Distances Among the Feuerbach Points

Transcription

1 Forum Geometricorum Volume ) FORUM GEOM ISSN Distances mong the Feuerbach Points Sándor Nagydobai Kiss bstract We find simple formulas for the distances from the Feuerbach points of a triangle to the vertices and among themselves Consider a triangle BC with the midpoints X Y Z of its sides BC C B respectively circumcenter O the incenter I and the excenters I a I b I c The radii of the circumcircle incircle and excircles are denoted by R r r a r b r c respectively The nine-point circle is the circle through X Y Z; it has center N and radius R By the famous Feuerbach theorem the nine-point circle is tangent to the incircle and each of the excircles The points of tangency are the Feuerbach points F e with the incircle and F a F b F c with the excircles I a ) I b ) I c ) respectively I b I c F e F c Z I F b Y N B X F a C I a Figure 1 In [] we have computed the distances from the Feuerbach points to X Y Z see Figure 1) Specifically if the lengths of the sides BC C B are a b c Publication Date: November Communicating Editor: Paul Yiu

2 37 S N Kiss then F e X F a X b c R OI F c a R ey OI F a b R ez OI ; 1) b c R c + a)r a + b)r F a Y F a Z ) OI a OI a OI a By Euler s formula OI RR r) and OI a RR +r a ) see [1 Theorems ]) these are equivalent to the following formulas F e X b c) R R r) F ey c a) R R r) F ez a b) R R r) ; 3) F a X b c) R R +r a ) F ay c + a) R R +r a ) F az a + b) R R +r a ) ) In this note we find simple formulas analogous to 1) ) for the distances among the Feuerbach points We begin with the distances to the vertices see Figure ) Proposition 1 The distances from the Feuerbach point F e to the vertices of triangle BC are given by F e s a) R rs R r BF e s b) R rs B R r CFe s c) R rs C R r where s is the semiperimeter of the triangle and S : b +c a S B c +a b and S C a +b c Proof We apply the median theorem for the triangles F e BC F e C F e B From 3) above we have F e X BF e + CFe F e Y CF e + Fe F e Z F e + BFe a 5) b 6) c 7) The combination 5)+6)+7) gives F e X + F e Y + F e Z Fe b + c a

3 Distances among the Feuerbach points 375 Hence Since Fe F e X + F e Y + F e Z + b + c a b c) +c a) +a b) )R + S R r) a b)a c)r R r) + R r)s R r) a b)a c)+s )R rs R r) a b)a c)+s a b)a c)+b +c a )b+c a) s a) we have F e s a) R rs R r The other two expressions follow similarly I b I c F e F c Z I F b Y N B X F a C I a Figure Proposition The distances from the Feuerbach point F a to the vertices of triangle BC are given by F a s R + r a S R +r a BF a s c) R + r a S B R +r a CF a s b) R + r a S C R +r a

4 376 S N Kiss Proof We applying the median theorem to triangles F a BC F a C F a B From ) above we have F a X BF a + CFa a 8) F a Y CF a + Fa b 9) F a Z F a + BFa c 10) The combination 8)+9)+10) gives F a X + F a Y + F a Z Fa b + c a Hence Fa F a X + F a Y + F a Z + S b c) +c + a) +a + b) )R + S R +r a ) a + b)a + c)r R +r a ) + R +r a)s R +r a ) a + b)a + c)+s )R +r a S R +r a ) Since a + b)a + c)+s a + b)a + c)+b + c a )a + b + c) s we have Fa s R+r as R+r a On the other hand the combination 8) 9)+10) gives F a X F a Y + F a Z BFa c + a b Therefore BFa F a X F a Y + F a Z + S B b c) c + a) +a + b) )R + S B R +r a ) a + b)b c)+s B)R +r a S B R +r a ) Since a+b)b c)+s B a+b)b c)+c +a b )a+b c) s c) we have BFa s c) R + r a S B R +r a The proof of the expression for CFa is similar

5 Distances among the Feuerbach points 377 Now we compute the distances among the Feuerbach points I b I c F e F c Z I F b Y N B X F a C I a Figure 3 Theorem 3 F b F c b+c)r OI b OI c F c F a c+a)r OI c OI a F a F b a+b)r OI a OI b Proof It is enough to prove the first formula Triangle NF b F c is isosceles with NF b NF c R ;wehave F b Fc R 1 cos F bnf c ) pplying the law of cosines to triangles NI b I c noting that I b I c Rcos we have R cos ) NIb + NI c NI b NI c cos I b NI c ) R ) R ) R R + r b + + r c + r b + r c [ ) )] R R R + r b + r c + + r b ) R + r c r b r c ) + 1 R +r b)r +r c )1 cos F b NF c ) ) cos I b NI c ) 1 cos F b NF c )

6 378 S N Kiss Therefore R R ) ) cos F b Fc rb r c ) OIb OI c Since s Rcos cos B cos C s tan B tan C ) R cos sin B C ) From this F b Fc 16R6 cos OIb OI c and F b F c R3 cos cos B C ) 1 sin B C )) 16R6 cos cos B C ) OIb OI c R3 cos B + C ) +cos + B C )) R3 cos π B) +cos π C)) R3 sin B +sinc) b + c)r Theorem F e F a b c R OI OI a F e F b c a R OI OI b F e F c a b R OI OI c Proof gain it is enough to prove the first formula Triangle NF e F a is isosceles with NF e NF a R see Figure 3); we have F e Fa R 1 cos F enf a ) pplying the law of cosines to triangle NII a we have noting that II a Rsin R sin ) NI + NIa NI NI a cos INI a ) R ) R ) ) R R r + + r a r + r a cos INI a [ ) )] R R ) ) R R r + r a + r + r a 1 cos F e NF a ) r + r a ) + 1 R r)r +r a)1 cos F e NF a ) Therefore R R sin ) ) ) b + c)tan R R sin ) ) ) b + c)tan F e Fa R r)r +r a ) OI OIa Since b + c)tan Rsin B +sinc)tan B+C Rsin cos B C tan R sin B C cos F e F a R R ) sin R sin OI OI a ) ) B C cos 16R6 sin OI OIa sin B C

7 Distances among the Feuerbach points 379 and F e F a R 3 sin B C sin OI OI a R 3 cos B + C ) cos + B C )) OI OI a R 3 cos π B) cos π C)) OI OI a R 3 sin B sin C) b c R OI OI a OI OI a References [1] N ltshiller-court College Geometry Dover Reprint 007 [] S N Kiss distance property of the Feuerbach point and its extension Forum Geom ) [3] M J G Scheer simple vector proof of Feuerbach s theorem Forum Geom ) S Nagydobai Kiss: Constatin Brâncuşi Technology Lyceum Satu Mare Romania address: dsandorkiss@gmailcom