A Nice Theorem on Mixtilinear Incircles
Khakimboy Egamberganov
Abstract
There are three mixtilinear incircles and three mixtilinear excircles in an arbitrary triangle.
In this paper, we will present many properties of mixtilinear incircles along with a famous
theorem involving concyclic points and its proof.
A mixtilinear circle of a triangle is a circle tangent to two sides and the circumcircle of the
triangle. If this circle is tangent to the circumcircle internally, it is called a mixtilinear incircle of
the triangle. Otherwise, it is called a mixtilinear excircle of the triangle.
THEOREM. Given a triangle ABC, let ω, Ω, ΩA , be the incircle, circumcircle, and mixtilinear
incircle opposite point A, respectively. Let ΩA be tangent to Ω at TA . Suppose that P is a point
on Ω such that the tangents from P to ω intersect BC at points X, Y . Then P , TA , X, Y are
concyclic.
Before the proof of the theorem, we need some theorems and properties about mixtilinear
incircles from [1], [2], [3]. We will present them with proofs. Suppose that ω, Ω, ΩA , are the incircle,
circumcircle, and mixtilinear incircle opposite A of a triangle ABC and TA is the mixtilinear point
opposite A. Let I, O be the incenter and the circumcenter of the triangle ABC respectively, and
let AI intersect Ω at points A and A1 . Let A2 be the diametrically opposite point to A1 on Ω.
Analogously, define TB , TC , ΩB , ΩC and B1 , B2 , C1 , C2 .
Theorem 1. Let ΩA be tangent to the sides AB, AC at points M, N , respectively. Then
(i) M, I, N are collinear;
(ii) A2 , I, TA are collinear;
Proof. There is a homothety centered at TA that sends Ω to ΩA . (See Picture 1 ). So B1 , N, TA
and C1 , M, TA are collinear and M N k B1 C1 . Since AM, AN are tangent lines to ΩA from A, we
get that TA A is the symmedian of triangle M TA N . We have
6
6
C1 TA A =
and
ACB
= 6 B1 TA A2
2
6
ABC
= 6 C1 TA A2 .
2
This shows that the line TA A2 is the isogonal conjugate of the line TA A with respect to 6 M TA N .
Hence TA A2 is the median line of triangle M TA N .
Let I 0 be the midpoint of M N . Then
6
6
B1 TA A =
BTA I 0 = 6 CTA I 0 = 90◦ −
6
BAC
= 6 AM N = 6 AN M
2
and the quadrilaterals BTA I 0 M and CTA I 0 N are cyclic. So
6
M BI 0 = 6 M TA I 0 = 6 C1 TA A2 =
6
ABC
,
2
6
1
N CI 0 = 6 N TA I 0 = 6 B1 TA A2 =
6
ACB
2
and I ≡ I 0 . Hence M, I, N are collinear. Since TA , I 0 , A2 are collinear, we get that the points
TA , I, A2 are also collinear. This proves Theorem 1.
Theorem 2. Suppose that M, N are defined in the same way as the previous problem (tangency
points of ΩA to AB, AC). Then the lines M N , BC and TA A1 are concurrent.
Proof. We have three circles (BIC), (TA IA1 ) and Ω. From Theorem 1 we get I is midpoint of
M N , IA1 ⊥ M N and ITA ⊥ TA A1 . We can see that the circumcenter of (BIC) is the point A1
and the circumcenter of (TA IA1 ) id the midpoint of IA1 . So, the circles (BIC) and (TA IA1 ) are
tangents each to other at point I. Thus, M N is the radical axis of the circles (BIC) and (TA IA1 ).
Moreover, BC is the radical axis of the circles (BIC), Ω and TA A1 is the radical axis of the circles
Ω, (TA IA1 ). Hence the lines M N , BC and TA A1 are concurrent at point U (See Picture 1 ). This
proves Theorem 2.
Proof of THEOREM. Consider an inversion with center I and radius r (the radius of the incircle
of triangle ABC). Suppose that TA inverts to the point K. Also, let D, E, F be the tangency
points of the incircle with sides BC, CA, AB respectively. We easily get that B1 C1 k EF and
AI ⊥ EF , so EF k M N . Moreover from Theorem 1 and Theorem 2, we get that the lines M N ,
BC, TA A1 are concurrent at a point U , that
6
ITA U = 6 IDU = 90◦
and that U , I, D, TA are concyclic (See Picture 2 ).
Now let H 0 be the orthocenter of triangle DEF . Then DH 0 ⊥ M N and since U IDTA is cyclic
we get
6 IDK = 6 IDH 0 = 90◦ − 6 DIU = 6 DU I = 6 ITA D.
And since IK · ITA = ID2 = r2 this implies that lines DH 0 and ITA intersect at K.
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Now, since 4IKD ∼ 4IDTA we have that
KD =
DTA
ID
· DTA = r ·
.
ITA
ITA
(∗)
The line A1 A2 passes through O and is perpendicular to BC. So, since U IDTA is cyclic
6
IA2 A1 = 6 TA A2 A1 = 6 TA U D = 6 TA ID
and since 6 A1 IU = 6 ITA U = 90◦ we get that
6
A2 IA1 = 180◦ − 6 TA IA1 = 180◦ − 6 TA U I = 6 TA DI.
Hence 4IDTA ∼ 4A1 IA2 .
From (∗) we have
KD = r ·
6 BAC
6 BAC IA1
DH 0
DTA
=r·
= r · sin
= r · cos 90◦ −
=
ITA
A1 A2
2
2
2
and thus K is the midpoint of DH 0 .
Lemma 1. Let ABC be an arbitrary triangle and let H be its orthocenter. Let O be the
circumcenter of triangle ABC and let O0 be the center of the nine-point circle of triangle ABC (the
midpoint of segment OH). Suppose that a point S lies on the nine-point circle of triangle ABC
and let l be the line that passes through S such that OS ⊥ l. If l cuts the circumcircle of triangle
ABC at points L, J then the nine-point circle of triangle ALJ passes through the midpoint of AH.
Proof of Lemma 1. Let ray HS intersect the circumcircle of triangle ABC at point T . Obviously,
HS = ST and LHJT is parallelogram. So
6
LHJ = 6 LT J = 180◦ − 6 BAC.
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Let A0 , J1 and L1 be the midpoints of segments AH, AL and AJ, respectively. Then J1 A0 k LH,
L1 A0 k JH and 6 J1 A0 L1 = 6 LHJ = 180◦ − 6 BAC = 180◦ − 6 J1 P L1 . Hence the points J1 , A0 , L1
and P are concyclic. Therefore, the midpoint of AH lies on the ninepoint circle of triangle ALJ,
as desired. (See Picture 3 ). This proves Lemma 1.
Now, we will use the Lemma 1 and prove the THEOREM. Let P Q∩(I) = R1 and P R∩(I) = Q1
(See the Picture 2 ). Suppose that after inversion about (I)
P → P 0,
X → X 0,
Y → Y 0,
A → A0 .
Then P 0 , X 0 , Y 0 , A0 are the midpoints of segments Q1 R1 , DR1 , DQ1 and EF , respectively. We have
1
1
TA → K, (O) → (DEF ) 2 where, K be the midpoint of DH 0 and (DEF ) 2 is nine-point circle of
1
the triangle DEF . Obviously, (DEF ) 2 passes through points K and P 0 . And, line Q1 R1 passes
through P 0 and satisfies IP 0 ⊥ Q1 R1 . Therefore from Lemma 1, the point K lies on ninepoint
circle of the triangle DQ1 R1 . So the points K, X 0 , Y 0 and P 0 are concyclic.
Hence the points P , TA , X and Y are also concyclic. This completes the proof of THEOREM.
References
[1] P.Yiu, Mixtilinear incircles, American Mathematical Monthly, Vol. 106, No. 10. (Dec.,1999),
pp. 952-955.
[2] K.L.Nguyen, J.C.Salazar, On Mixtilinear Incircles and excircles, Forum Geometricorum, Vol.
6, 2006.
[3] C.Pohoata, V.Zajic, On a Mixtilinear Coaxality, Mathematical Reflections, Iss. 1, 2012.
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[4] http://www.artofproblemsolving.com/community/c2335h1043724
Khakimboy Egamberganov
National University of Uzbekistan
Tashkent, Uzbekistan
E-mail:
[email protected]
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