarly define the circles (Kbc),(Kba);(Kca),(Kcb) H a = the 2nd
Message 18 of 20 , Feb 7, 2002
On Thursday, February 7, 2002, at 02:00 PM, Antreas P. Hatzipolakis
wrote:

> Another variation:
>
> AHa, AHb, AHc = altitudes.
BHb, CHc
>
> Hab = Orth. Proj. of Ha on AB,
> Hac = Orth. Proj. of Ha on AC.
>
>
> (Kab) = the circle with diameter HaHab
> (Kac) = the circle with diameter HaHac

Similarly define the circles (Kbc),(Kba);(Kca),(Kcb)

H'a = the 2nd intersection of (Kab),(Kac)
[the first is Ha]
that is, HaH'a is the common chord of (Kab),(Kac)

H'b = the 2nd intersection of (Kbc),(Kba)
[the first is Hb]

H'c = the 2nd intersection of (Kca),(Kcb)
[the first is Hc]

Are the triangles

1. HaHbHc, H'aH'bH'c

2. ABC, H'aH'bH'c

perspective?

Antreas

>
> A'= (2nd tangent to (Kab) from B) /\ (2nd tangent to (Kac) from C)
> [the 1st tangent is AB] [the 1st tangent is AC]
>
> Similarly B', C'.
>
> A
> /\
> / \
> / \
> / \
> / \
> / \
> / \
> Hab \
> / Hac
> / Kab \
> / Kac \
> B-----------Ha----------C
>
> A'
>
>
>
> The triangles ABC, A'B'C' are perspective.
>
> 1
> Perspector: ( ----------------- ::) = (tanAsin^2A ::) in Barycentrics.
> cotA*(1 + cot^2A)
>
>
> Antreas
>
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