Erdos mordell inequality TA

Manh Dung Nguyen, High School for Gifted Students, HUS, Việt Nam Brief history In geometry, the Erdős–Mordell inequality states that for any triangle ABC and point M inside ABC, the sum of the distances from M to the sides is less than or equal to half of the sum of the distances from M to the vertices The inequality was conjectured by Erdős as problem 3740 in the American Mathematical Monthly, 42 (1935) A proof was offered two years later by Mordell and Barrow These solutions were however not very elementary Subsequent simpler proofs were then found by Kazarinoff (1957) and Bankoff (1958) The inequality can be seen as a generalization of the classical Euler inequality, by taking M the circumcenter of the triangle ABC I’ve collected 20 solutions for Erdős–Mordell inequality If you have others solutions, please send to me in the email : nguyendunghus@gmail.com or post it to my blog: http://www.mathlinks.ro/weblog.php?w=1139 Hanoi, 08 / 08 / 2008 Manh Dung Nguyen, Student, 12A2 Mathematics, Hanoi University of Science, Vietnam Email: nguyendunghus@gmail.com or nguyendungtn@hotmail.com Blog: http://www.mathlinks.ro/weblog.php?w=1139 -1- Manh Dung Nguyen, High School for Gifted Students, HUS, Việt Nam Erdos-Mordell inequality: Louis Joel Mordell (1888 - 1972) Paul Erdős (1913 - 1996) Let ABC be a triangle and an interior point M We denote by AM = x, BM = y, CM = z The distances of the point M from BC, CA, AB are denoted by p, q, r Erdos-Mordell inequality asserts x + y + z ³ 2( p + q + r ) We have two ways to prove Erdos – Mordell inequality -2- Manh Dung Nguyen, High School for Gifted Students, HUS, Việt Nam Way 1: b c c a a b +q , y ³ p +r ,z ³q + p a a b b c c Then applying the AM – GM inequality: ỉa bư x + y + z r ỗ + ÷ ³ å 2r = ( p + q + q ) , a ø cyclic cyclic è b We will prove that x ³ r A First proof E x F r Mq y z p B C D Denote is the distances of the point A from BC We have: 2S ( ABC ) = aha = ap + bq + cr Because £ x + p so a ( x + p ) ³ aha b c Þ ax ³ bq + cr Þ x ³ q + r etc a a Second proof A F E B' M B C D C' Then DAFM ~ DAB ' B and DAEM ~ DAC ' C so: r BB ' q CC ' = Þ r.c = x.BB ' and = Þ qb = x.CC ' x c x b It follows that r.c + q.b = x ( BB '+ CC ' ) £ x.a -3- Manh Dung Nguyen, High School for Gifted Students, HUS, Việt Nam A Third proof F E M C' B C D B' x Let Ax the symmetric of the AM relative the bissectrice of the angle A and  = BAB ', MAB  = CAC ' BB’, CC’ the distances of B andC from Ax Þ MAC We have ' + AC.sin CAC ' = c.sin MAC  + b.sin MAB  = c q + b r a ³ BB '+ CC ' = AB.sin BAB x x Or ax ³ cq + br ; b c x ³ r + q , etc a a A Fouth proof E F M E' F' B C D P  = PAC  We have: Let P be a point on the side BC so that BAD AP.BC ³ S ( ABC ) = S ( PAB ) + S ( PAC ) = PE '.b + PF '.c where by E’,F’ are the feets of the perpendiculars from P to AB, AC respectively PE ' b PF ' c PE ' r We easily see that: ³ + But, DAPE ' ~ DAMF so = AP a AP a AP x PF ' q rb qc b c = And finally: ³ + Þ x ³ r + q , etc Similarly, AP x xa xa a a -4- Manh Dung Nguyen, High School for Gifted Students, HUS, Việt Nam Fifth Proof A E F M C' B' B C D The circle BMC intersects the line AM to the point A’ and the sides AB and AC to the points B’ and C’ respectively We will have: AB ' AC ' AM B ' C ' ³ S ( AMB ') + S ( AMC ') = r AB '+ q AC ' Þ x ³ r +q (1) B 'C ' B 'C ' The triangles AB’C’ and ACB are similar Therefore AB ' AC b AC ' c = = , = (2) B ' C ' BC a B ' C ' a b c From (1) and (2) : x ³ r + q , etc a a A Sixth proof E' F F' E M' B M D C We consider the circle AFME The parallel line from m to EF intersects the circle to the point M’ Then we drop the perpenticulars M’E’, M’F’ to M ' F ' ME = AC,AB respectively Easy to see that DAME ~ DAM ' F ' follows AM ' AM M ' E ' MF Similarly, = On the other hand: AM '.BC ³ AB.M ' F '+ AC ME so AM ' AM M ' E ' c M ' F ' b qc rb b c 1³ + = + or x ³ r + q , etc AM ' a AM ' a xa xa a a -5- Manh Dung Nguyen, High School for Gifted Students, HUS, Việt Nam A Seventh proof B' E F C' M D B C Let B’C’ denote the orthogonal projections of BC on the line FE Then we have: BC ³ B ' C ' = B ' F + FE + EC ' Because is cyclic so ME q ÐBFB ' = ÐAFE = ÐAME Þ B ' F = BF cos ÐBFB ' = BF = BF MA x r Similarly, EC ' = CE By the Ptolemy theorem, we have x AF q + AE.r AM EF = AF ME + AE.MF Þ EF = x q r AF q + AE.r It follows that: BC ³ BF + CE + x x x Or ax ³ q.BF + r.CE + AF q + AE.r = qc + rb Etc Eighth proof A F E M C' B B' C A' The circle ABC intersects the line AM to the point A’ The Ptolemy’s theorem to the inscribed ABA’C is: AA '.BC = A ' C AB + A ' B AC Let A’D and A’E the distances of the point A’ from AC and AB respectively A' D c A' E b Obviously A ' C ³ A ' C ', A ' B ³ A ' B ' so AA '.a ³ A ' C '.c + A ' B '.b Þ ³ + AA ' a AA ' a A' D q A' E r b c But by the Talet theorem = , = So x ³ r + q ,etc AA ' x AA ' x a a -6- Manh Dung Nguyen, High School for Gifted Students, HUS, Việt Nam Ninth proof A x r C' q M y E A' F B' z p B D C We drop the perpenticulars BB’, AA’, CC’ to the line EF Obviously we have: B ' C ' = c.cosAFA'+b.cosAEA'=c.sinMFA'+b.sinAEA' £ a By the Extended Law of Sines, we have sin MFA ' sin AEA ' sin A = = = q r EF x It follows that q r b c c + b £ a Û r + q £ x , etc x x a a Tenth Proof Let O be the center of circumcircle.Consider the bisector of the angle at A, and repale the given triangle by its mirror image B’AC’ Now apply Papus’s theorem to B’AC’, noting that OA is perpendiacular to B’C’, with the result that AM cos ( AM,AO ) B ' C ' = AC '.zc + AB ' y Or a.MA.cos ( AM,AO ) = bz + cy It follows that a.x ³ bz + cy Etc -7- Manh Dung Nguyen, High School for Gifted Students, HUS, Việt Nam Eleventh proof A x E F q r M y z p B D1 D D2 C Let D1D2, E1E2, F1F2 denote the respective orthogonal projection of FE, DF, ED on BC, CA, AB Then DD x + y + z ³ å x (1) EF cyc Because ÐFMB = ÐFDD1 in the cyclic quadrilateral BDMF, the right triangles DFD1 MF and MBF are similar and D1 D2 = FD Similarly and from (1), we obtain MB æ FD.z DE y x + y + z p ỗ + ÷ ³ 2( p + q + r) FD.z ø è DE cyc Twelfth proof From figure 2, we get that ax ³ bz + cy , etc -8- Manh Dung Nguyen, High School for Gifted Students, HUS, Việt Nam Way 2: sin B sin C +q ,… sin A sin A Then applying the AM – GM inequality We will prove that x ³ r Thirteenth proof A E F M B C D Apply the Since Law and the Cosine Law to obtain MA.sin A = EF = q + r - 2qrcos (p - A ) Similarly, we need to prove that å sin A q + r - 2qrcos (p - A ) ³ ( p + q + r ) On the other hand, we have FE = q + r - 2qr cos (p - A ) = q + r - 2qr ( cosBcosC-sinBsinC ) Using: sin B + cos B = sin C + cos 2C = we get 2 FE = ( q sin C + r sin B ) + ( qcosC - rcosB ) ³ ( q sin C + r sin B ) It follows that FE ³ q sin C + r sin B So q sin C + r sin B æ sin B sin C = ồỗ + sin A q + r - 2qrcos (p - A ) ³ å sin A ÷p è sin C sin B ø Etc -9- Manh Dung Nguyen, High School for Gifted Students, HUS, Việt Nam Fourteenth proof A x E F q r M y z p B C D Denote P is the angle between the distance y, z, and similarly for Q, R It suffices to show that P p £ yz cos (1) Let OD be the bisector of the angle P Then from area of the triangle MBC, P P P yz sin P = MD ( y + z ) sin ³ p ( y + z ) sin ³ p yz sin 2 With equality only when y = z, and (1) follows We need to prove that P Q R x + y + z - yz cos - zx cos - xy cos ³ 2 2 R Qư ỉ R Qử ổ ỗ x - y cos - z cos ữ + ỗ y sin - z sin ữ ³ 2ø è 2ø è We are done - 10 - Manh Dung Nguyen, High School for Gifted Students, HUS, Việt Nam Fifteenth proof A x E F r q M y z p B C E' D F' We drop the perpenticulars EE’, FF’ to BC We have: EF=AMsinA=xsinA Obviously EF ³ E ' F ' = F ' D + DE ' = r sin MFF '+ q sin MEE ' = q sin B + r sin C That is sin B sin C x³r +q sin A sin A Similarly, we obtain æ sin B sin C ö å x ³ å p ỗố sin C + sin B ữứ ( p + q + r ) A Sixteenth proof x E F r q M y z p B D C Using the same notation as in the sixteenth proof, we have To the triangle DEF p.EF ³ ([DMF]+[DME]) = pr sin B + pq sin C Therefore EF ³ r sin B + q sin C Û x sin A ³ r sin B + q sin C sin B sin C It follows that x³r +q Etc sin A sin A - 11 - Manh Dung Nguyen, High School for Gifted Students, HUS, Việt Nam A Seventeenth proof x E F r y M q z p B C E' D F' We drop the perpendiculars EE’, FF’ to BC We will have: But So åx³å E ' F ' = r.sin B + q.sin C , EF = x sin A sin C ö ổ sin B x ỗố r sin A + q sin A ÷ø , etc Eighteenth proof E 'F ' x EF A E' F r y x E F' p q M B D z C From E,F we drop the perpendiculars EE’ and FF’ to DM We obviously have: EF ³ EE '+ FF ' = r sin B + q sin C That id x sin A ³ r sin B + q sin C , etc - 12 - Manh Dung Nguyen, High School for Gifted Students, HUS, Việt Nam A Nineteenth proof x E F B' y C' q r M p z B C D The antiparallel from the point M intersects AB to the oint B’ and the side AC to the point C’ We have AB′.AB = AC′.AC We also easily see: AC ' AB ' sin C sin B x.B ' C ' ³ qAC '+ rAB ' Þ x ³ q +r =q +r B 'C ' B 'C ' sin A sin A Twentieth proof A' A E F M B D C The line DM intersects the circle AFE at the point A’ The triangle A’FE is similar to the triangle ACB Let de, df , the distances of the points E and F from A’M We obviously have: A ' E.ME = Rd e , A ' F FM = Rd f , where R is the radius of the circle AEF Adding we take A' E A' F sin C sin B q A ' E + r A ' F ³ x.FE Þ x ³ q + r =q + r FE FE sin A sin A Etc The end!!! - 13 - ... point M from BC, CA, AB are denoted by p, q, r Erdos- Mordell inequality asserts x + y + z ³ 2( p + q + r ) We have two ways to prove Erdos – Mordell inequality -2- Manh Dung Nguyen, High School... Việt Nam Erdos- Mordell inequality: Louis Joel Mordell (1888 - 1972) Paul Erdős (1913 - 1996) Let ABC be a triangle and an interior point M We denote by AM = x, BM = y, CM = z The distances of... applying the AM – GM inequality: ỉa bư x + y + z ³ å r ỗ + ữ 2r = ( p + q + q ) , a ø cyclic cyclic è b We will prove that x ³ r A First proof E x F r Mq y z p B C D Denote is the distances of the