Ann. Univ. Paedagog. Crac. Stud. Math. 15 (2016), 51-68
DOI: 10.1515/aupcsm-2016-0005
OPEN
FOLIA 182
Annales Universitatis Paedagogicae Cracoviensis
Studia Mathematica XV (2016)
Naga Vijay Krishna Dasari, Jakub Kabat
Several observations about Maneeals - a peculiar
system of lines
Abstract. For an arbitrary triangle ABC and an integer n we define points
Dn , En , Fn on the sides BC, CA, AB respectively, in such a manner that
|AC|n
|CDn |
=
,
|AB|n
|BDn |
|AB|n
|AEn |
=
,
|BC|n
|CEn |
|BC|n
|BFn |
=
.
|AC|n
|AFn |
Cevians ADn , BEn , CFn are said to be the Maneeals of order n. In this
paper we discuss some properties of the Maneeals and related objects.
1.
Introduction
Given an arbitrary triangle ABC we consider certain cevians [1] defined (for
given integral n) in the following way.
Definition 1.1
Let ABC be a triangle and let n be an integer. There are points Dn , En , Fn on
the sides BC, CA, AB respectively, satisfying
|CDn |
|AC|n
=
,
n
|AB|
|BDn |
|AB|n
|AEn |
=
,
n
|BC|
|CEn |
|BC|n
|BFn |
=
.
n
|AC|
|AFn |
We call the cevians ADn , BEn , CFn the order n Maneeals of the triangle ABC.
It is easy to check, that medians, bisectors and symmedians of a triangle are
examples of the Maneeals with integer n = 0, n = 1, n = 2 respectively. Furthermore, for the given triangle ABC, the points Dn , En , Fn (uniquely determined by
the integer n), are vertices of a new triangle, which is said to be the Maneeal’s
AMS (2010) Subject Classification: 51M04, 51M15, 51A20.
Keywords and phrases: Maneeals, Maneeal’s Points, Maneeals triangle of order n, Maneeal’s
Pedal triangle of order n, Cauchy-Schwarz inequality, Lemoine’s Pedal Triangle Theorem.
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[52]
Naga Vijay Krishna Dasari, Jakub Kabat
triangle of order n. For Maneeals ADn , BEn , CFn we can use the Ceva’s theorem
in order to prove that they are concurrent. Their point of intersection Mn is said
to be the Maneeal’s point of order n. For given Mn we can choose points Pn , Qn ,
Rn on the sides BC, CA, AB respectively, in such a manner that line segments
Mn Pn , Mn Qn , Mn Rn are perpendicular to the corresponding sides of the triangle
ABC. Points Pn , Qn , Rn are the vertices of a next triangle [11], which is said to
be the Maneeal’s pedal triangle of order n.
In the present note we investigate properties of Maneeal’s points and pedal
triangles.
In this paper, for a given triangle ABC and an integer n the points Dn , En ,
Fn , Pn , Qn , Rn should be always understood in accordance with the definitions
given above. Furthermore, we will use the following notation:
|AB| = c, |BC| = a, |CA| = b, qn = an + bn + cn ,
∆XY Z - the area of triangle XY Z,
∆ = ∆ABC , ∆n = ∆Dn En Fn , ∆0n = ∆Pn Qn Rn ,
R - the radius of the circumcircle of triangle ABC,
S - circumcenter of triangle ABC.
2.
First properties
Theorem 2.1 (Ceva’s Theorem [4, 5])
For a given triangle with vertices A, B, and C and non-collinear points D, E,
and F on lines BC, AC, AB respectively, a necessary and sufficient condition for
the cevians AD, BE, and CF to be concurrent (intersect in a single point) is that
|BD| · |CE| · |AF | = |DC| · |EA| · |F B|.
C
D
E
A
F
B
Theorem 2.2 (Maneeal’s points)
Maneeals of order n are always concurrent.
Proof. Since the points Dn , En , Fn lie always between the vertices A, B, C, they
cannot be collinear. The equality
bn cn an
|BDn | |CEn | |AFn |
·
·
= n · n · n =1
|CDn | |AEn | |BFn |
c
a
c
together with Theorem (2.1) proves the assertion.
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Several observations about Maneeals
[53]
In the next Lemma we collect a number of properties connected with Maneeals
and related objects.
Lemma 2.3
The lenghts of the segments of Maneeal’s triangle of order n are given by the
following formulas:
|En Fn |2
=
b2n c2 (an + cn )2 + b2 c2n (an + bn )2 − bn cn (an + bn )(an + cn )(b2 + c2 − a2 )
,
(an + bn )2 (an + cn )2
|En Dn |2
=
a2n b2 (cn + bn )2 + a2 b2n (cn + an )2 − an bn (cn + an )(cn + bn )(a2 + b2 − c2 )
,
(cn + bn )2 (cn + an )2
|Fn Dn |2
=
a2n c2 (bn + cn )2 + a2 c2n (bn + an )2 − an cn (bn + an )(bn + cn )(a2 + c2 − b2 )
.
(bn + cn )2 (bn + an )2
Lemma 2.4
The lenghts of the segments in which the Manneals of order n divide the sides of
the triangle are given by the following formulas:
cn · a
,
+ cn
an · b
|CEn | = n
,
c + an
bn · c
,
|AFn | = n
b + an
|BDn | =
bn
bn · a
,
+ cn
cn · b
|AEn | = n
,
c + an
an · c
|BFn | = n
.
b + an
|CDn | =
bn
Lemma 2.5
The Manneal’s point Mn of order n divides the Manneal’s segments of order n in
the following ratios:
|AMn |
cn + bn
=
,
|Mn Dn |
an
|BMn |
cn + an
=
,
|Mn En |
bn
|CMn |
an + bn
=
.
|Mn Fn |
cn
Lemma 2.6
It is also convenient to keep, for further reference, record of the following identities:
|AMn |
cn + bn
=
,
|ADn |
qn
|Mn Dn |
an
=
,
|ADn |
qn
|BMn |
cn + an
=
,
|BEn |
qn
|Mn En |
bn
=
,
|BEn |
qn
|CMn |
an + bn
=
,
|CFn |
qn
|Mn Fn |
cn
=
,
|CFn |
qn
(1)
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[54]
Naga Vijay Krishna Dasari, Jakub Kabat
and
b2 c2
[(bn + cn )(bn−2 + cn−2 ) − a2 bn−2 cn−2 ],
(bn + cn )2
a2 c2
|BEn |2 = n
[(an + cn )(an−2 + cn−2 ) − b2 an−2 cn−2 ],
(a + cn )2
b2 a2
|CFn |2 = n
[(bn + an )(bn−2 + an−2 ) − c2 bn−2 an−2 ].
(b + an )2
|ADn |2 =
(2)
Lemma 2.7
The Maneeal’s point of order n and every pair of vertices of the given triangle
ABC determine new triangles. Areas of these triangles are related to the areas of
the triangle ABC in the following way:
∆BMn C =
an
· ∆,
qn
∆CMn A =
bn
· ∆,
qn
∆BMn A =
cn
· ∆.
qn
(3)
We can also consider the triangles determined by pairs of points Dn , En , Fn and
one of the vertices of the triangle ABC. Their areas are given by the following
formulas:
an cn
∆BDn Fn = n
· ∆,
(b + cn )(bn + an )
an bn
∆CDn En = n
· ∆,
(4)
(c + bn )(cn + an )
bn cn
· ∆.
∆AEn Fn = n
(a + cn )(an + bn )
Lemma 2.8
It is also worth to note some properties connected with the lengths of line segments
between the vertices of pedal Maneeals triangles of order n, the Maneeals points of
order n, and vertices of the triangle ABC.
|Mn Pn | =
|BPn | =
|BRn | =
|CPn | =
|CQn | =
|AQn | =
|ARn | =
2∆an−1
,
qn
1
qn
1
qn
1
qn
1
qn
1
qn
1
qn
|Mn Qn | =
2∆bn−1
,
qn
|Mn Rn | =
2∆cn−1
.
qn
·
p
a2 c2 (an + cn )(an−2 + cn−2 ) − b2 an cn − 4a2n−2 ∆2 ,
·
p
a2 c2 (an + cn )(an−2 + cn−2 ) − b2 an cn − 4c2n−2 ∆2 ,
·
p
a2 b2 (an + bn )(an−2 + bn−2 ) − c2 an bn − 4a2n−2 ∆2 ,
·
p
a2 b2 (an + bn )(an−2 + bn−2 ) − c2 an bn − 4b2n−2 ∆2 ,
·
p
c2 b2 (cn + bn )(cn−2 + bn−2 ) − a2 cn bn − 4b2n−2 ∆2 ,
·
p
c2 b2 (cn + bn )(cn−2 + bn−2 ) − a2 cn bn − 4c2n−2 ∆2 .
(5)
(6)
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Several observations about Maneeals
[55]
Corollary 2.9
By definition we have |BPn | + |CPn | = a, |CQn | + |AQn | = b, |BRn | + |ARn | = c.
Hence, by adding all equations in (6), we get
(a + b + c) · (an + bn + cn )
p
= a2 c2 (an + cn )(an−2 + cn−2 ) − b2 an cn − 4a2n−2 ∆2
p
+ a2 c2 (an + cn )(an−2 + cn−2 ) − b2 an cn − 4c2n−2 ∆2
p
+ a2 b2 (an + bn )(an−2 + bn−2 ) − c2 an bn − 4a2n−2 ∆2
p
+ a2 b2 (an + bn )(an−2 + bn−2 ) − c2 an bn − 4b2n−2 ∆2
p
+ c2 b2 (cn + bn )(cn−2 + bn−2 ) − a2 cn bn − 4b2n−2 ∆2
p
+ c2 b2 (cn + bn )(cn−2 + bn−2 ) − a2 cn bn − 4c2n−2 ∆2 .
Now we are in a position to prove the following new identity.
Theorem 2.10
∆n = 2 ·
(an
+
an bn cn
· ∆,
+ cn )(cn + an )
bn )(bn
∆−n = ∆n .
Proof. We will start with simple consequences of the definition of Maneeals:
A
Fn
En
Mn
B
Dn
|AEn | =
cn
|En C|,
an
|BFn | =
C
an
|Fn A|.
bn
From Lemma 2.4 we obtain
|AEn |
cn
= n
,
|AC|
c + an
|BFn |
an
= n
.
|BA|
a + bn
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On the other hand we can observe, that
∆ABEn
|AEn |
=
|AC|
∆
and
|BFn |
∆BEn Fn
=
.
|BA|
∆BEn A
Finally we can express the area of the triangle BEn Fn by the formula
|BFn |
an
· ∆BEn A
· ∆BEn A = n
|BA|
a + bn
|AEn |
cn
an
an
·
· n
·∆
= n
·∆= n
n
n
a +b
|AC|
a +b
c + an
an cn
= n
· ∆.
(a + bn )(cn + an )
∆BEn Fn =
Strictly analogously, we can provide the following formulas:
cn an
· ∆,
+ bn )(an + cn )
an cn
= n
· ∆.
(a + bn )(cn + bn )
∆BEn Dn =
∆BFn Dn
(cn
To end the first part of the theorem, we only need to note, that
∆n = ∆BEn Dn + ∆BEn Fn − ∆BFn Dn .
Finally we get
an cn
an cn
cn an
·∆+ n
·∆− n
·∆
n
n
n
n
n
n
+ b )(a + c )
(a + b )(c + a )
(a + bn )(cn + bn )
an cn
= n
[(an + bn ) + (bn + cn ) − (cn + an )] · ∆.
(a + bn )(bn + cn )(cn + an )
∆n =
(cn
n n n
a b c
Above consideration implies that ∆n = 2 · (an +bn )(b
n +cn )(cn +an ) · ∆.
The last part of the proof is quite formal.
a−n b−n c−n
·∆
(a−n + b−n )(b−n + c−n )(c−n + a−n )
a−n b−n c−n
a2n b2n c2n
= 2 · −n
·
∆
·
(a + b−n )(b−n + c−n )(c−n + a−n )
a2n b2n c2n
n n n
a b c
=2· n
·∆
(a + bn )(bn + cn )(cn + an )
= ∆n .
∆−n = 2 ·
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Several observations about Maneeals
[57]
Now we pass to the Maneeal’s pedal triangle Pn Qn Rn .
Lemma 2.11
The area of Maneeals pedal triangle of order n is given by the following formula
∆0n =
22n−2 ∆n+1 Rn−2 2−n
[a
+ b2−n + c2−n ].
(an + bn + cn )2
Furthermore, sides of Maneeals pedal triangle have the following lengths:
2∆ p n−2
(a
+ bn−2 )(an + bn ) − an−2 bn−2 c2 ,
qn
2∆ p n−2
|Qn Rn | =
(b
+ cn−2 )(bn + cn ) − bn−2 cn−2 a2 ,
qn
2∆ p n−2
|Rn Pn | =
(c
+ an−2 )(cn + an ) − cn−2 an−2 b2 .
qn
|Pn Qn | =
Theorem 2.12
For any point X in the plane we have the following formula
|Mn X|2 =
an |AX|2 + bn |BX|2 + cn |CX|2
an + bn + cn
2 2 2
a b c
− n
(an−2 bn−2 + bn−2 cn−2 + cn−2 an−2 ).
(a + bn + cn )2
X
B
Dn
C
Proof. From the Stewart Theorem [3] for triangle XBC, we get
|XB|2 · |Dn C| + |XC|2 · |Dn B| = |BC| · [|XDn |2 + |Dn C| · |Dn B|],
or equivalently
|XB|2 ·
|Dn C|
|Dn B|
+ |XC|2 ·
= |XDn |2 + |Dn C| · |Dn B|.
|BC|
|BC|
If we use formulas from Lemma 2.4 and make a few simple transformations, then
we will get
|XDn |2 =
cn
bn
bn cn
2
2
|XC|
+
|XB|
−
|BC|2 .
c n + bn
c n + bn
(cn + bn )2
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A
X
Mn
Dn
B
C
Furthermore, an analogous consideration for the triangle AXDn shows that
|AMn |
|Dn Mn |
|Dn X|2 +
|AX|2 − |AMn | · |Dn Mn |
|ADn |
|ADn |
|Dn Mn |
|AMn | |Dn Mn |
|AMn |
|Dn X|2 +
|AX|2 −
·
· |ADn |2
=
|ADn |
|ADn |
|ADn | |ADn |
cn + bn cn
bn
bn cn a2 2
2
=
|XC|
+
|XB|
−
an + bn + cn cn + bn
c n + bn
(cn + bn )2
n n
n
n
a (c + b )
a
|AX|2 − n
|ADn |2
+ n
a + bn + cn
(a + bn + cn )2
an |AX|2 + bn |BX|2 + cn |CX|2
a2 bn cn
=
− n
n
n
n
n
a +b +c
(a + b + cn )(bn + cn )
an (cn + bn )
|ADn |2 .
− n
(a + bn + cn )2
|XMn |2 =
Now we will use formula
|ADn |2 =
(bn
b2 c2
[(bn + cn )(bn−2 + cn−2 ) − a2 bn−2 cn−2 ]
+ cn )2
to simplify the following part of the main formula
−
an (cn + bn )
a2 bn cn
− n
|ADn |2
n
n
n
+ + c )(b + c ) (a + bn + cn )2
a2 bn cn
=− n
n
(a + b + cn )(bn + cn )
an (cn + bn )
b2 c2
− n
[(bn + cn )(bn−2 + cn−2 ) − a2 bn−2 cn−2 ]
(a + bn + cn )2 (bn + cn )2
(an
bn
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Several observations about Maneeals
=−
[59]
a2 b2 c2
[an−2 bn−2 + bn−2 cn−2 + cn−2 an−2 ].
(an + bn + cn )2
Finally we have
|Mn X|2 =
an |AX|2 + bn |BX|2 + cn |CX|2
a n + bn + c n
2 2 2
a b c
− n
(an−2 bn−2 + bn−2 cn−2 + cn−2 an−2 ).
(a + bn + cn )2
(7)
Corollaries 2.13
From (7) we obtain for n = 0, 1:
a2 + b2 + c2
|AX|2 + |BX|2 + |CX|2
−
,
3
9
a|AX|2 + b|BX|2 + c|CX|2
abc
|M1 X|2 =
−
.
a+b+c
(a + b + c)
|M0 X|2 =
Note that M0 is the centroid of the triangle ABC and M1 is the incenter of ABC.
From (7), (1), (2) we obtain the last corollary, which states, that the distance
between two Maneeal’s points, of order m and n, is given by the following formula
|Mm Mn |2 =
1
· V,
(am + bm + cm )2 (an + bn + cn )2
where
V = a2 [bm cn − bn cm ]2 − [bm cn − bn cm ][am (bn − cn ) − an (bm − cm )]
− [am bn − an bm ][am cn − an cm ]
+ b2 [cm an − cn am ]2 − [cm an − cn am ][bm (cn − an ) − bn (cm − am )]
− [bm cn − bn cm ][bm an − bn am ]
+ c2 [am bn − an bm ]2 − [am bn − an bm ][cm (an − bn ) − cn (am − bm )]
− [cm an − cn am ][cm bn − cn bm ] .
In particular if we let m = 1, n = 0, we will get
|M1 M0 |2 =
1
[a|AM0 |2 + b|BM0 |2 + c|CM0 |2 − abc].
(a + b + c)
Corollary 2.14
From (2.12) for X=S we have
an R2 + bn R2 + cn R2
an + bn + cn
a2 b2 c2
− n
(an−2 bn−2 + bn−2 cn−2 + cn−2 an−2 )
(a + bn + cn )2
≥ 0.
|Mn S|2 =
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Therefore, we have
R2 ≥
a2 bn cn + b2 an cn + c2 an bn
.
qn2
In particular, for n = 1, since
R2 ≥
abc
= 2Rr
(a + b + c)
therefore
R ≥ 2r.
Other proofs of Euler’s inequality you can find at [2, 13, 14, 6, 7, 8, 9, 10].
Now we can obtain a few relationships from the Cauchy-Schwarz inequality.
They will be important part of proofs of several subsequent theorems.
Lemma 2.15
Any non-zero real numbers a, b, c satisfy the following inequalities:
a2n
a2
+
b2n
c2n (an + bn + cn )2
+
,
≥
b2
c2
a2 + b2 + c2
(a2n + b2n + c2n ) ≥
a2
an
+
(8)
1 n
(a + bn + cn )2 ,
3
(9)
b2
c2 (a + b + c)2
+ n ≥ n
.
n
b
c
a + bn + cn
(10)
Proof. Indeed, these are special cases of the Cauchy-Schwarz inequality
(x21 + x22 + x23 )(y12 + y22 + y32 ) ≥ (x1 y1 + x2 y2 + x3 y3 )2
with the substitutions
x1 =
bn
cn
an
, x2 = , x3 = , y1 = a, y2 = b, y3 = c
a
b
c
x1 = an , x2 = bn , x3 = cn , y1 = y2 = y3 = 1
for (8)
for (9)
and
r
x1 =
a2
, x2 =
an
r
b2
, x3 =
bn
r
√
√
√
c2
, y1 = an , y2 = bn , y3 = cn
n
c
for (10).
Lemma 2.16 (Inequality of arithmetic and geometric means)
For any real numbers x1 , . . . , xn there is
√
x1 + . . . + xn
≥ n x1 · . . . · xn .
n
(11)
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Several observations about Maneeals
[61]
The Symmedian point M2 has a special feature, which we will describe by
following
Theorem 2.17
Let S(n) = |Mn Pn |2 + |Mn Qn |2 + |Mn Rn |2 . Then S(2) ≤ S(n) for all n ∈ Z.
Proof. Using (5) we get
S(n) =
b2n
c2n i
4∆2 h a2n
+
+
.
qn2 a2
b2
c2
A
Rn
Qn
Mn
B
Pn
C
Using (8) we get
S(n) ≥
4∆2
.
q2
Now, an easy computation shows, that
S(2) = |M2 P2 |2 + |M2 Q2 |2 + |M2 R2 |2 =
4∆2
.
q2
This finishes the proof.
Not only the Symmedian point, but also the Centroid M0 , has a special feature:
Theorem 2.18
Let T (n) = a2 |Mn Pn |2 + b2 |Mn Qn |2 + c2 |Mn Rn |2 . Then T (0) ≤ T (n) for all
n ∈ Z.
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Proof. Using (5) we obtain
T (n) = 4∆2 ·
q2n
.
qn2
By (9) we get
T (n) ≥
4∆2
.
3
Now an easy computation shows, that
a2 |M0 P0 |2 + b2 |M0 Q0 |2 + c2 |M0 R0 |2 =
4∆2
.
3
This finishes the proof.
Theorem 2.19
Let W (n) = |MnaPn | +
b
|Mn Qn |
+
c
|Mn Rn | .
Then W (1) ≤ S(n) for all n ∈ Z.
Proof. Using (5) we get
W (n) =
qn a2
b2
c2 + n+ n .
n
2∆ a
b
c
Using (10) we get
W (n) =
(an + bn + cn ) a2
b2
c2 (a + b + c)2
+
+
≥
.
2∆
an
bn
cn
2∆
Now an easy computation shows, that
a
b
c
(a + b + c)2
+
+
=
.
|M1 P1 | |M1 Q1 | |M1 R1 |
2∆
This finishes the proof.
Theorem 2.20
Let K(n) = |Mn Pn | · |Mn Qn | · |Mn Rn |. Then K(0) ≥ K(n) for all n ∈ Z.
Proof. Using (5) we obtain
K(n) =
8an−1 bn−1 cn−1
· ∆3 .
(an + bn + cn )3
By a special case of 2.16 we get
√
3
(an + bn + cn ) ≥ 3 an bn cn .
Equivalently,
(an
1
1
≤
.
n
n
3
n
+b +c )
27a bn cn
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Several observations about Maneeals
[63]
Using this inequality, we get
K(n) ≤
8
8an−1 bn−1 cn−1
· ∆3 =
· ∆3 .
27an bn cn
27abc
Furthermore,
√
3
(an + bn + cn ) = 3 an bn cn
for n = 0.
Hence,
|M0 P0 | · |M0 Q0 | · |M0 R0 | =
8a−1 b−1 c−1
8
· ∆3 =
· ∆3 .
27
27abc
This equality finishes the proof.
Theorem 2.21
Let π be the circumcircle of the triangle ABC. For any n ∈ Z we choose points
Xn , Yn , Zn on π in such a manner, that chords AXn , BYn , CZn contain the
Maneeals ADn , BEn , CFn respectively. Let Dn0 , En0 , Fn0 be intersection points of
AXn , BYn , CZn and Yn Zn , Zn Xn , Xn Yn respctively.
00
00
00
be order m Maneeals of the
, Z n Fm
, Yn Em
Let finally m be an integer, Xn Dm
triangle Xn Yn Zn . Then the following conditions hold:
• D200 = D20 , E200 = E20 , F200 = F20 ,
• if Gm is an m-order Maneeal’s point of triangle Xn Yn Zn , then G2 = M2 .
A
0
Dn
Zn
Yn
En
Fn
Mn
0
En
Fn0
B
Dn
C
Xn
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[64]
Naga Vijay Krishna Dasari, Jakub Kabat
Proof. Line segments BC and AXn are the chords of circle π, and Dn is their
point of intersection. Hence,
|BDn | · |Dn C| = |ADn | · |Dn Xn |.
Using 2.4 we get
|Dn Xn | =
(bn
a2 bn cn
.
+ cn )2 |ADn |
Strictly analogously we get
|En Yn | =
(an
b2 an cn
,
+ cn )2 |BEn |
|Fn Zn | =
(an
c2 an bn
.
+ bn )2 |CFn |
Now we can use (1) and (2) to obtain
a2 bn cn
an |ADn |
+
(an + bn + cn ) (bn + cn )2 |ADn |
2 n n
2 n n
a b c + b c a + c2 an bn
= n
.
(a + bn + cn )(bn + cn )|ADn |
|Mn Xn | = |Mn Dn | + |Dn Xn | =
Strictly analogously we get
a2 bn cn + b2 cn an + c2 an bn
,
(an + bn + cn )(cn + an )|BEn |
a2 bn cn + b2 cn an + c2 an bn
|Mn Zn | = n
.
(a + bn + cn )(an + bn )|CFn |
|Mn Yn | =
Using (2) again, we get
|AXn | = |ADn | + |Dn Xn | =
|ADn |2
a2 bn cn
c2 bn + b2 cn
+ n
=
.
|ADn |
(b + cn )2 |ADn |
(bn + cn )|ADn |
Analogously we get
|BYn | =
c2 an + a2 cn
,
(an + cn )|BEn |
|CZn | =
b2 an + a2 bn
.
(an + bn )|CFn |
Now we observe, that triangles Xn Mn Zn and CMn A are similar. Indeed, we
only need to note, that Mn is the intersection point of chords AXn and CZn and
that respective angles are right.
From the similarity, we derive that
∆Xn Mn Zn
|Xn Mn |2
|Xn Zn |2
|Zn Mn |2
=
=
=
.
∆CMn A
|CMn |2
|CA|2
|AMn |2
Since
|Xn Zn |
|CA|
=
|Zn Mn |
|AMn | ,
|Xn Zn | =
therefore if we use (1), we get
|AC| · |Zn Mn |
a2 bn cn + b2 cn an + c2 an bn
=b· n
.
|AMn |
(a + bn )(bn + cn )|ADn ||CFn |
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Several observations about Maneeals
[65]
Strictly analogously we get
a2 bn cn + b2 cn an + c2 an bn
,
+ cn )(bn + cn )|ADn ||BEn |
a2 bn cn + b2 cn an + c2 an bn
|Yn Zn | = a · n
.
(a + cn )(an + bn )|CFn ||BEn |
|Xn Yn | = c ·
(an
By (3) and similarity of respective triangles we have
|Xn Zn |2
· ∆CMn A
|CA|2
bn (a2 bn cn + b2 cn an + c2 an bn )2
= n
· ∆.
n
(a + b + cn )(an + bn )2 (bn + cn )2 |ADn |2 |CFn |2
∆Xn Mn Zn =
By the same taken we get
an (a2 bn cn + b2 cn an + c2 an bn )2
· ∆,
(an + bn + cn )(an + bn )2 (an + cn )2 |BEn |2 |CFn |2
cn (a2 bn cn + b2 cn an + c2 an bn )2
∆Xn Mn Yn = n
· ∆.
(a + bn + cn )(cn + bn )2 (cn + an )2 |ADn |2 |BEn |2
∆Yn Mn Zn =
No we can determine the area of the triangle Zn BYn . Since
|Yn Mn |
∆Zn Mn Yn
=
,
∆Zn BYn
|BYn |
therefore
∆Zn BYn =
BYn
an (a2 bn cn + b2 cn an + c2 an bn )(c2 an + a2 cn )
· ∆.
· ∆Zn Mn Yn =
Yn Mn
(an + bn )2 (an + cn )2 |BEn |2 |CFn |2
Strictly analogously we get
∆Xn BYn =
cn (a2 bn cn + b2 cn an + c2 an bn )(an c2 + cn a2 )
· ∆.
(cn + bn )2 (cn + an )2 |BEn |2 |ADn |2
Furthermore we have the following relation
|Zn En0 |
∆Zn BYn
an (bn + cn )2 |ADn |2
=
=
.
|Xn En0 |
∆Xn BYn
cn (bn + an )2 |CFn |2
On the other hand since
|Yn Zn |
a(bn + cn )|ADn |
=
,
|Yn Xn |
c(bn + an )|CFn |
we can conclude, that
|Yn Zn |2
|Zn En0 |
=
2
|Yn Xn |
|Xn En0 |
if and only if n = 2.
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[66]
Naga Vijay Krishna Dasari, Jakub Kabat
00
On the other hand, by definition, the cevian Yn Em
, is an order m Maneeal of
triangle Xn Yn Zn if, and only if, the following condition holds
00
|Zn Em
|
|Yn Zn |m
.
=
00 |
|Xn Em
|Yn Xn |m
Now it is easy to see, that the above condition is satisfied for n = m = 2.
Hence, E200 = E20 . We can provide strictly analogously, that D200 = D20 , F200 = F20 .
In particular, cevians X2 D20 , Y2 E20 , Z2 F20 are Symmedians of the triangle X2 Y2 Z2 .
The fact, that G2 = M2 we conclude by the definition (construction) of points X2 ,
Y2 , Z2 .
Theorem 2.22 (Lemoine’s Pedal Triangle Theorem [15])
Let Dn En Fn be the order n Maneeal’s triangle and Pn Qn Rn be the order n pedal
Maneeal’s triangle of the given triangle ABC. We can choose points Tn , Un , Wn ,
on sides Rn Pn , Qn Pn , Qn Rn , respectively, in such a manner that cevians Rn Un ,
Pn Wn , Qn Tn contain the line segments Rn Mn , Pn Mn , Qn Mn respectively. Then
the following conditions hold:
0
0
• if Rn Rm
, Qn Q0m , Pn Pm
are order m Maneeals of the triangle Pn Qn Rn , then
0
0
U2 = R0 , W2 = P0 , T2 = Q00 ,
• Symmedian point of triangle ABC is the centroid of its pedals triangle Pn Qn Rn ,
•
|AF1 |
|F1 B|
=
|P1 U1 | |BD1 |
|U1 Q1 | , |D1 C|
=
|Q1 W1 | |CE1 |
|W1 R1 | , |E1 A|
|R1 T1 |
|T1 P1 | .
=
A
Fn
Qn
Wn
Rn
Mn
En
Un
Tn
B
Dn
C
Pn
Proof. We use (5) to make simple computations
|Pn Un |
∆Rn Mn Pn
=
=
|Un Qn |
∆Rn Mn Qn
1
2 |Rn Mn ||Mn Pn | sin (∠Rn Mn Pn )
1
2 |Rn Mn ||Mn Qn | sin (∠Rn Mn Qn )
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Several observations about Maneeals
=
=
[67]
2cn−1 ∆
2an−1 ∆
(an +bn +cn ) (an +bn +cn )
2cn−1 ∆
2bn−1 ∆
(an +bn +cn ) (an +bn +cn )
n−2
sin (∠180 − B)
sin (∠180 − A)
=
an−1 sin B
bn−1 sin A
a
bn−2
and get
|Pn Un |
an−2
= n−2 .
|Un Qn |
b
(12)
In particular, if we choose n = 2, we obtain |Pn Un | = |Un Qn |. Therefore, by
definition of order m Maneeals of the triangle Pn Qn Rn the cevian R2 U2 is a median
of this triangle. Finally R2 U2 = R2 R00 . Strictly analogously we can show, that
P2 W2 = P2 P00 , Q2 T2 = Q2 Q00 .
On the other hand, if we put n = 1 into (12), we obtain
|P1 U1 |
b
|AF1 |
= =
.
|U1 Q1 |
a
|F1 B|
And analogously
|Q1 W1 |
c
|BD1 |
= =
,
|W1 R1 |
b
|D1 C|
,
|R1 T1 |
a
|CE1 |
= =
.
|T1 P1 |
c
|E1 A|
For further properties see [12].
Final remarks. In this article we introduced Maneeal’s points, which to the
best of our knowledge, have non been studied in the literature before. We shared
a number of properties of these points, related lines and triangles. There is surely
much more to discover. We hope, this note will sparkle some interest in the
construction and will lead further research in this area of very classical triangle
geometry.
Acknowledgement. The authors would like to thank Prof. Tomasz Szemberg
for his help while preparing this manuscript. Additional thanks go to the referee
for valuable comments and suggestions.
References
[1] Altshiller-Court, Nathan. "College geometry. An introduction to the modern geometry of the triangle and the circle." Reprint of the second edition. Mineola, NY:
Dover Publications Inc.: 2007. Cited on 51.
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[3] Chau, William. "Applications of Stewart’s theorem in geometric proofs." Accessed
April 7, 2016. http://www.computing-wisdom.com/papers/stewart_theorem.
pdf. Cited on 57.
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[68]
Naga Vijay Krishna Dasari, Jakub Kabat
[4] Coxeter, Harold S.M. Introduction to Geometry. New York-London: John Wiley
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[7] Dasari, Naga Vijay Krishna. "The distance between the circumcenter and any
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[8] Dasari, Naga Vijay Krishna. "A few minutes with a new triangle centre coined as
Vivya’s point." Int. Jr. of Mathematical Sciences and Applications 5, no. 2 (2015):
357–387. Cited on 60.
[9] Dasari, Naga Vijay Krishna. "The new proof of Euler’s Inequality using Spieker
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[12] Honsberger, Ross. "Episodes in nineteenth and twentieth century Euclidean geometry." Vol 37 of New Mathematical Library. Washington, DC: Mathematical
Association of America, 1995. Cited on 67.
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[14] Pohoaţă, Cosmin. "A new proof of Euler’s Inradius — Circumradius Inequality."
Gazeta Matematica Seria B no. 3 (2010): 121–123. Cited on 60.
[15] Pohoaţă, Cosmin. "A short proof of Lemoine’s Theorem." Forum Geometricorum
8 (2008): 97—98. Cited on 66.
Naga Vijay Krishna Dasari
Department of Mathematics
Kesava Reddy Educational Instutions
Machilipatnam, Kurnool, Andhra Pradesh
India
E-mail: [email protected]
Jakub Kabat
Institute of Mathematics
Pedagogical University of Cracow
Podchorazych 2
30-084 Kraków
Poland
E-mail: [email protected]
Received: August 31, 2015; final version: September 5, 2016;
available online: October 13, 2016.
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