Geometric Inequalities in Pedal Quadrilaterals

Transcription

1 Forum Geometricorum Volume 8 (08) FORUM GEOM ISSN Geometric Inequlities in edl Qudrilterls Şhlr Meherrem Gizem Günel çıksöz Sereny Şen Zeynep Sezer nd Güneş şkes bstrct. The im of this pper is to investigte the generl properties of the pedl qudrilterl of point with respect to convex nd closed qudrilterl D. In prticulr we present n nlogue of Erdős-Mordell Inequlity stting tht for ny tringle nd point inside the sum of the distnces from to the sides is less thn or equl to hlf of the sum of the distnces from to the vertices of for n inscribed qudrilterl.. Introduction Let be tringle nd be point in plne. If the feet of the perpendiculrs drwn from to sides of the tringle [] [] [] re respectively X Y nd Z then tringle XY Z is clled the pedl tringle of with respect to nd is clled the pedl point ([3 7]). If the point lies on the circumcircle of the tringle then the points X Y nd Z re colliner. The line pssing through the points X Y nd Z is known s the Simson line of ([3]). Z Y Z Y X X Figure similr notion hs been generlized for polygons with n sides ([3]). Let 3 n be polygon nd be point inside 3 n. If the feet of the perpendiculrs from to the line segments [ ] [ 3 ]...[ n ] re respectively H H... H n then the polygon H H H n is clled the pedl polygon of with respect to 3 n nd is clled the pedl point ([3]). lthough there re lots of investigtions on pedl tringles so little investigtions hve been studied on pedl qudrilterls in the literture ([ ]). ubliction Dte: pril ommunicting Editor: ul Yiu.

2 04 M. Şhlr et l. In this pper we study on the generl properties of the pedl qudrilterl of point with respect to convex nd closed qudrilterl D. We investigte some interesting geometric inequlities on pedl qudrilterls. In prticulr we stte n nlogue of Erdős-Mordell Inequlity ([ 4 8 0]) for qudrilterls.. Geometric inequlities in pedl qudrilterls roposition. Let D be n inscribed qudrilterl nd be point inside D. If the feet of perpendiculrs from to the line segments [] [] [D] nd [D] re respectively K M N L nd the rdius of the circumcircle of D is R then KM = R KL = D R MN = D R LN = D R. K M L N D Figure roof. s seen in Figure nd since K = LD = N = M = 90 the qudrilterls KM LDN NM nd LKre inscribed qudrilterls. y the lw of sines we hve = Hence we get It cn be esily seen tht KL = D R KM sin KM nd R = sin(π KM) = KM =. R MN = D R sin KM. LN = D R.

3 Geometric inequlities in pedl qudrilterls 05 roposition. Let D be qudrilterl nd be point inside D.If the feet of perpendiculrs from to the line segments [] [] [D] nd [D] re respectively D then r + r + r 3 + r 4 M where r r r 3 nd r 4 re respectively the rdii of the incircles of the tringles D D D nd { M =mx cos + cos + cos + } cos D + nd is the perimeter of the qudrilterl D. The equlity holds if D is chosen s squre nd is chosen t the center of the squre. O r O r r 3 O 3 D r 4 O 4 D Figure 3 roof. y the lw of cosines we hve = + cos. () If S is the re of the tringle then S = sin. () Since S =( + + )r we get r = y the equlities () nd (3) S + +. (3) r = sin + +. (4) It follows from rithmetic-geometric Men Inequlity tht +. (5)

4 06 M. Şhlr et l. If the inequlities () nd (5) re considered together then = + cos cos cos (6) It is lso obvious from rithmetic-geometric Men Inequlity tht +. (7) Therefore by the inequlities (6) nd (7) we observe tht cos. (8) Hence it follows from the inequlities (4) nd (8) tht sin r cos + sin cos + ) sin ( cos + ) + ( cos + (9) ) { Let M = mx cos + cos + cos + cos D+ }. y the inequlity (9) r M ( + ). (0) Similrly we get r M ( + D ) r 3 M ( + ) () r 4 M ( D + DD ). If the inequlities (0) nd () re dded side by side it is esy to see tht r + r + r 3 + r 4 M where is the perimeter of the qudrilterl D. Now suppose tht D is squre with side length nd is t the center of the D (Figure 4). If the re of the tringle D is S D then S D = + r =.

5 Geometric inequlities in pedl qudrilterls 07 O O 3 r r 3 r O r 4 O 4 D D Figure 4 Hence we get r = +. Since r = r = r 3 = r 4 =8nd M = then the following equlity holds: ( ) + r + r + r 3 + r 4 =4 8 + = + = M. roposition 3. Let 3 n be polygon nd be point inside 3 n. If the feet of perpendiculrs from to the line segments [ ] [ 3 ]... [ n ] re respectively H H...H n then r + r + + r n M where r r... r n re respectively the rdii of the incircles of the tringles H H n H H...H n n H n nd { M =mx cos + cos +... cos n + } nd is the perimeter of the polygon n. The equlity holds if n is chosen s squre nd is chosen s the centroid of the squre. This cn be esily shown s in roposition. roposition 4. Let D be circumscribed qudrilterl with re S. If the center nd the rdius of the incircle of D re respectively the point nd r nd the feet of perpendiculrs from to the line segments [] [] [D] nd

6 08 M. Şhlr et l. [D] re respectively K L M nd N then S pedl S 4r + b + c + d where = L b = L c = MD d = N nd S pedl is the re of the pedl qudrilterl KLMN. The equlity holds if D is squre. L b d K S S S 4 S 3 c b M d N c D Figure 5 roof. s seen in Figure 5 if S S S 3 nd S 4 re respectively the res of tringles KL L M MN nd NK then we get Hence it is obvious tht S = r sin S = b r b sin S 3 = c r c sin D S 4 = d r d sin. S +S +S 3 +S 4 = r(+b+c+d) ( sin +b sin +c sin D+d sin ). Since we lso hve S =( + b + c + d)r weget S pedl S = S + S + S 3 + S 4 S = sin + b sin + c sin D + d sin. () r( + b + c + d)

7 Geometric inequlities in pedl qudrilterls 09 y the uchy-schwrz Inequlity we get + b + c + d = +b +c +d + b + c + d = + b + c + d. Thus + b + c + d + b + c + d. y the equlity () S pedl S = sin + b sin + c sin D + d sin r( + b + c + d) + b + c + d 4r + b + c + d = 4r + b + c + d. L r K r r M r N D Figure 6 Now suppose tht D is squre nd = = D = D =. s seen in Figure 6 since r = S =4 S pedl = wehve S pedl = S 4 = = = 4r + b + c + d. roposition 5. Let be point inside polygon 3 n where = 3 =... n = n. If the lengths of the perpendiculrs from to the sides of the polygon 3 n re respectively h h...h n then S h + + n h n

8 0 M. Şhlr et l. where S nd re respectively the re nd the perimeter of 3 n. The equlity holds if = = = n = nd h = h = = h n = h. 3 3 h h h 3 4 h Figure 7 roof. s seen in Figure 7 it is obvious tht S = h + h + + n h n.y the uchy-schwrz Inequlity we get ( ( h + h + + n h n ) ) n ( h h h n). n Since ( S ) n ( h h h n) n we hve Hence S ( n ) h + h + + n h n = h +. h + + n h n S h +. h + + n h n Now suppose tht = = = n = nd h = h = = h n = h. Since S = n h = n we get S = nh n = h n = h + h + + h = h +. h + + n h n

9 Geometric inequlities in pedl qudrilterls roposition 6. Let be point inside polygon 3 n where = 3 =... n = n. If the lengths of the perpendiculrs from to the sides of the polygon 3 n re respectively h h...h n then n h h n h n S where S is the re of 3 n. The equlity holds if = = = n = nd h = h = = h n = h. roof. y the uchy-schwrz Inequlity we hve ( ( h + h + + n h n ) ) ( h h n h n }{{} ). n times Since S = h + h + + n h n it is esy to see tht n h h n h n S. Now suppose tht = = = n = nd h = h = = h n = h. Since S = n h wehve = n h h n h n h = n nh = n S. Theorem 7. Let D be n inscribed qudrilterl nd be point inside D. If the feet of perpendiculrs from to the line segments [] [] [D] nd [D] re D respectively then D > D Figure 8

10 M. Şhlr et l. roof. Let = h = h 3 = h 3 nd 4 = h 4. Then by the lw of cosines we hve = h + h h h cos(π ) 3 = h + h 3 h h 3 cos(π ) 3 4 = h 3 + h 4 h 3h 4 cos(π D) 4 = h 4 + h h h 4 cos(π ). Hence = h + h h h cos(π ) = h + h h h cos( + + D π) = h + h +h h cos( + + D) = h + h +h h cos( + )cosd h h sin( + )sind = (h sin D h sin( + )) +(h cos D + h cos( + )). Since + = π wehve =(h sin D) +(h cos D h ). Now suppose tht (h cos D h ) =0. Then cos D = h h. Since = D by the lw of cosines = h + h h h cos D = h + h h h h h = h h. Hence we get h = + h implying tht =90 which is contrdiction. Therefore (h cos D h ) 0. Then it is esy to verify tht Similrly we hve >h sin D (3) 3 > h 3 sin (4) 3 4 > h 4 sin (5) 4 > h sin (6) Since 3 3 D 4 nd 4 re inscribed qudrilterls if we pply the lw of sines to the tringles 4 3 D 3 4

11 Geometric inequlities in pedl qudrilterls 3 then we respectively get = 4 sin (7) = sin (8) = 3 sin (9) D = 3 4 sin D (0) If the inequlities (3) (4) (5) nd (6) re respectively considered together with the inequlities (7) (8) (9) nd (0) then = sin = 3 sin D = 3 4 sin D = 4 sin Therefore it is esy to see tht D > h sin sin > 4 4 h sin sin by rithmetic-geometric Men Inequlity. References > h sin D sin > h 3 sin sin > h 4 sin sin D > h sin sin. + h sin D sin h sin D sin + h 3 sin sin h3 sin sin = h 4 sin sin D h4 sin sin D []. lsin nd R.. Nelsen visul proof of the Erdős-Mordell inequlity Forum Geom. 7 (007) [] L. nkoff n elementry proof of the Erdős-Mordell theorem mer. Mth. Monthly 65 (958) 5. [3] H. S. M. oxeter nd S. L. Greitzer Geometry Revisited Mth. ssoc. meric 967. [4]. Erdős roblem 3740 mer. Mth. Monthly 4 (935) 396. [5] D. Ferrrello M. F. Mmmn nd M. ennisi edl poygons Forum Geom. 3 (03) [6] R. Honsberger Episodes In Nineteenth nd Twentieth entury Eucliden Geometry Mth. ssoc. meric 995. [7] R.. Johnson dvnced Eucliden Geometry Dover ublictions 007. [8] D. K. Kzrinoff simple proof of the Erdős-Mordell inequlity for tringles Michign Mth. Journl 4 (957) [9] M. F. Mmmn. Micle nd M. ennisi Orthic qudrilterls of convex qudrilterl Forum Geom. 0 (00) 79 9.

12 4 M. Şhlr et l. [0] L. J. Mordell nd D. F. rrow (937). Solution to roblem 3740 mer. Mth. Monthly 44 (937) [] M. Şhin Mtemtik Olimpiytlrın Hzırlık Geometri - I lme Yyınlrı nkr 000. Şhlr Meherrem: Ysr University Fculty of Science nd Letters Deprtment of Mthemtics Izmir Turkey E-mil ddress: shlr.meherrem@ysr.edu.tr Gizem Günel çıksöz: Ozel Ege Lisesi Izmir Turkey E-mil ddress: gizem.ciksoz@egelisesi.k.tr Sereny Şen : Ozel Ege Lisesi Izmir Turkey E-mil ddress: sereny.sen3@gmil.com Zeynep Sezer : Ozel Ege Lisesi Izmir Turkey E-mil ddress: gunes bskes@hotmil.com Güneş şkes: Ozel Ege Lisesi Izmir Turkey E-mil ddress: zeynepsezer99@gmil.com