53 Observations about MANEEAL’S - a peculiar system of lines Dasari Naga Vijay Krishna Department of Mathematics, Keshava Reddy Educational Intuitions Machiliaptnam, Kurnool, INDIA [email protected]Abstract: In any triangle MANEEAL’S are the system of lines or cevians which divides the opposite sides in the ratio of nth powers (where n € Z) of other two sides and the point of intersection of Maneeal’s are called as MANEEAL POINTS OF ORDER n. Now in this paper let us discuss the properties of these peculiar type of lines. Keywords: Maneeal’s, Maneeal points of order n, Maneeal’s triangle of order n, Maneeal’s pedal triangle, Centroid, in center, symmedian point, Stewarts theorem, Cauchy Schwarz inequality, AM-GM inequality Introduction In any triangle, if we draw the cevians which exactly divides the opposite sides in the ratio of integral powers of remaining two sides then these cevians are called as Maneeal’s. For example, if we draw the cevians such that they divide the opposite sides in the ratio of zeroth powers of other two sides clearly these lines are medians so the medians are the examples of Maneeal’s. Similarly, If we draw the cevians such that they divide the opposite sides in the ratio of 1 st powers of the remaining sides, clearly these lines are angular bisectors. So angular bisectors are also the examples of Maneeal’s. In the similar way Symmedians are also examples of Maneeal’s. Formal definitions: Maneeal’s point of order n: Let ∆ABC is a given reference triangle and AD, BE, CF are the cevians which divides opposite sides BC,CA,AB in the ratio of nth powers (where n € Z) of the remaining sides then the lines AD,BE,CF are called as Maneeals and their point of intersection Mn is called as ‘Maneeal’s point of order n’.
Transcript
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Observations about MANEEAL’S - a peculiar system of lines
Dasari Naga Vijay Krishna
Department of Mathematics, Keshava Reddy Educational Intuitions Machiliaptnam, Kurnool, INDIA
Abstract: In any triangle MANEEAL’S are the system of lines or cevians which divides the opposite sides in the ratio of nth powers (where n € Z) of other two sides and the point of intersection of Maneeal’s are called as MANEEAL POINTS OF ORDER n. Now in this paper let us discuss the properties of these peculiar type of lines. Keywords: Maneeal’s, Maneeal points of order n, Maneeal’s triangle of order n, Maneeal’s pedal triangle, Centroid, in center, symmedian point, Stewarts theorem, Cauchy Schwarz inequality, AM-GM inequality
Introduction In any triangle, if we draw the cevians which exactly divides the opposite
sides in the ratio of integral powers of remaining two sides then these cevians are called as Maneeal’s. For example, if we draw the cevians such that they divide the opposite sides in the ratio of zeroth powers of other two sides clearly these lines are medians so the medians are the examples of Maneeal’s. Similarly, If we draw the cevians such that they divide the opposite sides in the ratio of 1st powers of the remaining sides, clearly these lines are angular bisectors. So angular bisectors are also the examples of Maneeal’s. In the similar way Symmedians are also examples of Maneeal’s. Formal definitions: Maneeal’s point of order n:
Let ∆ABC is a given reference triangle and AD, BE, CF are the cevians which divides opposite sides BC,CA,AB in the ratio of nth powers (where n € Z) of the remaining sides then the lines AD,BE,CF are called as Maneeals and their point of intersection Mn is called as ‘Maneeal’s point of order n’.
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Maneeal’s triangle of order n:
Let ∆ABC is a given reference triangle and AD, BE, CF are the cevians which divides opposite sides BC,CA,AB in the ratio of nth powers(where n € Z) of the remaining sides then the lines AD,BE,CF are called as Maneeals and the triangle formed by joining the feet of Maneeal’s AD,BE,CF is called as ‘Maneeal’s triangle of order of n’. Maneeal’s pedal triangle of order n: Let Mn is Maneeal’s point of order n of ∆ABC and MnP, MnQ and MnR are perpendiculars drawn from Mn to the sides BC,CA,AB then the triangle formed by joining the points P,Q and R is called as ‘Maneeal’s pedal triangle of order n’. Basic Lemmas:
1. If AD, BE,CF are Maneeal’s which intersect opposite sides BC,CA,AB at D,E,F respectively then BD: DC = bn : cn CE: EA = cn : an AF: FB = an : bn
since Clearly by the converse of Ceva’s theorem, all the Maneeal’s of n-th order are concurrent and the point of concurrency Mn is called as Maneeal’s point of order n.
2. If AD, BE, CF are Maneeal’s which intersect opposite sides BC, CA,AB at
D,E,F respectively then: BD = bna / bn + cn and CD = cna / bn + cn CE = cnb / cn + an and EA = anb / cn + an AF = anc / an + bn and FB = bnc / an + bn
3. If AD,BE,CF are Maneeal’s which intersect opposite sides BC,CA,AB at
D,E,F respectively and If Mn is the Maneeal’s point of order n then
1
nn
nn
nn
ba
ac
cb
FBAF
EACE
DCBD
nnn
n abc
DMAM n n
nn
n bca
EMBM n n
nn
n cba
FMCM n
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4. If AD,BE,CF are Maneeal’s which intersect opposite sides BC,CA and AB at
D,E,F respectively and If Mn is the Maneeal’s point of order n then
5. If a, b and c are the lengths of sides of triangle ABC, and AD,BE,CF are
Maneeal’s of nth order which intersect opposite sides BC,CA,AB at D,E,F respectively then the lengths of Maneeal’s are given by
6. If a, b and c are the lengths of sides of triangle ABC, and AD,BE,CF are
Maneeal’s of nth order which intersect opposite sides BC,CA,AB at D,E,F respectively then
Now let us prove some theorems related to Maneeal’s
THEOREM 1: If a, b, c are the sides of given reference triangle ABC and if ∆n is the area of Maneeal’s triangle of order n then prove that
and also prove that
PROOF: A
F E
CDB
Mn
Let AD,BE and CF are Maneeal’s and let Mn is Maneeal’s point of order n and let ∆DEF is Maneeal’s triangle of order n whose area is ∆n and ∆ is area of ∆ABC. That is
Now we have
nnnnnnnnn
n accbbacba
2
nn
nDEFABC , nn
n
bccABEABC
ABEACAE
nnnn
nn
bacacaBEFBEA
BEFABBE
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In the similar manner
And
and we have
In the similar manner
now area of ∆DEF = [∆DEF] = [∆BDE] + [∆BEF] – [∆BDF]
Now put n = -n then
nnn
caaBECBAC
BECCACE
nnnnnn
cacbcaBEDBEC
BEDBCBD
nnn
cacABDABC
ABDCBBD
nnnnnn
bcbacaBDFBDA
BDFABBF
nnnnnn
nnnnnn
nnnnnn
cbabca
bacaca
cacbcaDEF
nnnnnnnnnnnn
nn cacbbabacacbca
nnnnnnnnn
accbbacba
2
nnnnnnnnn
n accbbacba
2
nnnnnn
nnnn
accbbacba
1111111112
nnnnnnnnnn
accbbacba 2
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Hence
COROLLORIES:
1. If P,Q and R are the feets of perpendiculars drawn from Mn then triangle formed by joining P,Q and R is called as Maneeal’s pedals triangle of order n. If ∆1n is its area then
Where R is circum radius
2. The sides of Maneeal’s pedals triangle ∆PQR is given by
THEOREM 2: If Mn is the Maneeal’s point of order n of ∆ABC and X be any point in the plane of triangle ABC then
PROOF: We have BD : DC = cn : bn Now from ∆BXC, DX is a cevian, so by Stewarts theorem we have
nn
nnnnnn
nnnn cbacba
R 222
221121 2
22222 )(2 cbababacbaPQ nnnnnnnnn
22222 )(2 acbcbcbcbaQR nnnnnnnnn
22222 )(2 bacacaccbaRP nnnnnnnnn
2222222
2222,,
2 nnnnnnnnncba nnn
nn accbbacba
cbaAXcbaaXM
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A
X
CDB
Mn
And We have AMn : MnD = cn + bn : an Now from ∆AXD, XMn is a cevian, so by Stewarts theorem we have
CDBDBXBCCDCXBC
BDDX .222
2222
nnnn
nnn
nnn
cbacbBXbc
bCXbcc
DMAMAXADDMDXAD
AMXM nnnnn .222
2222
nnnn
nnn
nnn
nnnnn
cbacbBXbc
bCXbcc
cbabc
222
22
nnnn
nnnnnn
nnnn
cbacbADcba
acbAXcbaa
22
22,, ADcba
acbbccba
cbaAXcbaa
nnnnnn
nnnnnnn
cba nnnn
nnnnnnn
cba nnnn
bccbacbaAXcba
a 22
,,
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COROLLORIES:
1. If n=0 then Mn = M0 = G = Centroid
2. If n=1 then Mn = M1 = I = In centre
3. If Mn is the Maneeal’s point of order n of ∆ABC then
And
222222
222
nnnnnnnnnnn
nnn cbacbcbcbcb
cbaacb
2222222
2222,,
nnnnnnnnncba nnn
n accbbacbacbaAXcba
a
2222222
2222,,
2 nnnnnnnnncba nnn
nn accbbacba
cbaAXcbaaXM
2020202020202000
2222,, 000
02 accbbacbacbaAXcba
aGX cba
2222,, 9
131 cbaAXcba
cbaabcAXcba
aIX cba 2,,
2
222nnn CMBMAM
nnnnnnnnnnnnnnn babacacacbcbcbacba 3331 222
222n
nn
nn
n CMcBMbAMa nnnnnn
nnn bacacbcbacba2221
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4. If Mm and Mn are Maneeal’s points of orders m and n respectively then the
distance between any two maneeal’s points is given by
Where
For example If m=1 and n=0, that is Mm = M1= I = in center and Mn = M0= G =Centroid Hence
Where
So
THEOREM 3: Prove that among all the Maneeal points (Mn) of order n, the point M2=Symmedian point(L) is the point which makes the sum MnP2+MnQ2+MnR2 minimum. Where P,Q and R are feet’s of perpendiculars drawn from Mn to the sides BC,CA and AB of ∆ABC
But by Angel form of Cauchy – Schwarz Inequality we have
So
That is MnP2+MnQ2+MnR2 takes minimum value ∆ at n=2 Since LP2+LQ2+LR2 = ∆ , where L is symmedian point. Hence the point Mn=M2=L=Symmedian point, minimizes the sum MnP2+MnQ2+MnR2.
nnnn
n cbaaPM
12nnn
nn cba
bQM 12
nnnn
n cbacRM
12
2
22
22
22
2222 4cc
bb
aa
cbaRMQMPM nnnnnnnnn
222
22
22
22
2
cbacba
cc
bb
aa nnnnnn
2222222 4
cbaRMQMPM nnn
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THEOREM 4: Prove that among all the Maneeal points(Mn) of order n, the point M0=Centroid(G) is the point which makes the sum a2 MnP2+ b2 MnQ2+ c2 MnR2 minimum. Where P,Q and R are feets of perpendiculars drawn from Mn to the sides BC,CA and AB of ∆ABC PROOF:
A
R Q
CPBMn
We have by lemma 9
So
But by Angel form of Cauchy – Schwarz Inequality we have So
Hence a2 MnP2+ b2 MnQ2+ c2 MnR2 takes minimum value ∆ when equality holds in Cauchy Schwarz inequality, This could be happen when an = bn = cn . That is either a = b = c or n=0. So a2 MnP2+ b2 MnQ2+ c2 MnR2 is minimum when n=0. Hence the point Mn=M0=G=Centroid, minimizes the sum a2 MnP2+ b2 MnQ2+ c2 MnR2.
nnnn
n cbaaPM
12nnn
nn cba
bQM 12
nnnn
n cbacRM
12
nnnnnn cbacbaRMcQMbPMa nnn
2222
2222222 4
2222 nnnnnn cbacba
34 2222222 RMcQMbPMa nnn
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THEOREM 5: Prove that among all the Maneeal points(Mn) of order n, the point M1=In center(I) is the point which makes the sum + + minimum. Where P,Q and R are feet’s of perpendiculars drawn from Mn to the sides BC,CA and AB of ∆ABC PROOF:
A
R Q
CPBMn
We have by lemma 9
So
But by Angel form of Cauchy – Schwarz Inequality we have
So
Hence + + takes the minimum value , when equality holds in the Cauchy Schwarz inequality, this could be happening when either a =b =c or n = 1. So + + is minimum when n = 1 Hence the point Mn=M1=I=In center, minimizes the sum + + .
nnnn
n cbaaPM
12nnn
nn cba
bQM 12
nnnn
n cbacRM
12
nnn
nnn
nnn cc
bb
aacba
RMc
QMb
PMa 222
2
nnnnnn cbacba
cc
bb
aa
2222
rscba
RMc
QMb
PMa
nnn
22
2
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THEOREM 6: Prove that among all the Maneeal points(Mn) of order n, the point M0=Centroid(G) is the point which makes the product (MnP)(MnQ)(MnR)maximum, Where P,Q and R are feets of perpendiculars drawn from Mn to the sides BC,CA and AB of ∆ABC PROOF:
A
R Q
CPBMn
We have by lemma 9
So
But by AM-GM inequality we have
So
So the product (MnP)(MnQ)(MnR) takes the maximum value as ∆ , when equality holds in AM-GM inequality. This could be happening when either a =b=c or n=0
nnnn
n cbaaPM
12nnn
nn cba
bQM 12
nnnn
n cbacRM
12
311138.. nnn
nnnnnn cba
cbaRMQMPM
nnnnnn
nnnnnn
nnnnnn
cbacba
cbacbacbacba
2711
273
3
33
cbaabccbacbaRMQMPM nnn
nnnnnn 3
232
32
278
278.. 31113
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Hence the point Mn=M0=G=Centroid, maximizes the product (MnP)(MnQ)(MnR). In the similar argument whatever we adopted to prove the theorems 3,4,5 and 6 we can also prove the following corollaries.
1. Among all the Maneeal points (Mn) of order n, the point M0 = G = Centroid is the point which make the sum ( ) + ( ) + ( ) minimum.
2. Among all the Maneeal points (Mn) of order n, the point M0 = G = Centroid is the point which make the sum ( ) + ( ) + ( ) minimum.
3. Among all the Maneeal points (Mn) of order n, the point M1 = I = In center is the point which make the sum a ( ) + ( ) + ( ) minimum.
4. Among all the Maneeal points (Mn) of order n, the point M2 = L = symmedian point is the point which make the sum ( ) + ( ) + ( ) minimum.
5. Among all the Maneeal points (Mn) of order n, the point M2 = L = symmedian point is the point which make the sum + + minimum, where P, Q and R are the feets of perpendiculars drawn from Maneeal’s point(Mn) of order n to the sides BC,CA and AB.
THEOREM 7: If ∆ABC is a given triangle and AD,BE and CF are Maneeal’s, Let X,Yand Z are the points of intersection of extensions of AD,BE and CF with circumcircle of ∆ABC then prove that the lines ADX, BEY and CFZ are also act as Maneeal’s of ∆XYZ with same order n as AD,BE and CF only when n=2. That is prove that the symmedian points of ∆ABC, ∆XYZ coincide each other.
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PROOF: A
Y
C
X
B
ZF D1
E
F1E1
DMn
Let AD, BE and CF are Maneeal’s of ∆ABC with order n and X, Y and Z are the points of intersections of extensions of AD, BE and CF with the circumcircle of ∆ABC. Now let D’, E’ and F’ are points of intersections of AX, BY and CZ with YZ, XZ and XY Now we know by chords property,
In the similar manner we can find
DXADDCBD .. ADcb
cbaDX nnnn
22
CFbabacFZandBEca
cabEY nnnn
nnnn
22
22
70
And we have
Similarly,
Now
Similarly,
Now clearly by angle chasing, we have ∆XMnZ ~ ∆ CMnA
Hence
ADcbcba
cbaADaDXDMXM nn
nnnnn
nnn 2
2
ADcbcbabacacbcba
nnnnnnnnnnn
222
BEaccbabacacbcbaYM nnnnn
nnnnnnn
222
CFbacbabacacbcbaZM nnnnn
nnnnnnn
222
ADcbcbaADDXADAX nn
nn2
2
ADcbbccb
nnnn
22
CFbaabbaCZandBEca
accaBY nnnn
nnnn
2222
222
nn
nn
nn
AMZM
CAXZ
CMXM
ACMZXMso
CFADcbbabacacbcbab
AMZMACXZ nnnn
nnnnnn
nn
.. 222
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Similarly,
And
Similarly,
And now we have
And we have
BECFabcabacacbcbaaYZandBEADcbca
bacacbcbacXY nnnnnnnnnn
nnnnnnnnnn
..222222
222222222
.CFADcbbacbabacacbcbabACMCA
XZZXM nnnnnnnnnnnnnn
nn
22222222
.BECFcaabcbabacacbcbaaZYM nnnnnnn
nnnnnnnn
22222222
.BEADcbcacbabacacbcbacYXM nnnnnnn
nnnnnnnn
YZMYMBYZBYBY
YMZBY
YZMn
nnn
222222222
.BECFcaabaccabacacbcbaa
nnnnnnnnnnnnn
222222222
.BECFcaabaccabacacbcbaaZBY nnnn
nnnnnnnnn
YXMYMBYXBYYM
BYYXM
BYXn
nnn
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Now clearly from the figure,
Hence YE’ is a symmedian if n=2 So the lines YE’, XD’ and ZF’ are symmedians of∆ XYZ if n=2 Hence if n=2 and if the lines AD, BE and CF are symmedians of ∆ ABC, then the lines XD’, YE’ and ZF’ are also acts as symmedians of ∆ XYZ. That is the symmedian point of ∆ ABC coincides with the symmedian point of ∆ XYZ. THEOREM 8: If MnP, MnQ and MnR are perpendicular s drawn from the Maneeal’s point(Mn) of order n to the sides BC,CA and AB then ∆ PQR is Maneeal’s pedals triangle of order n. If L,M and N are the points of intersection of extensions of lines RMn , PMn and QMn respectively with the sides PQ, QR and RP of Maneeal’s pedals triangle ∆ PQR of order n. then prove that
1. Symmedian point acts as Centroid of its pedals triangle. 2. If n=1 then the points F,D and E which are the feets of Maneeal’s on the
sides AB,BC and CA and the points L,M and N which are the feets of lines
222222222
.BEADcbacaccabacacbcbacXBY nnnn
nnnnnnnnn
2222
''
CFabcADcba
BXYZBY
XEYE
nnnnnn
CFabcADcba
YXYZbut nn
nn
222222
22
CFabcADcba
YXYZ
nnnn
2''
22 nYX
YZXE
ZEhence
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drawn through R,P and Q and through Mn divides AB,BC ,CA and PQ, QR, RP in the same ratio in anti-parallel way. That is
PROOF :
Mn
A
R Q
CPBL
M
N
F
E
D We have
Similarly
Now to prove 1 Consider
So L is the midpoint of PQ Hence if n=2, the points L, M and N are acting as mid points of sides of ∆ PQR That is Mn=M2= Symmedian point is acting as Centroid of its pedals triangle’ Now to prove 2
NPRN
EACEandMR
QMDCBD
LQPL
BFAF ,
22
nn
nn
ba
QRMPRM
LQPL
22
22
n
nnn
ac
NPRNandc
bMRQM
22
n
n
ba
LQPL
LQPLba
LQPLnif
12 2222
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Consider
Similarly,
Hence proved These are very few properties of Maneeal’s Acknowledgment The author is grateful to the creators of the free Geogebra software, without which this work would have been impossible and the author is would like to thank an anonymous referee for his/her kind comments and suggestions, which lead to a better presentation of this paper. References : 1. Applications of Stewart's theorem in Geometric Proofs. — WILLIAM CHAU 2. College Geometry: An Introduction to the Modern Geometry of the Triangle and the Circle (Dover Books on Mathematics) by Nathan Altshiller-Court 3. H.S.M Coxeter, Introduction to Geometry, John Wiley 8 Sons, NY, 1961 4. H.S.M Coxeter and S.L.Greitzer, Geometry Revisited, MAA, 1967 5. MODERN GEOMETRY OF A TRIANGLE by WILLIAM GALLATLY 6. Ross Honsberger: Episodes in Nineteenth and Twentieth Century Euclidean Geometry, USA 1995. 7. Weitzenbock inequality – 2proofs in a more geometrical way using the idea of “lemoine point” and “fermat point” , ggijro2.files.wordpress.com/2015/04/art68
