X(37):
Let {cA} be the circle with center on (BC) and tangent to (AB) and (AC). {cA} cuts
(BC) at Ab, Ac. Build Ba, Bc, Ca, Cb cyclically. Points Ab, Ac, Ba, Bc, Ca, Cb
lie on an ellipse with center X(37). [03-13-2014]
X(57):
Let A' be the antipode of the A-extouch in the A-excircle and (a') the tangent
to the A-excircle through A'. Build (b') and (c') cyclically. The triangle
bounded by (a'), (b'), (c') and the triangle ABC are perspective with
perspector X(57).
X(100):
If A'B'C' is the excentral triangle of ABC, then the Euler lines of triangles
A'BC, B'CA and C'AB concur at X(100) [04-23-2014]
X(140):
Let A'B'C' and A"B"C" be the medial and the orthic triangle of
ABC, respectively. Let A*, B*, C* be the midpoints of [A'A"], [B'B"]
and [C'C"], respectively. The Euler lines of AB*C*, BC*A* and CA*B* concur
at X(140). Also ABC and A*B*C* are orthologic with centers X(54) and X(5). [01-28-2014]
X(264):
Let A'B'C' be the orthic triangle of ABC and H its orthocenter. The
perpendicular line to (B'C') through H
cuts (BC) in A". Built B", C" cyclically. Lines
(AA"), (BB"),(CC") concur at X(264) [01-22-2014]
X(278):
The circle centered at B with radius AC=b cuts (BC) in two points, being Ab the
nearest one to C; the circle centered at C with radius AB=c cuts (BC) in two
points, being Ac the nearest one to B. The perpendicular to (BC) through Ab
cuts (AB) at Ab* and the perpendicular to (BC) through Ac cuts (AC) at Ac*. Let
(La) be the line joing Ab* and Ac*. Build (Lb) and (Lc) cyclically. ABC and the
triangle bounded by (La), (Lb) and (Lc) are perspective with perspector X(278).
(06-11-2013)
X(281):
The circle centered at B with radius AC=b cuts (BC) in two points, being Ab the
fartest one to C; the circle centered at C with radius AB=c cuts (BC) in two
points, being Ac the farthest one to B. The perpendicular to (BC) through Ab
cuts (AB) at Ab* and the perpendicular to (BC) through Ac cuts (AC) at Ac*. Let
(La) be the line joing Ab* and Ac*. Build (Lb) and (Lc) cyclically. ABC and the
triangle bounded by (La), (Lb) and (Lc) are perspective with perspector X(281).
(06-11-2013)
X(371): Let
O be the circumcenter of ABC. The line through O perpendicular to(AO) cuts the
circumcircle ABC at Ab and Ac, where Ab is on the small arc AB and Ac is on the
small arc AC. Let A' be the intersection of lines (BAb) and (CAc). Build B', C'
cyclically. Lines(AA'), (BB') and (CC') meet at X(371)=Kemnotu point. [02-13-2014]
X(381):
Let Ab,Ac be the reflections of A on B and C, respectively. Let A' be the
intersection of the diagonals of BAbAcC. Build B',C' cyclically. The center of
circle {A'B'C'} is X(381)=midpoint[G,H]. [01-26-2014]
X(479):
Let {a'} be the circle through B and C and tangent to the incircle {i} of ABC. Let A"={a'}/\{i}.
Build B" and C" cyclically. Lines (AA"), (BB") and
(CC") concur at X(479). [11-29-2013]
X(505):
Let A' be the A-excenter of ABC. Let A* be the nearest point to B where line
(A'B) cuts the A-excircle and A** the nearest point to C where line (A'C) cuts
the A-excircle. Build B*, B**, C*, C** cyclically. The triangle bounded by
lines (A*A**), (B*B**) and (C*C**) is perspective to ABC and the perspector is
X(505) (3rd isoscelizer point) Same result taking antipodes of A*,A** in A-excircle and so on [05-30-2013]
X(523):
Let A'B'C' and A"B"C" be the medial and the orthic triangle of
ABC. Through B' and C" are drawn the tangents to the NPC of ABC and they
meet at A*. B* and C* are built cyclically. Triangles ABC and A*B*C* have
perspector X(523). This point lies on the the line at infinity. Same result is
found when the the tangents are drawn through B" and C'. [06-25-2013]
X(631):
Let A’B’C’ be the cevian triangle of O in ABC. Let A”,B”,C” be the midpoints of
[OA’], [OB ’], [OC’], resp. Let
A*=(BC”)/\(CB”), B*=(CA”)/\(AC”), C*=(AB”)/\(BA”). A*B*C* and the
anticomplementary triangle of ABC are perspective with perspector X(631)=3*OG/5.
[03-08-2014]
X(1001):
Let A'B'C' and A"B"C" be the medial and excentral triangles of
ABC. The lines A"B' and A"C' cut (BC) in A1 and A2. Build B1,B2,C1,C2
cyclically. The points A1, A2, B1, B2, C1, C2 lie on a conic with center
X(1001). (06-07-2013)
X(1118):
Let A'B'C' be the orthic triangle of ABC. Let A" be the contact point of
the A-excircle and its tangent through A'. Build B" and C"
cyclically. The perspector of ABC and A"B"C" is X(1118).
X(1157):
Let A'B'C' be the reflection triangle of ABC. Circles {AB'C'}, {BC'A'} and
{CA'B'} concur at X(1157) [01-29-2014]
X(1263):
Let H and A'B'C' be the orthocenter and orthic triangle of ABC and
A"B"C" the reflecion triangle of A'B'C'. Circumcircles of
HAA", HBB" and HCC" concur at H and X(1263). [02-11-2014]
X(2051):
Let (a) be the line containing the polar of the Spieker center X(10) of ABC in
its A-excircle. Build (b) and (c) cyclically. The triangle bounded by the lines
(a), (b), (c) is perspective to ABC with perspector X(2051). (06-01-2013)
X(2184):
Let (a) be the line containing the polar of the orthocenter X(4) of ABC in its
A-excircle. Build (b) and (c) cyclically. The triangle bounded by the lines
(a), (b), (c) is perspective to ABC with perspector X(2184). (06-01-2013)
X(2646):
Let Ia be the A-excenter of triangle ABC: Denote by Aa, Ab, Ac the tangency
points of the A-excircle with the sides (BC), (CA), (AB), respectively. The
circumcircles of triangles AIaAa, BIaAb, CIaAc have a common point A",
different from Ia.
X(3062):
Let (a'), (b'), (c') be the polars of the incenter in the excircles and
A'=(b')/\(c'). Build B'and C' cyclically.
ABC and A'B'C' are perspective with perspector X(3062)
X(3062):
Let Ba, Ca be the inverses of B and C in A-excircle. Let (a')=(BaCa) and build (b'), (c') cyclically. ABC and the triangle bounded by (a'), (b'), (c') are
perspective with perspector X(3062) [11-21-2013]
X(3062): Let
A' be the center of the circle which is tangent to (AB), (AC) and the circumcircle
(externally). Let Ab, Ac be the orthogonal projections of A' on (AC), (AB),
respectively. Let
(a')=(AbAc). Build (b'), (c') cyclically. Then
ABC and the triangle bounded by (a'), (b'), (c') have perspector X(3062).
Note:
A'=[(1/4)*(-2*a*b*c-b*c^2-c*a^2+a^3-a*c^2+c^3-b*a^2+b^3-a*b^2-c*b^2)/(a*b*c),
1, 1] [01-20-2014]
X(3068):
Let Ba, Ca, C'a, B'a be the vertex of the A-inner-inscribed-square of ABC, with
Ba,Ca on (BC), C'a on (AC) and B'a on (AB). Let B"a=(B-C'a)/\(Ba-B'a) and
C"a=(C-B'a)/\(Ca-C'a) and (a")=(B"a-C"a). Build (b")
and (c") cyclically. ABC and the triangle bounded by (a"), (b"),
(c") are perspective with perspector X(3068) [12-22-2013]
X(3069):
Let Ba, Ca, C'a, B'a be the vertex of the A-outer-inscribed-square of ABC, with
Ba,Ca on (BC), C'a on (AC) and B'a on (AB). Let B"a=(B-C'a)/\(Ba-B'a) and
C"a=(C-B'a)/\(Ca-C'a) and (a")=(B"a-C"a). Build (b")
and (c") cyclically. ABC and the triangle bounded by (a"), (b"),
(c") are perspective with perspector X(3069) [12-22-2013]
X(3090):
Let A'B'C' be the orthic triangle of ABC, H its orthocenter and A", B", C" the
midpoints of [HA'], [HB'], [HC'], respectively. The Euler lines of triangles
A"BC, B"CA, C"AB and ABC concur at X(3090) [03-08-2014]
Antreas Hatzipolakis has pointed me
out that this result can be generalized: if A", B", C"
are such that
HA"/HA'=HB"/HB'=HC"/HC'=t then the Euler lines of triangles
A"BC, B"CA, C"AB
concur at Z(t) for all t.
That is true. For any t:
Z(t) = (SB*SC*(1-t)+S^2)/a
: : (trilinears)
= 3*(t-2)*X(2)-2*(t-1)*X(3)
and Z(t) lies on the Euler line of ABC.
Here is a small list of pairs
[t,Z(t)] (for |t|<=12 and computed with step=1/60):
[-12, X(3861)],
[-8, X(3845)], [-6, X(3843)], [-5, X(3839)], [-4, X(546)], [-3, X(3832)],
[-8/3, X(3858)],[-12/5,
X(3856)], [-2, X(381)], [-5/3, X(3854)], [-8/5, X(3857)], [-3/2, X(3855)],
[-4/3, X(3850)], [-1,
X(3091)],[-4/5, X(5066)], [-2/3, X(3851)], [-1/2, X(3545)], [-2/5,
X(5072)],
[-1/3, X(5068)],
[-1/4, X(3544)], [0, X(5)], [1/4, X(5071)], [1/3, X(5056)], [2/5,
X(5055)],
[1/2, X(3090)],
[2/3, X(1656)], [3/4, X(5067)], [4/5, X(3628)],[1, X(2)], [7/6, X(3533)], [6/5,
X(3526)],
[5/4, X(3525)],
[4/3, X(140)], [3/2, X(631)], [8/5, X(549)], [5/3, X(3523)],[7/4, X(3524)], [2,
X(3)],
[9/4, X(3528)],
[7/3, X(3522)], [12/5, X(548)], [5/2, X(376)], [8/3, X(550)], [14/5, X(3534)],
[3, X(20)], [10/3,
X(1657)], [7/2, X(3529)], [11/3, X(5059)], [4, X(30)], [14/3, X(5073)], [5,
X(3146)],
[6, X(382)],[7,
X(3543)], [8, X(3627)], [10, X(3830)], [12, X(3853)]
X(3579):
Let A'B'C' be the excentral triangle of ABC. Let Ab, Ac be the midpoints of
[A'B] and [A'C], respectively. Build Ba,Bc,Ca,Cb cyclically. The perpendicular
bisectors of [AbAc], [BaBc] and [CaCb] meet in X(3579). [08-24-2013]
This result can be generalized:
if
A", B", C" are such that A’Ab/A'B=A’Ac/A'C=B’Bc/B'C=B’Ba/B'A=C’Ca/C'A=C’Cb/C'B=t
then the perpendicular bisectors of [AbAc], [BaBc] and [CaCb] concur at
Z(t) for all t.
Z(t) = (SB*b+c*SC-2*S*R)*t-SA*a+SB*b+c*SC-2*S*R
: : (trilinears)
= (1-t)*X(1)-(2-t)*X(3)
and lies on line (IO) of ABC.
Pairs [t,Z(t)] (for |t|<12 and
computed with step=1/60):
[0, X(40)], [1/2, X(3579)], [2/3,
X(165)], [1, X(3)], [4/3, X(3576)], [3/2, X(1385)], [2, X(1)], [3, X(1482)]
X(5423):
Let {a'} be the A-excircle of ABC and {a"} the circle through B and C and
tangent to {a'}. If A'={a'}/\{a"} and B', C' are built cyclically, then
(AA'), (BB') and (CC') concur at X(5423) [11-13-2013]
X(5462): Let
A'B'C' and A"B"C" be the medial and orthic triangle of ABC. Let
(ra) be the radical axis of the NPC's of A'B"C" and A"B'C'. Build
(rb) and (rc) cyclically. The lines (ra), (rb), (rc) concur at X(5462)
[08-18-2013]
X(5542):
Let A'B'C' be the cevian triangle of the Gergonne point Ge in ABC. The line
thorugh Ge parallel to (B'C') cuts [A'C'] and [A'B'] at Ba and Ca,
respectively. Build Ab, Cb, Ac, Bc cyclically. The point Ba, Ca, Ab, Cb, Ac, Bc
are concyclic on a circumference with center X(5542) and radius
(1/2)*r*sqrt(r^2+s^2+8*r*R+16*R^2)/(r+4*R) [11-21-2013]
No comments:
Post a Comment