## Re: A locus problem

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• ... ABCP. [snip] ... That s the locus Antreas asked for. The locus of the center of this reflection (as P varies on the circumcircle) is also interesting,
Message 1 of 13 , Feb 28, 2001
jean-pierre.ehrmann wrote:
> Dear Antreas, John and other Hyacinthists
> [Antreas]
> > > Let ABC be a triangle and P a point on its circumcircle.
> > > It is well known
> > > that the four orthocenters Hi, i = 1,2,3,4, of the triangles
> > > ABP, PAC, CPB, CAB, resp.,form a quadrilateral congruent to
ABCP.

[snip]
> If Ha is the orthocenter of PBC,... the segments AHa, BHb, CHc, PH
> have the same midpoint.
>
> Hence the congruence is the reflection w.r.t. this point and the
> locus of the the circumcenter of (Ha,Hb,Hc,H) is the circle with
> Best regards. Jean-Pierre

That's the locus Antreas asked for. The locus of the center of this
reflection (as P varies on the circumcircle) is also interesting,
because it is a trivial result that the midpoint of PH lies on the
nine-point circle.

Barry Wolk <wolkb AT ccu.umanitoba.ca>
• Let ABC be a triangle, P a point and A B C the pedal triangle of P. Denote: Ba, Ca = the orthogonal projections of B , C on AB, AC, resp. Which is the locus
Message 2 of 13 , Jul 17, 2016

Let ABC be a triangle, P a point and A'B'C' the pedal triangle of P.

Denote:

Ba, Ca = the orthogonal projections of B', C' on AB, AC, resp.

Which is the locus of P such that the Euler line La of ABaCa is perpendicular
to BC?

N lies on the locus.

We have three loci corresponding to A,B,C

Which points, other than N, are their intersections?

APH
• [APH] ... The same question with the Euler line of ABaCa parallel to BC instead of perpendicular. That is: Let ABC be a triangle and P a point. Denote: Ba, Ca
Message 3 of 13 , Jul 17, 2016

[APH]

Let ABC be a triangle, P a point and A'B'C' the pedal triangle of P.

Denote:

Ba, Ca = the orthogonal projections of B', C' on AB, AC, resp.

Which is the locus of P such that the Euler line La of ABaCa is perpendicular
to BC?

N lies on the locus.

We have three loci corresponding to A,B,C

Which points, other than N, are their intersections?

APH

The same question with the Euler line of ABaCa parallel to BC instead of perpendicular.

That is:

Let ABC be a triangle and P a point.

Denote:

Ba, Ca = the orthogonal projections of B', C' on AB, AC, resp.

Which is the locus of P such that the Euler line La of ABaCa is parallel
to BC?

Note: One point of the locus we can get by Gossard theorem:

Let Ba, Ca be the intersections of the Euler line of ABC with AB, AC, resp.

Denote:

B' = (perpendicular to AC at Ca) intersection (AB)

C' = (perpendicular to AB at Ba) intersection (AC)

P  = (perpendicular to AB at C') intersection (perpendicular to AC at B')
P lies on the locus.

APH

• [APH]: Let ABC be a triangle, P a point and A B C the pedal triangle of P. Denote: Ba, Ca = the orthogonal projections of B , C on AB, AC, resp. Which is the
Message 4 of 13 , Jul 17, 2016

[APH]:

Let ABC be a triangle, P a point and A'B'C' the pedal triangle of P.

Denote:

Ba, Ca = the orthogonal projections of B', C' on AB, AC, resp.

Which is the locus of P such that the Euler line La of ABaCa is perpendicular

to BC?

N lies on the locus.

We have three loci corresponding to A,B,C

Which points, other than N, are their intersections?

For A,B,C the required loci are the correspondent cevian of N, together with the line at infinity.

For P=N the Euler lines La, Lb, Lc concur at:

Z= (cos(2*A)-2)*cos(B-C)+cos(A) :: (trilinears)

= X(4)+3*X(51)

= midpoint of X(i),X(j) for these {i,j}: {4,389}, {5,5446}, {52,5907}, {143,546}, {1112,7687}, {5480,9969}

= reflection of X(i) in X(j) for these (i,j): (5447,3628), (9729,5462)

= On lines:

(3,5943), (4,51), (5,141), (6,1598), (20,5640), (25,578), (30,5462), (49,7545),

(52,381), (54,1495), (64,3531), (68,3818), (140,6688), (143,546), (155,576),

(181,3073), (182,7387), (373,631), (382,9730), (403,3574), (428,6146), (517,5795),

(550,5892), (568,3843), (569,7517), (575,7530), (970,3560), (973,1112), (1092,1995),

(1154,3850), (1173,1199), (1181,5198), (1597,3357), (1656,3819), (1843,3089),

(1864,1871), (1872,2262), (2818,7686), (2979,5056), (3060,3091), (3072,3271),

(3090,3917), (3098,7393), (3627,5946), (3628,5447), (3832,5889), (3845,6102),

(3851,5891), (3858,5876), (3861,5663), (5067,5650), (5071,7999), (5752,6913),

(6403,6622), (6530,6750), (6995,9833), (7486,7998), (7529,9306)

= {X(i),X(j)}-Harmonic conjugate of X(k) for these (i,j,k):

(4,51,389), (4,1093,8887), (4,3567,185), (4,9781,51), (6,1598,6759), (51,185,3567),

(52,381,5907), (185,3567,389), (1597,9786,3357), (1598,3527,6), (3060,3091,5562),

(3851,6243,5891), (5198,9777,1181)

= [ -1.328685665701012, -1.74991120893009, 5.465381010721065 ]

• ... Dear César, Thanks ! A natural question is which is the locus of P such that the Euler lines are concurrent. That is: Let ABC be a triangle, P a point and
Message 5 of 13 , Jul 18, 2016

[APH]:

Let ABC be a triangle, P a point and A'B'C' the pedal triangle of P.

Denote:

Ba, Ca = the orthogonal projections of B', C' on AB, AC, resp.

Which is the locus of P such that the Euler line La of ABaCa is perpendicular

to BC?

N lies on the locus.

We have three loci corresponding to A,B,C

Which points, other than N, are their intersections?

For A,B,C the required loci are the correspondent cevian of N, together with the line at infinity.

For P=N the Euler lines La, Lb, Lc concur at:

Z= (cos(2*A)-2)*cos(B-C)+cos(A) :: (trilinears)

= X(4)+3*X(51)

= midpoint of X(i),X(j) for these {i,j}: {4,389}, {5,5446}, {52,5907}, {143,546}, {1112,7687}, {5480,9969}

= reflection of X(i) in X(j) for these (i,j): (5447,3628), (9729,5462)

= On lines:

(3,5943), (4,51), (5,141), (6,1598), (20,5640), (25,578), (30,5462), (49,7545),

(52,381), (54,1495), (64,3531), (68,3818), (140,6688), (143,546), (155,576),

(181,3073), (182,7387), (373,631), (382,9730), (403,3574), (428,6146), (517,5795),

(550,5892), (568,3843), (569,7517), (575,7530), (970,3560), (973,1112), (1092,1995),

(1154,3850), (1173,1199), (1181,5198), (1597,3357), (1656,3819), (1843,3089),

(1864,1871), (1872,2262), (2818,7686), (2979,5056), (3060,3091), (3072,3271),

(3090,3917), (3098,7393), (3627,5946), (3628,5447), (3832,5889), (3845,6102),

(3851,5891), (3858,5876), (3861,5663), (5067,5650), (5071,7999), (5752,6913),

(6403,6622), (6530,6750), (6995,9833), (7486,7998), (7529,9306)

= {X(i),X(j)}-Harmonic conjugate of X(k) for these (i,j,k):

(4,51,389), (4,1093,8887), (4,3567,185), (4,9781,51), (6,1598,6759), (51,185,3567),

(52,381,5907), (185,3567,389), (1597,9786,3357), (1598,3527,6), (3060,3091,5562),

(3851,6243,5891), (5198,9777,1181)

= [ -1.328685665701012, -1.74991120893009, 5.465381010721065 ]

Dear César,

Thanks !

A natural question is which is the locus of P such that the Euler lines are concurrent.

That is:

Let ABC be a triangle, P a point and A'B'C' the pedal triangle of P.

Denote:

Ba, Ca = the orthogonal projections of B', C' on AB, AC, resp.

La = the Euler line of ABaCa.

Smilarly Lb, Lc.

Which is the locus of P such that La, Lb, Lc are concurrent?

APH

• ... Let A ,B , C be the intersections of La, Lb, Lc and BC, CA, AB, resp. ABC, A B C are orthologic. Which is the orthologic center (ABC, A B C ) ? (the
Message 6 of 13 , Jul 18, 2016

[APH]:

Let ABC be a triangle, P a point and A'B'C' the pedal triangle of P.

Denote:

Ba, Ca = the orthogonal projections of B', C' on AB, AC, resp.

Which is the locus of P such that the Euler line La of ABaCa is perpendicular

to BC?

N lies on the locus.

We have three loci corresponding to A,B,C

Which points, other than N, are their intersections?

For A,B,C the required loci are the correspondent cevian of N, together with the line at infinity.

For P=N the Euler lines La, Lb, Lc concur at:

Z= (cos(2*A)-2)*cos(B-C)+cos(A) :: (trilinears)

= X(4)+3*X(51)

= midpoint of X(i),X(j) for these {i,j}: {4,389}, {5,5446}, {52,5907}, {143,546}, {1112,7687}, {5480,9969}

= reflection of X(i) in X(j) for these (i,j): (5447,3628), (9729,5462)

= On lines:

(3,5943), (4,51), (5,141), (6,1598), (20,5640), (25,578), (30,5462), (49,7545),

(52,381), (54,1495), (64,3531), (68,3818), (140,6688), (143,546), (155,576),

(181,3073), (182,7387), (373,631), (382,9730), (403,3574), (428,6146), (517,5795),

(550,5892), (568,3843), (569,7517), (575,7530), (970,3560), (973,1112), (1092,1995),

(1154,3850), (1173,1199), (1181,5198), (1597,3357), (1656,3819), (1843,3089),

(1864,1871), (1872,2262), (2818,7686), (2979,5056), (3060,3091), (3072,3271),

(3090,3917), (3098,7393), (3627,5946), (3628,5447), (3832,5889), (3845,6102),

(3851,5891), (3858,5876), (3861,5663), (5067,5650), (5071,7999), (5752,6913),

(6403,6622), (6530,6750), (6995,9833), (7486,7998), (7529,9306)

= {X(i),X(j)}-Harmonic conjugate of X(k) for these (i,j,k):

(4,51,389), (4,1093,8887), (4,3567,185), (4,9781,51), (6,1598,6759), (51,185,3567),

(52,381,5907), (185,3567,389), (1597,9786,3357), (1598,3527,6), (3060,3091,5562),

(3851,6243,5891), (5198,9777,1181)

= [ -1.328685665701012, -1.74991120893009, 5.465381010721065 ]

Let A",B", C" be the intersections of La, Lb, Lc and BC, CA, AB, resp.

ABC, A"B"C" are orthologic.

Which is the orthologic center (ABC, A"B"C") ?
(the other one is the point of concurrence of La, Lb, Lc)

APH

• ... [César Lozada]: Answer: cevians of X(523) (which lies in the infinity) Note:
Message 7 of 13 , Jul 18, 2016

The same question with the Euler line of ABaCa parallel to BC instead of perpendicular.

That is:

Let ABC be a triangle, P a point and A'B'C' the pedal triangle of P.

Denote:

Ba, Ca = the orthogonal projections of B', C' on AB, AC, resp.

Which is the locus of P such that the Euler line La of ABaCa is parallel
to BC?

Note: One point of the locus we can get by Gossard theorem:

Let B', C' be the intersections of the Euler line of ABC with AC, AB, resp.
Let P be the intersection of the perpendiculars to AC, AB at B', C', resp.

Then P lies on the locus.

APH

Answer: cevians of X(523) (which lies in the infinity)

Note: One point of the locus we can get by Gossard theorem:

Let B', C' be the intersections of the Euler line of ABC with AC, AB, resp.
Let P be the intersection of the perpendiculars to AC, AB at B', C', resp.

Then P lies on the locus.

Sorry. I can´t find this.

**********************

Dear César,

> Sorry. I can´t find this.

It iwas wrong (incomplete) so I deleted it and sent a new message:
Anopolis #3207 = Hyacinthos #23818

Note: One point of the locus we can get by Gossard theorem:

Let Ba, Ca be the intersections of the Euler line of ABC with AB, AC, resp.

Denote:

B' = (perpendicular to AC at Ca) intersection (AB)

C' = (perpendicular to AB at Ba) intersection (AC)

P  = (perpendicular to AB at C') intersection (perpendicular to AC at B')
P lies on the locus.

P lies on the A cevian of X(523)
(Let E be the Euler line reflection point, ie the point the Euler line reflections in the sidelines concur at = X(110).
The reflection of AE in the A bisector is the A cevian of X(523))

APH

• ... [...] Variation: Let ABC be a triangle, P a point and B , C the orthogonal projections of P on AC, AB, resp. The locus of P such that the Euler line of
Message 8 of 13 , Aug 9, 2016

[APH]:

Let ABC be a triangle, P a point and A'B'C' the pedal triangle of P.

Denote:

Ba, Ca = the orthogonal projections of B', C' on AB, AC, resp.

Which is the locus of P such that the Euler line La of ABaCa is perpendicular

to BC?

N lies on the locus.

We have three loci corresponding to A,B,C

Which points, other than N, are their intersections?

For A,B,C the required loci are the correspondent cevian of N, together with the line at infinity.

For P=N the Euler lines La, Lb, Lc concur at:

Z= (cos(2*A)-2)*cos(B-C)+cos(A) :: (trilinears)

[...]

Variation:

Let ABC be a triangle, P a point and B', C' the orthogonal projections of P on AC, AB, resp.

The locus of P such that the Euler line of AB'C' is perpendicular to BC is the the a-cevian of the isogonal conjugate of N = X54

Application:

Let ABC be a triangle and A'B'C' the pedal triangle of N.

Denote:

A", B", C" = the reflections of A in B'C', C'A', A'B', resp.
(AA", BB", CC" concur at X54 = isog. conjugate of N)

Ab, Ac = the orthogonal projections of A" on AC, AB, resp.

L1 = the Euler line of AAbAc. Similarly L2, L3.

A*B*C* = the triangle bounded by L1, L2, L3.

ABC, A*B*C* are parallelogic with parallelogic center (ABC, A*B*C*) = H of ABC (since L1, L2, L3 are perpendiculars to BC, CA, AB, resp.)

APH

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