Fontene theorem’s application
Mr.Linh Nguyen Van
11/2010
Problem: Given a triangle ABC with its circumcircle (O), incenter I, excenters Ia , Ib , Ic . Let Ta Tb Tc be
the triangle formed by three tangencies of (O) through A, B, C.
a, Prove that Ib Tb , Ic Tc and BC concur at A1 . Similarly we define B1 , C1 . Prove that AA1 , BB1 , CC1 concur
at L which lies on (O).
b, Let da be the reflection of d wrt BC, similarly with db , dc ; A2 B2 C2 be the triangle formed by da , db , dc .
Prove that AA2 , BB2 , CC2 concur at L.
Solution:
a, Lemma 1(1st Fontene’s theorem): Given triangle ABC. Let P be an arbitrary point in the plane.
A1 , B1 , C1 are the midpoints of BC, CA, AB; A2 B2 C2 is the pedal triangle of P with respect to triangle ABC.
Let X, Y, Z be the intersections of B1 C1 and B2 C2 , A1 C1 and A2 C2 , A1 B1 and A2 B2 . Then A2 X, B2 Y, C2 Z
concur at the intersection of (A1 B1 C1 ) and (A2 B2 C2 ).
Proof
A
L
B2
Q
X
C1
B1
O
C2
P
B
A2
F
E
O'
A1
C
Let E be the center of (A1 B1 C1 ), O0 be the center of (A2 B2 C2 ), F be the intersection of OP and the circle
with diameter OA, L be the reflection of A2 with respect to B1 C1 then AL//BC. We obtain ∠ALP = 90o .
Since ∠AF P = ∠AB2 P = ∠AC2 P = ∠ALP = 90o we get L, F, B2 , C2 are on (AP ).
∠F C1 X = ∠F AB1 = ∠B2 C2 F then F XC1 C2 is a cyclic quadrilateral.
Denote L0 the intersection of F X and (AP ). We have AL0 C2 F is a cyclic quadrilateral. But F XC1 C2 is
also cyclic therefore AL0 //B1 C1 or L0 ≡ L, which follows that L, X, F are collinear.
Denote Q the intersection of A2 X and (E).F 0 is the reflection of Q with respect to B1 C1 .
Consider the Symmetry SB1 C1 : (AO) 7→ (E), but Q ∈ (E) hence F 0 ∈ (AO).
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On the other side, SB1 C1 maps A2 to L. Furthermore A2 , X, Q are collinear so L, X, F 0 are collinear, which
is equivalent to F 0 ≡ F .
We deduce that A2 LQF is a isosceles trapezoid.
This means XQ.XA2 = XL.XF = XB2 .XC2 .
Therefore Q lies on (O0 ). Similarly B2 Y, C2 Z also pass through Q. We are done.
Back to our problem:
We can change the problem above to the equivalent one as below:
Given a triangle ABC with its orthocenter H, circumcenter O. Let A1 B1 C1 be the orthic triangle of ∆ABC;
A2 B2 C2 be the median triangle of ∆ABC; (E) be the 9-point circle of triangle ABC, A0 B 0 C 0 be the tangentinal
triangle of ∆A1 B1 C1 .
a, Prove that BB 0 , CC 0 , BC concur at A3 and A1 A3 passes through the Jerabek point L of triangle ABC.
b, Let XY Z be the triangle formed by the reflections of OH with respect to B1 C1 , C1 A1 , A1 B1 . Prove that
A1 X, B1 Y, C1 Z concur at L.
Proof:
A
B3
Y
L
B1
K
C2
A3
B2
G
O
X
E
P
C1
Z'
Y'
X'
H
B'
B
A2
A1
C
Z
Part a:
Let Y be the intersection of A1 C1 and A2 C2 then applying lemma 1 we obtain B3 , L, B1 are collinear. Let
A03 be the intersection of B2 C2 and B1 C1 then A1 , A03 , L are collinear.
Note that A1 L ∩ C1 B1 = {A3 }, C1 C2 ∩ A1 A2 = {B}, B 0 is the intersection of the tangencies of (E) through
A1 , C1 then A03 , B, B 0 lies on the polar of B3 wrt (E) or A03 ≡ A3 . We are done.
Part b:
Denote K = OA2 ∩ (E), G = C2 B2 ∩ OH, X 0 = OH ∩ B1 C1 , similarly with Y 0 , Z 0 ; P = XZ ∩ AA1 .
Note that the Jerabek point of triangle ABC is the Anti-Steiner point of triangle A2 B2 C2 hence L, K, G are
collinear.
Since A1 K is the diameter of (E) we get ∠A1 LK = 90o . So we only need to prove that A1 X ⊥ LG.
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iff ∠LGC2 = ∠AA1 X iff ∠(OH, BC) = ∠AA1 X. (*)
1
On the other side, A1 is the X-excenter of triangle XY 0 Z 0 then ∠Y 0 A1 Z 0 = 90o − ∠Y 0 XZ 0 or ∠Y 0 XA1 =
2
90o − ∠C1 A1 B1 .
∠XP A1 = ∠P Y 0 A1 + ∠Y 0 A1 A = 180o − ∠HY 0 A1 + ∠C1 A1 A = 180o − (∠(OH, BC) + ∠Y 0 A1 B) + ∠C1 A1 A.
= 180o − ∠(OH, BC) − 90o + 2∠C1 A1 A = 90o − ∠(OH, BC) + ∠C1 A1 B1 .
Therefore ∠AA1 X = 180o −∠XP A1 −∠Y 0 XA1 = 180o −(90o −∠(OH, BC)+∠C1 A1 B1 )−(90o −∠C1 A1 B1 )
= ∠(OH, BC) or (∗) is obviously true. Our problem is proved.
But, it is not enough at all, we can generalize the part b as below:
Generalization: Given a triangle ABC with its circumcircle (O).Let d be an arbitrary line in the plane,
A0 B 0 C 0 be the triangle formed by the reflection of d wrt 3 sides of triangle ABC. Then AA0 , BB 0 , CC 0 concur
at I which lies on (O).
Proof:
A
E
D
F
C
B
A'
I
C'
B'
Denote D, E, F the intersections of d and BC, CA, AB. Note that A be the A’-excenter of triangle DA0 E
then A0 A is the bisector of angle C 0 A0 B 0 . Similarly we get AA0 , BB 0 , CC 0 concur at the incenter I (or excenter)
of triangle A0 B 0 C 0 .
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Moreover, ∠B 0 IC 0 = 90o + ∠DA0 E = 180o − ∠BAC then I lies on (O). Our problem is proved.
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3