Alexey a zaslavsky, one property of the jerabek hyperbola and its corollaries

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ONE PROPERTY OF THE JERABEK HYPERBOLA AND ITS
COROLLARIES
ALEXEY A. ZASLAVSKY
Abstract. We study the locus of the points P having the following property:
if A
1
B
1
C
1
is the circumcevian triangle of P with respect to the given triangle
ABC, and A
2
, B
2
, C
2
are the reﬂections of A
1
, B
1
, C
1
in BC, CA, AB,
respectively, then the triangles ABC and A
2
B
2
C
2
are perspective. We show

that this locus consists of the inﬁnite line and the Jerabek hyperbola of ABC.
This fact yields some interesting corollaries.
We start with the following well-known fact [1, p. 4.4.5].
Statement 1. Let the tangents to the circumcircle of ABC at A and B meet
in C
0
. The line CC
0
meets the circumcircle of ABC for the second time in C
1
,
and C
2
is the reﬂection of C
1
in AB. Then CC
2
is a median in ABC.
Proof. Let C

be the common point of CC
1
and AB, and A

, B

be the common
points of AC
2
, BC

2
with BC, AC, respectively. Since ∠C

CB = ∠C
1
AB =
∠BAA

, the triangles BCC

and BAA

are similar; therefore, BA

=
AB
BC
· BC

.
But CC

is a symedian in ABC, so BC

=
BC
2
BC
2
+AC

2
· AB. Therefore,
BA

BC
=
AB
2
AC
2
+BC
2
. Analogously, the ratio
AB

AC
has the same value, yielding the claim. 
A B
C
C
0
C
1
C
2
A

A

A


A

A

A

A

A

A

A

A

A

A

A

A

A

A

A


A

A

C

C

C

C

C

C

C

C

C

C

C

C

C


C

C

C

C

C

C

C

C

C

C

C

C

C

C

C


C

C

C

C

C

C

C

Fig. 1.
This fact yields the following corollary: let A
1
B
1
C
1
be the circumcevian tri-
angle of the Lemoine point, L, and let A
2
, B
2
, C
2
be the reﬂections of A

1
, B
1
,
53
54 ALEXEY A. ZASLAVSKY
C
1
in BC, CA, AB, respectively. Then the triangles ABC and A
2
B
2
C
2
are
perspective.
Our goal is to ﬁnd the locus of the points sharing this property. To this end,
ﬁrst we formulate the following assertion.
Lemma 1. Let CC
1
divide AB in ratio x : y. Then CC
2
divides AB in ratio
x(b
2
(x + y) − c
2
x) : y(a
2
(x + y) − c

2
y).
In order to prove this, it suﬃces to repeat the argument by which we demon-
strated the previous assertion and then to apply Ceva’s theorem.
Now let P be the point with barycentric coordinates (x : y : z). Using Lemma 1
and Ceva’s theorem, we see that a point P has the property in question iﬀ it lies
on some cubic c. From the following assertion, we infer that c is degenerated.
Statement 2. Let three parallel lines passing through the vertices of ABC meet
its circumcircle in A
1
, B
1
, C
1
. The points A
2
, B
2
, C
2
are the reﬂections of
A
1
, B
1
, C
1
in BC, CA, AB, respectively. Then the lines AA
2
, BB

2
, CC
2
are
concurrent.
Proof. Consider the three lines which pass through A
1
, B
1
, C
1
and are parallel
to BC, CA, AB, respectively. It is easy to see that they meet at a point on the
circumcircle of ABC. The points A
2
, B
2
, C
2
are the reﬂections of this point in
the midpoints of the sides of ABC. Therefore, the triangles ABC and A
2
B
2
C
2
are centrosymmetric. 
A B
C
A

1
B
1
C
1
A
2
B
2
C
2
Fig. 2.
Note also that the center of perspective in this claim lies on the Euler circle.
And so, c consists of the inﬁnite line and some conic k. In order to determine k
completely, it suﬃces to indicate ﬁve point lying on it. We already know that k
contains the Lemoine point, L. Furthermore, k contains the vertices of ABC
as well as its orthocenter H (in this case, all of A
2
, B
2
, C
2
coincide with H).
Therefore, k is the Jerabek hyperbola.
Here follow some corollaries of this fact.
ONE PROPERTY OF THE JERABEK HYPERBOLA AND ITS COROLLARIES 55
Statement 3. Let P be a point on the Euler line of ABC, A
1
B
1

C
1
be the cir-
cumcevian triangle of P , and A
2
, B
2
, C
2
be the reﬂections of A
1
, B
1
, C
1
in the
midpoints of BC, CA, AB, respectively. Then the lines AA
2
, BB
2
, CC
2
are
concurrent.
Proof. The isogonal conjugate, Q, of P lies on the Jerabek hyperbola. Let CQ
meet the circumcircle of ABC for the second time in C
3
. Then C
2
and C

3
are
symmetric with respect to AB. 
A B
C
A
1
B
1
C
1
A
2
B
2
C
2
C
2
C
2
C
2
C
2
C
2
C
2
C

2
C
2
C
2
C
2
C
2
C
2
C
2
C
2
C
2
C
2
C
2
C
2
C
2
C
2
C
2
C

2
C
2
C
2
C
2
C
2
C
2
C
2
C
2
C
2
C
2
C
2
C
2
C
2
P
P
P
P
P

P
P
P
P
P
P
P
P
P
P
P
P
P
P
P
P
P
P
P
P
P
P
P
P
P
P
P
P
P
P

Fig. 3.
Statement 4. Let P be a point on the Euler line of ABC, A
0
, B
0
, C
0
be the
midpoints of BC, CA, AB, respectively, and A
1
, B
1
, C
1
be the projections of the
circumcenter O of ABC onto AP , BP , CP , respectively. Then the lines A
0
A
1
,
B
0
B
1
, C
0
C
1
are concurrent.
A B

C
A
1
A
1
A
1
A
1
A
1
A
1
A
1
A
1
A
1
A
1
A
1
A
1
A
1
A
1
A

1
A
1
A
1
A
1
A
1
A
1
A
1
A
1
A
1
A
1
A
1
A
1
A
1
A
1
A
1
A

1
A
1
A
1
A
1
A
1
A
1
B
1
C
1
A
0
B
0
C
0
C
0
C
0
C
0
C
0
C

0
C
0
C
0
C
0
C
0
C
0
C
0
C
0
C
0
C
0
C
0
C
0
C
0
C
0
C
0
C

0
C
0
C
0
C
0
C
0
C
0
C
0
C
0
C
0
C
0
C
0
C
0
C
0
C
0
C
0
P

P
P
P
P
P
P
P
P
P
P
P
P
P
P
P
P
P
P
P
P
P
P
P
P
P
P
P
P
P
P

P
P
P
P
O
O
O
O
O
O
O
O
O
O
O
O
O
O
O
O
O
O
O
O
O
O
O
O
O
O

O
O
O
O
O
O
O
O
O
O
O
O
O
O
O
O
O
O
O
O
O
O
O
O
O
O
O
O
O
O

O
O
O
O
O
O
O
O
O
Fig. 4.
Proof. It suﬃces to apply homothety of center the centroid of ABC and coeﬃ-
cient −
1
2
to the conﬁguration of the previous claim. 
Statement 5. Let O, I be the circumcenter and incenter of ABC. An arbitrary
line perpendicular to OI meets BC, CA, AB in A
1
, B
1
, C
1
, respectively. Then
the circumcenters of the triangles IAA
1
, IBB
1
, ICC
1
are collinear.

Proof. Applying an inversion of center I and the previous assertion we see that
the circumcircles of the three triangles in question have a common point other
than I. 
56 ALEXEY A. ZASLAVSKY
A B
C
A
1
A
1
A
1
A
1
A
1
A
1
A
1
A
1
A
1
A
1
A
1
A
1