Journal of Science and Arts Year 18, No. 2(43), pp. 381-396, 2018

ORIGINAL PAPER THE METRIC RELATION OF FEUERBACH POINT & ITS APPLICATIONS

DASARI NAGA VIJAY KRISHNA1 ______Manuscript received: 11.11.2017; Accepted paper: 18.03.2018; Published online: 30.06.2018.

Abstract. In this article we study the metric relation of Feuerbach point and its applications in proving distance properties related to this point, we also study about Feuerbach Conic and some concurrencies and collinarities concerned about this point. Keywords: Inner Feuerbach Point, Outer Feuerbach Point, Feuerbach Conic.

1. INTRODUCTION

The famous Feuerbach theorem ([1] and [2] ) states that “The nine-point circle of a triangle is tangent internally to the incircle and externally to each of the excircles”. Given triangle ABC, the inner Feuerbach point Fe is the point of tangency with the incircle and exterior Feuerbach points Fa , Fb, Fc are the points of tangency with three excircles. In this note we give the metric relations of these four points (inner and outer Feuerbach points) which are useful in giving the minimal proof for all the fundamental distance properties of these points. Using this relation we can also investigate some new properties of these points. In the conclusion of the article we discuss about a new proof of Feuerbach conic mentioned in the article [3]. In the articles [4] and [5], S´andor Nagydobai Kiss gave the synthetic proof of distance properties of the Feuerbach point, but in this short note we prove all those distance properties using a simple metric relation of this point which is the consequence of Feuerbach theorem. The metric relation whatever we deal throughout the article is actually not a new one, basing on the works of Feuerbach and Euler we just formulated it to the present form. We make use of standard notations in triangle geometry (see [6]). Given triangle ABC, denote by a, b, c the lengths of the sides BC, CA, AB respectively, s the semiperimeter, Δ the area, and R, r, ra , rb , rc the circumradius, inradius and ex radii respectively. Its classical centers are circumcenter O, incenter I, three excenters Ia, Ib, Ic, centroid G, Nine point center N and orthocenter H. bca222 Working with the distance formula, we also make use of SbcAcos , A 2 cab222 cab222 SacB cos Β and SacB cos Β 2 2 it is clear that

aSABC bS cS abccos A  cos B  cos C  4  R r .

1Narayana Educational Instutions, Machilipatnam, Bengalore, India.E-mail: [email protected].

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22 22 22 And also using Euler’s formula we haveOI R2 Rr , OIaa R2 Rr , OIbb R2 Rr and OI22 R2 Rr cc

2. MAIN RESULTS

Theorem 2.1. If Fe is the inner Feuerbach point, Fa, Fb , Fc are exterior Feuerbach points of non equilateral triangle ABC and let M be a finite point in the plane of the triangle then 22 2 R IM 2 r NM Rr (A). FMe  Rr 22 22 2 R IMaa 2 rNM Rr a (B). FMa  Rr 22a 22 2 R IMbb 2 rNM Rr b (C). FMb  Rr 22b R IM22 2 rNM Rr (D). FM2 cc c c Rr 22 c

Proof:

Figure 1.

For proving (A), we apply Stewart’s theorem to triangle MFeN with Cevian IM (provided M doesn’t lie on line join of N and Fe) (Fig. 1).

We get,

222 NFee .IM = F I.NM + NI.F eee M F I.. IN NF (1)

Using Feuerbach theorem the following metric relations are well known

R NFeabc NF NF NF 2 IF r , IF r, IF r, IF r e aa a bb b cc c R  2r R  2ra R  2rb R  2rc IN  , NIa  , NIb  and NIc  2 2 2 2

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By replacing these metric relations in (1) and further simplification proves (A).

Figure2.

Now for (b), we consider the triangle MNIa in which MFa is a Cevian (Fig. 2). By applying Stewart’s theorem to triangle MNIa, we get

NI .F M222 = F I .NM + NF .I M NF.. F I NI aa aa aa aaa a

Further simplification by replacing the metric relations obtained from Feuerbach theorem gives (B). Similarly we can prove (C) and (D)

Remark:The metric relations presented in Theorem-2.1 are true even if the point M is collinear with the lines formed by join of (N, Fe), (N, Fa), (N,Fb), (N, Fc) and its proof is quite obvious. In the relations we used the directed line segments only.

Lemma 2.1.If P is any point on the nine point circle then

222R 222R (a). FPe  IP r (b). FaaaPIPr Rr 2 Rr 2 a

222R 222R (c). FbbbPIPr (d). FcccPIPr Rr 2 Rr 2 b c

Proof: For proving (a), we proceed as follows: By fixing M as P in theorem – 2.1(A), we get

22 2 R IP 2 r NP Rr FeP  (2) Rr 22 R R Since P is a point on the nine point circle, we have NP  . By replacing NP as in 2 2 (2) and further simplification gives (a). Similarly we can prove (b), (c) and (d).

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3. APPLICATIONS OF THEOREM 2.1

Theorem2.2. If Feis the inner Feuerbach point of triangle ABC, and X, Y, Z are the midpoints of the sides BC, CA, AB, respectively, then one of the distances Fe X, FeY , Fe Z is equal to the sum of other two distances[4, 5, 7].

Proof:

Figure3.

222R Using Lemma 2.1(a), we have FPe  IP r . Let us fix P as X, Y and Z Rr 2 (since X, Y and Z are the mid points of the sides so they lie on the nine point circle).

We get,

222R 222R 222R FXe  IX r , FYe  IY r and FZe  IZ r Rr 2 Rr 2 Rr 2

Let D, E, F are the points of contact of incircle with the sides BC, CA and AB (Fig. 3). So BD=BF=s-b, CD=CE=s-c and AE=AF=s-a Hence from triangle DIX, by Pythagoras theorem IXIDDX22  2

22 22 2 22aa 1 2 IX r DX BX  BD  CD  CX  s b s c b c 22 4

2 2 222RRbcR 2 So FXe  IX r 2  b c . It implies Rr2224 Rr OI

R FXe  b c 2OI

R R Similarly we can prove FY c a and FZ a b. Without loss of e 2OI e 2OI generality let us consider abc. It implies one out of the three ab , bc and ca is

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equal to the sum of the remaining two.It follows one of the distances FeX, FeY , FeZ is equal to the sum of other two distances.

Theorem2.3. If the nine-point circle touches the A-excircle at Fa , then one of FaX, FaY , FaZ is the sum of the remaining two [4, 5, 7].

Proof: Using Lemma 2.1(b) , we have

222R FaaaPIPr (3) Rr 2 a

Let us fix P as X, since X is also lies on nine point circle, we get

222R FaaaXIXr Rr 2 a

Let DA, EA, FA are the points of contact of A-excircle with the sides BC, CA and AB(Fig. 4). So BDA =BFA=s-c, AEA =AFA = s, CDA =CEA =s-b and AEA =AFA = s. Hence from triangle D I X, by Pythagoras theorem I XIDDX222 A a aaAA

22 22 2 22aa1 2 IaaAX r D X BX BD A  CD A  CX  s c s b b c 224

2 2 222RRbcR 2 So FXaaa I X r 2  b c . It implies Rr2224aaa Rr OI

R F Xbc  a 2OI a

In the similar manner to get the values FaY and Fa Z , replace P in (3) with Y and Z

222R (since Y and Z also lie on nine point circle), we get FaaaYIYr and Rr 2 a

222R FaaaZIZr Rr 2 a

Now it is very clear that for triangles EAIaY, FAIaZ, using Pythagoras theorem, we have

I YIEEY222 and I ZIFFZ222 aaAA aaAA

22 22 2 22bb1 2 IaaAY r E Y CY CE A  AE A  AY  s b s a c 224

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Figure 4.

Similarly

22 22 2 22cc1 2 IaaAZ r F Z BZ BF A AF A AZ s c s a b 224

So

2 2 222RRabR 2 R FYaaa I Y r 2  a b FaYab  Rr2224 Rr OI 2OI aaa a

2 2 222RRacR 2 R and FZaaa I Z r 2  a c Fa Zac . Rr2224 Rr OI 2OI aaa a

Without loss of generality let us consider abc. It implies one out of the three

bc , ab and ac is equal to the sum of the remaining two. It follows one of FaX, FaY ,

FaZ is the sum of the remaining two.

Remark: The similar results mentioned in Theorem 2.3 are also holds true, for the other two excircles. For proof we can proceed as we dealt in Theorem2.3.

Theorem 2.4. The distances from the Feuerbach point Feto the vertices of triangle ABC are given by

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saRS2 sbRS2 scRS 2  AF 2  A , BF 2  B and CF 2  C [4, 5, 7] e Rr 2 e Rr 2 e Rr 2

Proof: Using Theorem2.1(A), we have

22 2 R IM 2 r NM Rr FMe  (2.1A) Rr 22

Since (2.1A) is true for any M let us fix M as A, B and C, we get,

22 2 RIA 2 rNA Rr  FeA  Rr 22 22 2 RIB 2 rNB Rr  FeB  (4) Rr 22 22 2 RIC 2 rNC Rr  FeC  Rr 22 and it is well known that

IA22 r() s a 2, IBr22() sb 2, ICr22() sc 2 and

42NA22222 b c a R  S  R 2, 42NB22222 a c b R  S  R 2, 42NC22 S R A B C

By replacing these relations in (4) and further simplification proves the conclusions

2 2 2 2 saRSA 2 sbRS  B 2 scRS  C AFe  , BFe  andCFe  Rr 2 Rr 2 Rr 2

Theorem2.5. The distances from the Feuerbach point Fa to the vertices of triangle ABC are 2 2 2 2 sR rS s cRrS s bRrS given by AF  aA, BF 2  aBand CF 2  aC[4, 5, 7]. a Rr 2 a Rr 2 a Rr 2 a a a

Proof: Using Theorem2.1(B), we have

22 2 R IMaa 2 rNM Rr a FMa  (2.1B) Rr 22a

Since (2.1B) is true for any M, let us fix M as A, B and C, we get

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22 2 RIaaA  2 r NA Rr a Fa A  Rr 22 a  22  2 RIaaB  2 r NB Rr a Fa B  (5) Rr 22a  22  2 RIaaC  2 r NC Rr a FaC  Rr 22a  and we are familiar with the results,

222 22 2 22 2 IaaAr s, IaaBr() sc, IaaCr() sb and

222222 42NA SA R , 42NB SB  R , 42NC SC  R

By replacing these relations in (5) and further simplification proves the conclusions

2 2 2 2 sR rSaA 2 s cRrSaB 2 s bRrSaC AFa  , BFa  and CFa  . Rr 2 a Rr 2 a Rr 2 a

Remark: The similar results mentioned in Theorem 2.5 also holds true for the other two excircles. For proof we can proceed as we dealt inTheorem 2.5.

Theorem2.6. If Fe, Fa, Fb and Fc are the inner and outer Feuerbach points then

b-c R2 c-a R2 a-b R2 FFea = , FFeb = and FFec = [4, 5, 7]. OI.OIa OI.OIb OI.OIc

Proof: For proving first formula we proceed as follows:

Clearly from Lemma 2.1(a), we have, If P is any point on the nine point circle then

222R FPe  IP r (2.1a) Rr 2

Since (2.1a) is true for any P, let us fix P as Fa , we get

222R FFea IF a r (6) Rr 2

22 2 RIaa M 2 r NM Rr a but using (2.1B), we have FMa  Rr 22a

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22 2 RIaa I 2 r NI Rr a By replacing M as I in (2.1B) we get FIa . Now from (6), Rr 22a

22 22R RIaa I 2 r NI Rr a FFea r (7) Rr222 Rra

A R  2r Further simplification by replacing IIa  4sin R  and NI  , gives 2 2

2 2322RARr  2  FFea32 R sin 4 r a   Rr a R 2 r a 2 r R 2 r a 22RrRr 2 2  2  a 

2 2222RA 2 It further implies FFea16 R sin  r r a . Further 22RrRr 2a  2 A BC  simplification by replacing rra 4sin R  cos  , gives 22  

42A 16R sin  222 B C FFea1cos RrRr22 2 a 

62A 2 BC  44sinR  sin   64 224RR22 FF2   sinBC sin  bc ea OI22. OI OI22.. OI OI 22 OI a aa

R2 c-a R2 It implies FFea b c. Similarly we can prove FFeb = and OI. OIa OI.OIb a-b R2 FF = . ec OI.OI c

Theorem2.7. If Fa, Fb and Fc are outer Feuerbach points then

abR 2 bcR  2 caR  2 FF= , FF = and FF = [4, 5, 7]. ab OI .OI bc OI .OI ca OI .OI ab bc ca

Proof: For proving first formula we proceed as follows: Clearly from Lemma 2.1(b), we have, If P is any point on the nine point circle then

222R FaaaPIPr (2.1b) Rr 2 a

Now since (2.1b) is true for any P, let us fix P as Fb , we get

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222R FabFIFr ab a (8) Rr 2 a

22 2 RIbb M 2 r NM Rr b But using (2.1C), we have FMb  (2.1C) Rr 22b 22 2 RIba I 2 r b NI a Rr b by replacing M as Ia in (2.1C) we get FI  ba Rr 22 b Now from (7),

22 22R RIab I 2 r b NI a Rr b FFab r a (9) Rr222 Rr ab

C R  2ra Further simplification by replacing IIab 4cos R  and NIa  , gives 2 2

2 2322RC Rr 2 a FabFRrRrRrrRr32 cos 4 b b 2 b 2 a 2 b 22RrRr 2 2 2 ab

2 2222RC 2 It further implies FFab16 R cos  r a r b . 22RrRr 2 2 ab CAB   Further simplification by replacing rrab4cos R  sin   gives 22  

42C 62CAB 2   16R cos  44cosR  cos   222 A B 2 22   FFab1sinFFab 22 RrRr22 2 OI. OI ab ab

64 4RR22 sinA sin Bab OI22.. OI OI 22 OI ab ab

R2 It implies FFab  a b . Similarly we can prove OIab. OI

bcR  2 caR  2 FF = and FF = bc OI .OI ca OI .OI bc ca

Theorem2.8. If Fe, Fa, Fb and Fc are the inner and outer Feuerbach points then

22 22 22 FIabc r FI r FI r  R 2222rRrRrRr   FI22 rabc FI 22 r FI 22 r ea a eb b ec c

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222R Proof: It is clear from Lemma 2.1(a), FPe  IP r . R  2r Fix P as Fa , we get

222R FFea IF a r (10) Rr 2

Similarly by fixing P as Fe in Lemma 2.1(b), we get

222R FaeFIFr ae a (11) Rr 2 a

Now from (10) and (11), it is clear that

22 FIa  r Rr 2 NIOI 22 (12) FIea r a R2 r a NI a OI a

Similarly we can prove

22 FIb  r Rr 2 NIOI 22 (13) FIeb r b R2 r b NI b OI b

22 FIc  r RrNIOI 2 and 22 (14) FIec r c R2 r c NI c OI c

Using (12), (13) and (14) we can prove the desired result

22 22 22 FIabc r FI r FI r  R 2222rRrRrRr   FI22 rabc FI 22 r FI 22 r ea a eb b ec c

Theorem2.9. If Fa, Fb and Fc are outer Feuerbach points then

F IrFIrFIr22 22 22 FIrFIrFIr 22 22 22 abb bcccaa baa cbb acc

222R Proof: It is clear from lemma-2.1(b), FaaaPIPr Rr 2 a Fix P as Fb , we get

222R FabFIFr ab a (15) Rr 2 a

Similarly by fixing P as Fain lemma-2.1(c), we get

222R FbaFIFr ba b (16) Rr 2 b

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Now from (15) and (16), it is clear that

22 FIab r b R2 r b NI b OI b 22 (17) FIba r a R2 r a NI a OI a

Similarly we can prove

22 FIbc r c R2 r c NI c OI c 22 . (18) FIcb r b R2 r b NI b OI b and

22 FIca r a R2 r a NI a OI a 22 (19) FIac r c R2 r c NI c OI c

By multiplying (17), (18) and (19) we get desired result

F IrFIrFIr22 22 22 FIrFIrFIr 22 22 22 abb bcccaa   baa cbb acc 

To discuss one more theorem related to these points (inner and outer Feuerbach points) we need a list of lemmas which we are presenting here without proofs.

Lemma 2.2. Let Aa, Ab, Ac are the feet of internal and external angular bisector of angle A( Ab closer to the vertex B and Ac closer to the vertex C) on the side BC respectively, similarly define the points Ba, Bb, Bc, Ca, Cb and Cc. . If Fe, Fa, Fb and Fc are the inner and outer Feuerbach points then the following pair of triangles are directly similar [8, 9].

(a) The triangle with feet of internal angular bisectors of the angles A, B and C as vertices is directly similar to the triangle formed by outer Feuerbach points.

That is AabBC c FFF abc.

(b) The triangle with feet of two external angular bisectors and one internal angular bisector as vertices is directly similar to the triangle formed by two outer Feuerbach points and one inner Feuerbach point.

That is AbaBC c FFF bae, AcbBC a FFF cea and AacBC b  FFF ecb.

Lemma 2.3. The lines AFa, BFb and CFc are concurrent at X(12), which is collinear with the points Fe , I and N. [10]

Lemma 2.4. The following set of points is collinear [8]: (a) The points Fe, Aa, Fa are collinear. (b) The points Fe, Bb, Fb are collinear. (c) The points Fe, Cc, Fc are collinear. (d) The points either Cb, Fa, Fb, or Ca, Fa, Fb are collinear. (e) The points either Bc, Fa, Fc, or Ba, Fa, Fc are collinear. (f) The points either Ac, Fb, Fc, or Ab , Fb, Fc,are collinear.

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That is the feet of internal and external bisectors lies in 6 lines defined by Fe, Fa, Fb, Fc [11- 13].

Lemma 2.5. Let Aa , Bb and Cc be the feet of internal bisectors on respective sides of triangle. Lines Aa Bb= l1 , Bb Cc= l2 , Cc Aa= l3 are called the axes of internal bisectors. Then each of these lines passes through the foot of the respective external bisector and that lkis perpendicular to line IOk(k = 1, 2, 3) [11].

The feet of external bisectors are collinear. (This line is called the axe of external bisectors, we note it l ). Line l is perpendicular to line IO [11].

Lemma 2.6. Let bFa, cFa are the points of intersection of the lines FaFc and FaFb with side BC, similiarly the points aFb, cFb, aFc, bFc are defined then (a) The lines through the points A, cFa and B, cFb are concurrent and the point of concurrency lies on the internal angular bisector (CCc) of angle C. (b) The lines through the points C, aFc and B, aFb are concurrent and the point of concurrency lies on the Internal angular bisector (AAa) of angle A. (c) The lines through the points A, bFa and C, bFc are concurrent and the point of concurrency lies on the internal angular bisector (BBb) of angle A. For (a), (b), (c) see Fig. 5.

Figure 5.

(d) The lines through the points A, bFa and B, aFb are concurrent and the point of concurrency let us call as VC . (e) The lines through the points C, bFc and B, cFb are concurrent and the point of concurrency let us call as VA . (f) The lines through the points A, cFa and C, aFc are concurrent and the point of concurrency let us call as VB. (g) The points VA, VB and VC are collinear. (h) The lines formed by join of (C, Vc ), ( B, Vb )and (A, Va) are concurrent at V. For (d), (e), (f), (g), (h) see Fig. 6.

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Figure 6.

Theorem 2.10. The six points bFa, cFa, cFb, aFb, aFc and bFc are defined as stated in lemma-2.5, then these six points lie on a conic , for the recognition sake let us call this conic as Feuerbach conic [3].

Proof: Consider the hexagon whose vertices are bFa, cFa, cFb, aFb, aFc and bFc. clearly these six points lies on lines FaFb, FbFc and FcFa in the particular order. Let us fix L  FF FF , M  FF FF and NFFFF  . bacaabac cacbacbc cbabbcba So as to prove the six points lie on a conic, it is enough to prove that using the converse of Pascal theorem [14], the points L, M and N are collinear (Fig. 7).

Figure 7.

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Clearly the line through bFa, cFa is BC and the line through aFb, aFc is FbFc. Hence their point of intersection L is the feet of external angular bisector (using Lemma2.5). Similarly M and N are the feet of external angular bisectors.That is L= Ab or Ac , M = Ca or Cb and N= Ba or Bc. So by Lemma2.6, the points L, M and N are lie on the axe of external bisectors.Hence they are collinear and the line through L, M and N acts as Pascal line (axe of external bisector) (Fig. 8). It proves that the six points lie on a conic (Feuerbach Conic).

Figure 8.

Acknowledgement: 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. The Author is declaring that there is no funding service to the research it is an independent research work.

REFERENCES

[1] Krishna, D.N.V., Global Journal of Science Frontier Research:F Mathematics and Decision Science., 4, 9, 2016. [2] Krishna, D.N.V., Universal Journal of Applied Mathematics & Computation., 4, 32, 2016. [3] Krishna, D.N.V., Forum Geometricorum., 17, 289, 2017. [4] Kiss, S.N., Forum Geometricorum, 16, 283, 2016. [5] Kiss, S.N., Forum Geometricorum, 16, 373, 2016. [6] Yiu, P., Introduction to the Geometry of the Triangle, Florida Atlantic University Course Notes, 2001 (with corrections 2013), available online at http://math.fau.edu/Yiu/Geometry.html. [7] Kodera, T., Tohoku Mathematical Journal, 41, 455, 1935. [8] Emelyanov, L., Emelyanov , T., Forum Geometricorum, 1, 121, 2001. [9] Sharygin, I.F., Problem № 586 (in Russian), Mir Publications, .

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[10] Kimberling, C., Encyclopedia of Triangle Centers, available online at http://faculty.evansville.edu/ck6/encyclopedia/ETC.html [11] Emelyanov, L., Emelyanov , T., Around of Feet of Bisectors - Introduction, Some Theoretical Facts, International Mathematical Tournament of Towns, 2017. [12] Emelyanov, L., Emelyanov , T., Around of Feet of Bisectors -Main Problems, Theory, International Mathematical Tournament of Towns, 2017. [13] Emelyanov, L., Emelyanov , T., Around of Feet of Bisectors -Main Problems, Using solutions by Bazhov I. And Chekalkin S, International Mathematical Tournament of Towns, 2017. [14] Palej, M., Journal for Geometry and Graphics, 1,1, 1997.

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