hinh học 2

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I.68)nh l Archimedesnh l:Cho INCLUDEPICTURE "http://uvyd.com/forums/cgi-bin/mimetex.cgi?%20M%20" \* MERGEFORMATINET l trung im , im chuyn ng ty trn .T k .Chng minh rng Chng minh:Trn tia dng im sao cho .Ta c: ng thi suy ra Vy ng thi l ng cao, ng trung tuyn tam gic nn tam gic cn ti M suy ra do l trung im cnh hay ni cch khc (dpcm)

I.69) nh l Urquhartnh l:Cho hai b ba im thng hng v , l giao im ca v .Chng minh rng khi v ch khi .Chng minh:u tin ta cn chng minh b sau:Trong tam gic ABC ta c vi p l na chu vi v a=BC.Ta c: mt khc .nn suy ra do nn nghch o hai v ta c dpcmTr li bi ton gi cc gc nh trn hnh v ta c:(dpcm)

I.70)nh l Mairon Waltersnh l:Cho tam gic ABC v cc ng thng chia 3 cnh i din nh hnh v.Chng minh rng Chng minh::Trc tin ta cn chng minh b sau:Cho tam gic im di ng trn ng thng sao cho .Gi s giao nhau ti th chia on theo t s .Gi s th ta c:m li do cng phng vi nn tn ti mt s sao cho Mt khc cch biu din ny l duy nht nn ta c ng nht thc:suy ra:.Quay li bi ton ban u ta p dng b nhiu ln lin tip ta c(y ch l k nng tnh ton nn mnh ch ghi kt qu cc bc thng cm)suy ra:(dpcm)

I.71)nh l Poncelet v bn knh ng trn ni tip,bng tip trong tam gic vung.nh l:Cho tam gic c ln lt l bn knh cc ng trn ni tip, bng tip gc .Chng minh rng: tam gic ABC vung ti A khi v ch khi .Chng minh::Ta c:tam gic vung ti (dpcm)

I.72)nh l Hansennh l:Cho tam gic .Chng minh rng cc iu kin sau tng ng:1)Tam gic vung2) 3) Chng minh:u tin ta chng minh mt vi h thc ph sau:1) 2)3)Ta c: Tr li bi ton ban u ta c:Do nn chia c hai v cho ta c"(1)t ta suy ra do d (1) tam gic vung.

(2)tng t thay nh trn vi ch v ta c:(2)

(3)do ln hn nn (3) tam gic vung.[RIGHT][I][B]Ngun: MathScope.ORG[/B][/I][/RIGHT][RIGHT][I][B]Ngun: MathScope.ORG[/B][/I][/RIGHT]

*ch thch:

I.73)nh l Steinbart m rngnh l:.Cho tam gic ni tip .Cc tip tuyn ca ng trn ti giao nhau ti .Trn (O) ly cc im .Chng minh rng ng quy khi v ch khi ng quy hoc cc giao im ca vi 3 cnh tam gic thng hng.Chng minh:Gi Ta c:m

nn

Tng t ta suy ra:do nu v phi bng th biu thc trong ngoc v tri bng hoc v ngc li hay ni cch khc ng quy khi v ch khi ng quy hoc thng hng

I.74)nh l Monge & d'Alembert Inh l:Cho 3 ng trn c bn knh khc nhau v khng cha nhau.Tip tuyn chung ngoi ca mi ng trn giao nhau ln lt ti .Chng minh rng: thng hng.Chng minh:V cc ng trn c vai tr nh nhau nn khng mt tnh tng qut ta gi s:.Khi ta c th chng minh c:Suy ra: Theo nh l Menelaus ta suy ra dpcm

*Ch thch:l php v t tm t s bin thnh [RIGHT][I][B]Ngun: MathScope.ORG[/B][/I][/RIGHT]

I.75)nh l Monge & d'Alembert IInh l:Cho 3 ng trn c bn knh khc nhau v khng cha nhau.Tip tuyn chung trong ca (A) v (C), (B) v (C) giao nhau ln lt ti , tip tuyn chung ngoi ca v giao nhau ti .Chng minh rng: thng hng.Chng minh:V cc ng trn c vai tr nh nhau nn khng mt tnh tng qut ta gi s:.Khi ta c th chng minh c:Suy ra: Theo nh l Menelaus ta suy ra dpcm

*Ch thch:l php v t tm t s bin thnh

I.76)nh l Steiner v bn knh cc ng trnnh l:Chng minh rng trong tam gic ta c: Chng minh:Ta c:(hin nhin) suy ra dpcm

I.77)nh l Bellavitisnh l::Cho t gic l t gic iu ho k hiu .Chng minh rng: Chng minh:Gi ng trn ng knh l ng trn Apollonius ca tam gic ng vi nh .V l t gic iu ho nn do thuc ng trn Apollonius ca tam gic .Dng i xng vi qua phn gic .Ta suy ra: Vy (dpcm)

I.78)nh l Feuer bach-Luchterhand: [RIGHT][I][B]Ngun: MathScope.ORG[/B][/I][/RIGHT]

II/Mt s im v ng c bit c xc nh duy nht vi tam gic v t gicII.1) ng thng Euler ca tam gic.nh l:Cho tam gic gi l trc tm, trng tm, tm ng trn ngoi tip.Chng minh rng: thng hng.Chng minh:Gi l trung im .Ta c do suy ra thng hng

II)ng trn v tm EulerKt qu : Trong mt tam gic ,trung im cc cnh ca tam gic ,chn cc ng cao v trung im cc on thng ni trc tm vi cc nh cng nm trn mt ng trn gi l ng trn Euler ca tam gic y.Ch dn chng minh:Thc ra y ch l mt trng hp c bit ca hai im ng gic (Xem mc I.50)

II.3)ng i trung, im LemoineKt quCho tam gic th 3 ng i trung ca tam gic ng quy ti im ca tam gic.Ch dn chng minh::ng i trung ca tam gic ng vi 1 nh l ng thng i xng vi trung tuyn qua phn gic tng ng ca nh .T nh ngha trn p dng nh l Xeva dng sin ta c:

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sin(

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Suy ra ng quy ti im ca tam gic.*Ch thch: k hiu tng ng vi

II.4)im Gergonne,im Nobb, ng thng Gergone1)Kt qu v im Gergonne:Tam gic ABC vi ng trn ni tip (I).Tip im ca (I) trn BC,CA,AB ln lt l D,E,F.Khi AD,BE,CF ng quy ti mt im gi l im Gergonne ca tam gic ABC.Ch dn chng minh:Ch cn dng nh l Ceva v cc kt qu n gin : DB=DC,EA=EC,FA=FB l ra.2)Kt qu v im Nobb v ng thng Gergonne(Vn vi cc k hiu trn)Mt tam gic khng cn c 3 im Nobb tng ng l giao im ca cc cp ng thng EF v CB ,DE v AB ,DF v AC. V 3 im Nobb cng nm trn mt ng thng gi l ng thng Gergonne ca tam gic ABC.Ch dn chng minh:Xt cc v i cc i vi (I). ng i cc ca A l EF i qua M,nn ng i cc ca M i qua A.Mt khc d thy ng i cc ca M i qua D nn suy ra ng i cc ca M l AD. Hon ton tng t ta c: ng i cc ca N l BE v ng i cc ca P l CF Theo trn ,do AD,BE,CF ng quy nn s c iu phi chng minh.Bnh lun: Kt qu trn c th m rng nh sau: Cho tam gic ABC v 3 im D,E,F theo th t thuc BC,CA,AB sao cho AD,BE,CF ng quy v D,E,F khc trung im on thng.Gi M,N,P ln lt l im chung ca cc cp ng thng (EF,BC) ,(DF,CA) ,(DE,AB).Khi M,N,P thng hng Bn c th chng minh kt qu trn bng nh l Menelaus nhng thm ch bi ton m rng ny cng ch l trng hp c bit ca nh l Desargues m thi!!!!

(Xem them hai file : FG200821.bdf ; jcgeg200722.bdf)

II.5)im NagelKt qu:.Cho tam gic . Cc ng trn bng tip xc vi 3 cnh tng ng nh ln lt ti th ta c 3 ng thng ng quy ti im ca tam gic.Ch dn chng minh:Ta c: Suy ra dpcm *Ch thch:

II.6) im Brocardnh ngha:Trong mt tam gic ABC cho trc c hai im Brocard M,N c xc nh sao cho:v .

II.7)im Schifflernh ngha:Cho tam gic c l tm ng trn ni tip tam gic.Khi 4 ng thng Euler ca tam gic v ng quy ti im ca tam gic.Ch dn chng minh:Gi l tm ng trn ngoi tip tam gic, l trng tm tam gic, l trung im l trng tm tam gic ct ti ct ti tip xc vi cnh ti ct ti , ct ti .R rng l tm ng trn ngoi tip tam gic .Do l ng thng Euler ca tam gic .p dng nh l Menelaus cho tam gic vi ct tuyn ta c:.Suy ra:Ap dng nh l Menelaus cho tam gic vi ct tuyn ta c:Do l trng tm tam gic IBC nn Do :Tng t ta thy cc ng thng Euler ca cc tam gic cng ct ti (c xc nh bi h thc )Vy cc ng thng Euler ca 4 tam gic v ng quy ti

(Xem them FG200312.bdf)

II.8)im FeuerbachKt qu:Trong mt tam gic ,ng trn Euler tip xc vi ng trn ni tip ca n,v tip im c gi l im Feuerbach ca tam gic trn.Ch dn chng minh:(leductam post)Gi l tm ng trn Euler, ngoi tip, ni tip ca tam gic . l trc tm, l ng knh vung gc vi l hnh chiu ca ln l bn knh ng trn ngoi tip, ni tip. l chn ng phn gic gc 1.ng trn Euler tip xc trong vi ng trn ni tip.Ta d dng chng minh c V suy ra (1)Chng minh tip: V nn m cn ti do Vy Chiu h thc trn ln theo phng vung gc vi ta c:(2)T (1) v (2) ta suy ra:(3)Ta c:(4)T (3) v (4) ta c: Vy suy ra dpcm.

Hon ton tng t ta cng c:

2.ng trn Euler tip xc ngoi vi cc ng trn bng tipT k vung gc vi do nn T (2) ta c: Vy (5)Mt khc: nn:(6)T (5) v (6) ta c: (7)T k . Trong tam gic vung ta c:

(8)T (7) v (8) ta c:Vy suy ra dpcm

(Xem them BalticFeuer; GenFeuerPDF; FG200117)

II.4)im Gergonne,im Nobb, ng thng Gergone1)Kt qu v im Gergonne:Tam gic ABC vi ng trn ni tip (I).Tip im ca (I) trn BC,CA,AB ln lt l D,E,F.Khi AD,BE,CF ng quy ti mt im gi l im Gergonne ca tam gic ABC.Ch dn chng minh:Ch cn dng nh l Ceva v cc kt qu n gin : DB=DC,EA=EC,FA=FB l ra.2)Kt qu v im Nobb v ng thng Gergonne(Vn vi cc k hiu trn)Mt tam gic khng cn c 3 im Nobb tng ng l giao im ca cc cp ng thng EF v CB ,DE v AB ,DF v AC. V 3 im Nobb cng nm trn mt ng thng gi l ng thng Gergonne ca tam gic ABC.Ch dn chng minh:Xt cc v i cc i vi (I). ng i cc ca A l EF i qua M,nn ng i cc ca M i qua A.Mt khc d thy ng i cc ca M i qua D nn suy ra ng i cc ca M l AD. Hon ton tng t ta c: ng i cc ca N l BE v ng i cc ca P l CF Theo trn ,do AD,BE,CF ng quy nn s c iu phi chng minh.Bnh lun: Kt qu trn c th m rng nh sau: Cho tam gic ABC v 3 im D,E,F theo th t thuc BC,CA,AB sao cho AD,BE,CF ng quy v D,E,F khc trung im on thng.Gi M,N,P ln lt l im chung ca cc cp ng thng (EF,BC) ,(DF,CA) ,(DE,AB).Khi M,N,P thng hng Bn c th chng minh kt qu trn bng nh l Menelaus nhng thm ch bi ton m rng ny cng ch l trng hp c bit ca nh l Desargues m thi!!!!

II.9)im Kosnitanh ngha.Cho tam gic l tm ng trn ngoi tip tam gic.Gi l tm cc ng trn ngoi tip tam gic .Khi ba ng thng v ng quy ti im ca tam gic.Ch dn chng minh:Gi tng ng l giao im ca vi Ta c:Tng t vi im v ri sau nhn cc t l thc vi nhau ta c dpcm

II.8)im FeuerbachKt qu:Trong mt tam gic ,ng trn Euler tip xc vi ng trn ni tip ca n,v tip im c gi l im Feuerbach ca tam gic trn.Ch dn chng minh:(leductam post)Gi l tm ng trn Euler, ngoi tip, ni tip ca tam gic . l trc tm, l ng knh vung gc vi l hnh chiu ca ln l bn knh ng trn ngoi tip, ni tip. l chn ng phn gic gc 1.ng trn Euler tip xc trong vi ng trn ni tip.Ta d dng chng minh c V suy ra (1)Chng minh tip: V nn m cn ti do Vy Chiu h thc trn ln theo phng vung gc vi ta c:(2)T (1) v (2) ta suy ra:(3)Ta c:(4)T (3) v (4) ta c: Vy suy ra dpcm.Hon ton tng t ta cng c:2.ng trn Euler tip xc ngoi vi cc ng trn bng tipT {I}_{a} k {I}_{a}{X}_{a} vung gc vi BC do {I}_{a}S = SI nn {X}_{a}M = MNT (2) ta c:Vy (5)Mt khc: nn:(6)T (5) v (6) ta c: (7)T k . Trong tam gic vung ta c:

(8)T (7) v (8) ta c:Vy suy ra dpcm

B sung mt chng minh khc bng php nghch o :Xt c:+ ng trn ni tip tip xc vi theo th t ti .+ ng trn bng tip trong gc tip xc vi ti .+ ng trn Euler qua trung im 3 cnh l .K tip tuyn chung ca v tip xc vi chng ln lt ti . (ch l v ). Gi l tm v t trong ca v .Ta c:Nu th hin nhin tip xc vi v nn ta ch quan tm n trng hp .Khi , nn theo h thc Newton, .t , li c nn suy ra .Cui cng, xt php nghch o cc , phng tch . Nhng tip xc vi v cn bn thn 2 ng trn ny bt bin qua php nghch o ang xt nn ta cng c cng tip xc vi v .Lp lun tng t cho thy tip xc vi .Kt thc chng minh!

B :1, tam gic l phn gic. im nm trong mt phng tam gic th2, t gic ni tip c, khi phn gic gc v ng quy. [RIGHT][I][B]Ngun: MathScope.ORG[/B][/I][/RIGHT]

II.10)im Musselman,nh l Paul Yiu v im MusselmanKt qu: Cho tam gic . Cc im ln lt l im i xng vi qua cc cnh i din v l tm ng trn ngoi tip. Khi v cng i qua 1 im l nh ca im qua php nghch o ng trn ngoi tip tam gic .Ch dn chng minh:Gi v ln lt l nh ca im v qua php nghch o ng trn ngoi tip tam gic .Khi ta c:Suy ra By gi ta cn chng minh 3 im , v thng hng.Ta c b tm t c ca im v tm ng trn chn im do im v tm ng trn chn im l 2 im ng gic ca tam gic Mt khc - tm ng trn ngoi tip v - trc tm cng l 2 im ng gic ca tam gic nn suy ra Ta cng d dng chng minh c Vy hay ni cch khc 4 im v cng nm trn 1 ng trnTng t vi cc ng trn cn li ta suy ra dpcm im c gi l im ca tam gic nh l Paul Yiu v im Musselman: Vi gi thit nh trn th 3 ng trn v cng i qua im .Ch dn chng minh:V nh l ny em xin trch dn trc tip li gii ca Darij Grinberg:Theo chng minh trn th im nm trn v nn ta c: Ta c:Mt khc: Vy hay v cng nm trn ng trn.Tng t vi cc ng trn khc ta suy ra dpcm

II.11)Khi nim vng cc ca tam gic.Khi nim::Cho tam gic t . Chn cc ng cao i din cc nh l v .Khi vng cc ca tam gic l ng trn c tm l v bn knh xc nh vi Trong : - bn knh ng trn ngoi tip- di 3 cnh tam gic- 3 gc tam gic Ngoi ra ta c 1 tnh cht ca vng cc l: cho 3 im bt k chuyn ng trn cc ng cnh ca tam gic ABC dng cc ng trn c ng knh l on thng ni 1 nh vi im chuyn ng trn cnh i din th khi vng cc ca tam gic trc giao vi tt c cc ng trn .

II.12)im GibertKt qu:: Cho tam gic v im - nh ca im qua php nghch o ng trn ngoi tip tam gic . Gi l im i xng vi im qua cc ng cnh ca tam gic. Khi cc ng thng v ng quy ti im nm trn ng trn ngoi tip tam gic Ch dn chng minh: Gi giao im ca v l th ta c:

C

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II13/Trc Lemoinenh l:Cho tam gic ABC ni tip ng trn .Tip tuyn ti A ca ng trn ct ng thng BC ti X.nh ngha tng t cho Y,Z.Chng minh rng X,Y,Z thng hng v ng thng cha X,Y,Z c gi l trc Lemoine ca tam gic ABCChng minh:Khng mt tnh tng qut gi s hai vecto v cng hng.Do hai tam gic ABX v CXA ng dng vi nhau suy ra:Nhn hai ng thc trn vi nhau suy ra:Do X nm ngoi tam gic nn suy ra:Tng t ta c hai ng thc sau:Nhn ba ng thc vi nhau ta c:Theo nh l Menelaus th ba im X,Y,Z ng quy.Chng minh (2):Ta s chng minh ng i cc ca X,Y,Z i vi ln lt l ba ng i trung ca tam gic ABC.Edit later... [RIGHT][I][B]Ngun: MathScope.ORG[/B][/I][/RIGHT]

II.14)Tm MorleyKt qu::Tm th nht c nh ngha l tm ng trn ngoi tip ca tam gic Morley th nht. Tm th hai c nh ngha l tm phi cnh ca tam gic vi tam gic Morley th nht.Ch dn chng minh:p dng nh l tri Xeva cho 3 ng thng ng quy l v ta c:m Suy ra

ACB

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Tng t vi cc ng cn li ta c dpcm

II.15) Tm Spieker v ng thng NagelKt qu:Cho tam gic . Gi ln lt l trung im ca v . Tm ng trn ni tip tam gic l tm ca tam gic ABC.Khi 4 im tm ng trn ni tip, trng tm, im v tm cng nm trn ng thng ca tam gic Ch dn chng minh:Ta c php v t tm t s bin tam gic thnh tam gic nn suy ra (1)Xt im c b s tm t c:(2)T (1) v (2) ta suy ra im v cng nm trn 1 ng thng.Ngoi ra ta cn 1 vi tnh cht ca im Spieker nh n l tm ng phng ca 3 ng trn bng tip tam gic ABC, tm Spieker, im Brocard th 3 v im ng c vi tm ng trn ni tip thng hng...

II.16)Hai im FermatKt qu:Cho tam gic . Dng ra pha ngoi(vo trong) cc tam gic u v . Khi tm phi cnh ca tam gic v tam gic c gi l im th nht(th hai) hay ngi ta cn gi l im dng(m).Ch dn chng minh:Gi giao im ca v ln lt vi v l ta c:Tng t vi v ta c:

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Tng t ta suy ra dpcm Ngoi ra: Trong : l ln din tch tam gic

Xem them

The Fermat Point and Generalizations

P. Fermat (1601-1665) challenged Evangelista Torricelli(*) (1608-1647), the inventor of barometer with the following question

Find the point such that the sum of its distances from the vertices of a triangle is a minimum.

Torricelli presented several solutions. In one he observed that the circumcircles of the equilateral triangles constructed externally on the sides of a given triangle meet in a point. Many more have been found since. I'll present several solutions and two quite surprising generalizations.

Solution 1

(Published by Joseph Ehrenfried Hofmann (1900--1973) in 1929. It was independently discovered by Tibor Galai and others [Honsberger, p. 26].)

In ABC, select a point P and connect it with vertices A, B, and C. Rotate ABP 60o around B into position C'BP'. By construction, BPP' is equilateral, PB = P'B, and PA = C'P'. We thus have PA + PB + PC = C'P' + P'P + PC. As the image of A under the rotation, position of C' does not depend on P. Also, PA + PB + PC CC' because the broken line CPP'C' is no shorter than the straight line CC'. Therefore, PA+PB+PC reaches its minimum iff P lies on CC'. For this P, BPC' = 60o. Had we rotated ABP around A, we would have found that APC' = 60o.

The result is clearly related to Napoleon's theorem. On the sides of ABC construct equilateral triangles ABC', ACB', and BCA'. We know that the Fermat point P that minimizes the sum PA + PB + PC lies on CC'. By the same token, it lies on AA' and BB'. Therefore, it lies on their intersection. As far as Napoleon's theorem goes, the three lines AA', BB', and CC' are concurrent. (This was already known to Thomas Simpson (1710--1761).) Not only that but they cross at angles equal 120o. Furthermore, AA' = BB' = CC' since each of them equals PA + PB + PC for the Fermat point P.

So the Fermat point is unique and lies at the intersection of three straight lines that connect vertices of the triangle with opposite vertices of Napoleon's triangles. The construction fails if one of the internal angles of ABC is 120o or more. In this case, the vertex corresponding to the largest angle of the triangle solves Fermat's problem.

(There is a dynamic illustration pertinent to the above proof.)

Solution 2

This solution only works under the assumption that the Fermat point exists and is unique. Let for a given ABC, P be that point. Let's move, say, A a little into position A1 on the ray PA. Will P move along into a different position? Assuming it will and denoting its new position as P1, we arrive at contradiction. Indeed, for ABC,

PA + PB + PC < P1A + P1B + P1C

Also, by our assumption for A1BC,

P1A1 + P1B + P1C < PA1 + PB + PC

Summing up and cancelling, we get

PA + P1A1 < PA1 + P1A

Recollect that we selected A1 on the ray PA so that PA1 + A1A = PA. Which leads to

AA1 + P1A1 < P1A

which is of course absurd. Therefore, when A slides along the ray PA, point P does not change. Slide then vertices A and B so as to make ABC equilateral. This is always possible. For example, start with sliding A so as to make AB = BC. The triangle ABC becomes isosceles. It is obvious that, for an isosceles triangle, the Fermat point lies on the axis of symmetry of the triangle. For this reason, the triangle remains isosceles wherever B is located on that axis. Slide it so as to make all three sides of ABC equal. For an equilateral triangle, its centroid that also serves as the incenter and the circumcenter, serves as the Fermat point as well.

Solution 3

([Honsberger, p. 26] ascribes this solution to Torricelli himself and mentions that it was rediscovered almost 300 years later by F. Riesz. In a private communication, Douglas Rogers mentioned that the solution has been also proposed by J. Steiner. Thus the same object that is most frequently referred to as Fermat's point, Torricelli's point and Fermat-Torricelli point, is sometimes also named after J. Steiner, as Steiner's point, especially in the framework of Steiner's networks. [Tikhomirov, p. 31] and [Courant and Robbins, p. 354] even refer to Fermat's problem as Steiner's. [Johnson, p. 221] attributes the present solution to Steiner.)

The following solution depends on explicit knowledge of the fact that, for Fermat's point, AFB = BFC = AFC = 120o. Without proving that such a point exists, it shows that, if it does, it solves Fermat's problem.

At the vertices of ABC draw perpendiculars to lines AF, BF, and CF. The new lines form an equilateral triangle RST. Let F1 be a point different from F. Drop perpendiculars F1A1, F1B1 and F1C1 to the sides of RST. It's a nice property (known as Viviani's theorem) of equilateral triangles that in RST,

FA + FB + FC = F1A1 + F1B1 + F1C1.

(The sum of the distances from a point inside an equilateral triangle to the sides of the triangle does not depend on the point. )

On the other hand, obviously,

F1A1 + F1B1 + F1C1 < F1A + F1B + F1C

All that remains is to combine the two.

(Douglas Rogers made a delightful observation. RST is the largest equilateral triangle circumscribing ABC. According to Viviani's theorem, FA, FB, FC add up to the altitude of RST. For any other equlateral LMN curcumscribing ABC, at least one of FA, FB, or FC is not perpendicular to the side and hence is longer than such perpenduclar. It follows that the sum FA + FB + FC is bound to be larger than the alititude of LMN. This implies that the side of LMN is smaller than that of RST. In passing, RST is known as the antipedal triangle of F with respect two ABC.)

Solution 4

(The proof by Lou Talman is found on one of the discussions at the mathforum. Still very simple, it uses a little bit of analytic geometry and calculus. This is probably very close to one of Torricelli's original proofs.)

One can see that the Fermat point does minimize FA + FB + FC as follows: Let z be the sum FA + FB, where F is chosen to minimize FA + FB + FC, and consider the locus of points P that satisfy PA + PB = z. This locus is an ellipse with foci at A and B. Because F minimizes z + FC, the line determined by C and F must be normal to the tangent to this ellipse at F. By the reflection property of the ellipse, the angles AFC and BFC must be equal. A similar argument shows that the angles AFC (say) and AFB must also be equal. From this it follows that the three angles are all 120 degrees.

Solution 5

(The argument is similar to that in Solution 4, but from a different perspective. [F. G.-M., p. 442] credits Lhuilier (1811) with the proof.)

Assume FC is constant. Then point F that minimizes FA + FB + FC, minimizes also FA + FB and lies on the circle with center C and radius FC. The minimum is achieved for a point F on the circle for which the angles AFC and BFC are equal.

For more detail, draw a tangent to the circle at F and choose any point G on the tangent other than F. Let H be the intersection of HC with the circle. Then, assuming angles AFC and BFC equal,

FA + FB < GA + GB < HA + HB

(The latter inequality warrants extra attention. It follows from the fact the various ellipses with foci at A and B do not intersect.) From here,

FA + FB + FC < HA + HB + FC = HA + HB + HC.

The argument can be repeated assuming either FA or FB constant. As a result, at Fermat's point F all three angles AFC, BFC and AFB are found to be equal.

Solution 6

Fermat's point can be located with the help of Euler's generalization of Ptolemy's Theorem, see [Pedoe, pp. 93-94].

Any triangle has at least two acute angles. Given ABC, let angles at B and C be acute. Form an equlateral triangle BCD, with D and A on opposite sides of BC. Let (O) be the circumcircle of BCD.

For a point Q, by Ptolemy's inequality

BQCD + CQBD DQBC,

with equality only if Q lies on (O) and such that the quadrilateral BQCD is convex. Note that, by construction, BC = CD = BD which reduces the above to

BQ + CQ DQ.

Therefore

AQ + BQ + CQ AQ + DQ AD.

Unless, Q lies on AD, AQ + BQ + CQ > AD. Let P be the intersection of AD with (O) other than D. BPCD is an inscribed convex quadrilateral and P lies on AD. So in this (and only in this) case AP + BP + CP = AD. For any other selection of Q,

AQ + BQ + CQ > AP + BP + CP.

Angle BPC is supplementary to BDC = 60 so that BPC = 120. Further, D is in the middle of the arc BDC so that angles BPD and CPD are 60 making angles APC and APB both equal 120.

If the triangle is acute the same construction applies to the other two sides which brings up the framework of Napoleon's theorem. The three circles intersect at Fermat's point and three lines joining the vertices of ABC with the opposite vertices of the Napoleon triangles concur at the point P.

The latter fact can be used for a more direct proof.

Solution 7

(This was a part of a solution by Grimbal to a problem posted at the wu::forum.)

Construct Napoleon's equilateral triangles ABC', AB'C', A'BC externally on the sides of ABC. Let P be the intersection of AA' and BB'. AA' is CC' rotated clockwise 60 about B, and BB' is CC' rotated counterclockwise 60 about A, it follows that angle APB' is 60. Let X be the point on BB' making triangle APX equilateral. Now, if BB' is rotated clockwise 60 about A, then X goes to P, B' to C, and B to C'. Hence P, C, and C' are collinear; and so AA', BB', and CC' are concurrent, and any two of these lines make an angle of 120.

Searching for the Fermat point we discovered a nice property of Napoleon's triangles. I found quite surprising generalizations of those properties in a column by David Gale in The Mathematical Intelligencer which was kindly pointed out to me by Professor McWorter.

Two lines passing through a vertex of a triangle are called isogonal with respect to that vertex if they form equal angles with its (internal) angle bisector. Following is the first generalization.

Theorem 1

As in the diagram, assume lines AB' and AC' are isogonal as are pairs CB', CA' and BA', BC'. Then three lines AA', BB', and CC' are concurrent, i.e., meet at a common point.

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Napoleon's theorem is obtained when all three angles involved are equal to 30o. The common point is then known as the First Napoleon point. (The Second Napoleon Point is obtained when the equlateral triangles are formed internally to the given triangle. Similarly, there are two Fermat's points, which are also known as the first and second isogonic centers.) If the base angles are just equal between themselves, the theorem bears the name of Ludwig Kiepert who replaced equilateral triangles with isosceles ones. An interesting specification of the theorem is obtained when the three angles add up to 180o. But the theorem also admits further generalization.

In Theorem 1, pairs of (isogonal) lines were related to angle bisectors. As is well known from elementary geometry, angle bisectors meet at a single point (incidentally, the incenter of the triangle.)

Two lines AB' and AC' through a vertex A of a triangle are said to be isotomic if they intersect the opposite side BC in points equidistant from its midpoint Ma. Theorem 1 remains valid for isotomic lines as well. And, in a certain sense, for any three concurrent lines.

A second generalization that I am about to formulate and then prove belongs to the realm of Projective Geometry.

Theorem 2

Let p, q, r be concurrent lines through the vertices A, B, and C, respectively, of ABC. Let PA be the pencil of lines at A and let TA be the (unique) projective mapping on PA which

1. interchanges lines AB and AC,

2. leaves p fixed.

Define PA, PB, and TA, TB similarly. For any line a in PA, let a'=TA(a). Similarly, for b in PB and c in PC, let b'=TB(b) and c'=TC(c). Let C'=a'b, B'=c'a, and A'=b'c.

Then AA', BB', and CC' are concurrent.

Note that (an apparently more general) Theorem 2 follows from Theorem 1 with a projective mapping that leaves vertices A,B,C fixed but takes the incenter into any point P (the intersection of lines p, q, r.)

To prove Theorem 2, perform the projective mapping that carries C to the origin, A to the point at infinity at y-axis, B to the point at infinity at x-axis, and P to the point (1, 1).

Lines p, q, and r are carried onto the lines x = 1, y = 1, and x = y, respectively. We thus have

PA is the set of all vertical lines,PB is the set of all horizontal lines,PC is the set of all lines through the origin.

Transformation TA preserves the vertical direction and, therefore, is in reality a 1-dimensional projective transformation. So it's a Mbius transformation which, in general, has the form f(x)=(ax+b)/(cx+d). Transformations that interchange x=0 with the point at infinity are given by f(x)=a+b/x(c=1 and d=0.) Among those, there is a single one that leaves x=1 fixed: f(x)=1/x. So TA maps x=a onto x=1/a.

To simplify the notations, we will denote the vertical line with x=a by a, the horizontal line y=b by b, and the line y=cx through the origin, by c. Then with similar definitions for TB and TC we have

a' = TA(a) = 1/a, b' = TB(b) = 1/b, and c' = TC(c)

Further

C'=a'b = (1/a, b), B'=c'a = (a, a/c), andA'=b'c = (1/(bc), 1/b).

Thus AA' has equation x=1/(bc), BB' has equation y=a/c, and CC' has equation y = (ab)x. Then, AA'BB' = (1/(bc), a/c) which lies on the line CC'.

II.17)im Parry reflection.Kt qu:: Cho tam gic . K qua cc ng thng song song vi nhau v song song vi ng thng Euler ca tam gic. Gi ln lt l cc ng i xng vi qua v . Khi cc ng ny ng quy ti im Parry reflection ca tam gic ABC.Ch dn chng minh:Gi l trc tm tam gic , l nh ca qua php i xng trc v . Php v t tm t s bin tam gic thnh tam gic khi 2 ng thng ca 2 tam gic ny trng nhau. Gi l giao im ca vi v Ta c:

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(Xem them File FG200806.bdf)

II.18)ng trn Taylor ,tm TaylorKt qu:.Cho tam gic cc ng cao . T k cc ng vung gc vi cc cnh khi chn cc ng vung gc ny nm trn cng 1 ng trn gi l ng trn ca tam gic Ch dn chng minh:Ta c: Vy 4 im cng nm trn 1 ng trn.Tng t vi 2 b 4 im v .Ta c: Vy 4 im cng cng nm trn 1 ng trn hay ta suy ra c dpcm

II.19)im Bevan Kt qu:. Cho tam gic , l tm cc ng trn bng tip. Khi tm ng trn ngoi tip tam gic c gi l im ca tam gic .Sau y ta s n vi 1 vi tnh cht cua im : *Ta thy rng l tm ng trn ni tip tam gic chnh l trc tm ca tam gic khi ng trn ngoi tip tam gic ABC l ng trn chn im ca tam gic suy ra l trung im ca *Tm l trung im ca on ni trc tm vi im .u tin ta chng minh 3 im ny thng hng.Xt tam gic ta c:

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Vy 3 im ny thng hng.Xt tam gic ta c 3 im thng hng suy ra:Vy ta c dpcm*im l trung im ca on ni im v im .im l im i xng ca trc tm qua tm ng trn ngoi tip.Khi tng t nh trn ta cng c c dpcm[RIGHT][I][B]Ngun: MathScope.ORG[/B][/I][/RIGHT]

II.20)im VectenKt qu:. Cho tam gic . Dng ra pha ngoi(hay vo trong) cc hnh vung v . Khi ng ni 1 nh ca tam gic vi tm hnh vung dng trn cnh i din ng quy ti im Vecten ca tam gic .Ch dn chng minh:Theo nh l th iu ny hin nhin v cng d dng suy ra c c 2 im trong v ngoi(hay m v dng).Ngoi ra ta cn c 1 tnh cht kh th v v im : Tm ng trn chn im v 2 im thng hng.

II.21)im MittenpunktKt qu: Cho tam gic , l tm cc ng trn bng tip, ln lt l trung im cc cnh . Khi cc ng thng ng quy ti im ca tam gic .Ch dn chng minh:Ta c:

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Ngoi ra ta cn c 1 vi tnh cht bn l kh th v: Vi gi thit nh trn gi l im tip xc ca ng trn ni tip tam gic vi 3 cnh th ng quy. Khi im ny v im l 2 im ng gic.*im , tm v trc tm thng hng*im , tm ng trn ni tip v im thng hng.

*im , trng tm va im thng hng vi

II.22)im NapoleonKt qu:.Cho tam gic , dng ra pha ngoi( hay vo trong) cc tam gic u v . Khi ng ni 1 nh ca tam gic vi trng tm tam gic u dng trn cnh i din ng quy ti im Napoleon ca tam gic .Ch dn chng minh:p dng nh l ta suy ra dpcm cng d dng suy ra c 2 im l trong v ngoi

II.23)ng trn AdamKt qu:Cho tam gic ABC vi im Gergonne G.ng thng qua G song song vi EF ct AB,AC S,P.ng thng qua G song song vi DE ct AC,BC Q,M.ng thng qua G song song vi DF ct BA,BC R,N.Khi cc im M,N,P,Q,R,S cng thuc mt ng trn gi l ng trn Adam ca tam gic ABC.Ch dn chng minh:Gi (I) l ng trn ni tip tam gic ABC,tip im ca (I) trn BC,CA,AB ln lt l D,E,F.ng thng qua A v G song song vi BC tng ng ct DE,DF (H,K),(V,T).Ta thy:.Do AH=AK nn GT=GV.By gi rng GTDN v GVDM l hai hnh bnh hnh nn ta cng c DM=DN.Kt hp vi ID vung gc vi BC ta thu c IM=IN.Tng t IP=IQ,IR=ISMt khc d thy :IM=IQ=IR.T cc khng nh trn ta d nhn c iu cn chng minh.

II.24)Tam gic Fuhrmann ,ng trn FuhrmannKt qu:Cho tam gic ni tip ng trn tm . Gi l trung im cc cung BC, CA, AB. Ly cc im trn i xng qua cc cnh tng ng ta c 3 im na l v .Khi y tam gic c gi l tam gic ca tam gic .ng trn ngoi tip tam gic c gi l ng trn Tnh cht:1)

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2)3)Trc tm ca tam gic trng vi tm ng trn ni tip ca tam gic .( chng minh ta c th s dng tch v hng)4)Tm ng trn chn im ca tam gic v tam gic trng nhau. Bn knh ng trn ngoi tip tam gic c di bng

II.25)Hnh luc gic v ng trn Lemoine th nhtKt qu:Cho tam gic v im . Qua k cc ng thng song song vi cc cnh ct cc cnh cn li ti . Khi y lc gic c gi l lc gic th nht ca tam gic .Tnh cht:1)Lc gic th nht l lc gic ngoi tip. ng trn ngoi tip lc gic ny c gi l ng trn th nht:Do l ng thng i trung i qua trung im ca nn l ng i song tng ng cnh . Suy ra t gic l t gic ni tip.2)Cc cnh b kp gia cc ng song song t l vi ly tha bc ba cnh tng ng:Gi x, y, z l khong cch t L ti 3 cnh tam gic. Khi :3)3 on trn cng 1 cnh t l vi bnh phng cc cnh ca tam gic 4)Tm ca ng trn Lemoine th nht l trung im on ni im Lemoine vi tm ng trn ngoi tip tam gic ABC: vung gc vi khi ta d dng suy ra c dpcm.5)Bn knh Ngoi ra ta cn c cch vit khc: Trong l bn knh ng trn Lemoine th hai(s c ni n trong phn sau)

II.26)Hnh lc gic v ng trn Lemoine th haiKhi nim v ng i song:Cho tam gic . Vi 2 im bt k thuc cc cnh v ta c 2 kiu chn sao cho tam gic ng dng vi tam gic . Th nht l song song vi Th hai l t gic ni tip nh hnh v. Khi y v c gi l cc ng i song tng ng vi gc Tnh cht:ng i trung lun i qua trung im ca cc ng i song tng ng vi cng 1 nh.Kt qu:Cho tam gic v im . Qua k cc ng i song tng ng vi cc cnh ca tam gic ct cc cnh cn li ti Khi lc gic c gi l lc gic th hai.Tnh cht:1)Lc gic th hai ni tip 1 ng trn v ng trn ny c gi l ng trn th hai ca tam gic ABC.2)Bn knh Thc cht lc gic v ng trn ch l trng hp c bit ca lc gic v ng trn . Xem thm trong

II.27)im Euler ca T gic ni tipKt qu:Cho t gic ni tip . Gi ln lt l trc tm cc tam gic v . Khi y th cc ng thng v ng quy. im ng quy c gi l im ca t gic ni tip.Ch dn chng minh:Ta c song song v cng bng 2 ln khong cch t ti nn t gic l hnh bnh hnh suy ra giao nhau ti trung im mi ng.Tng t ta suy ra bn ng thng ng quy ti trung im mi ng.T y ta suy ra c nhiu tnh cht th v ca im 1)im nm trn ng vung gc h t trung im mt cnh ti cnh i din(hoc trung im ng cho ti ng cho cn li).2)ng thng ca nh vi tam gic th i qua im . Tng t vi cc nh cn li.3)ng trn chn im ca cc tam gic v ng quy ti im Xem thm

II.28)ng thng Steiner ca t gic ton phnKt qu:Cho t gic ton phn Khi trc tm ca cc tam gic v cng nm trn 1 ng thng c gi l ng thng ca t gic ton phn.Ch dn chng minh:Gi ln lt l trc tm cc tam gic v Gi l trung im cc ng cho Khi : Vy nm trn trc ng phng ca 2 ng trn v Tng t ta cng c 3 trc tm cn li cng nm trn trc ng phng ca hai ng trn ny suy ra dpcm.

II.29)ng thng Gauss ca t gic ton phn.Kt qu:Cho t gic ton phn . Khi trung im cc ng cho cng nm trn mt ng thng c gi l ng thng ca t gic ton phn.Ch dn chng minh:Gi ln lt l trung im cc ng cho v l tam gic trung bnh ca tam gic Khi cc im nm trn cc cnh ca tam gic .Ta c:Nhn cc v cc ng thc trn ta c:

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II.30) im Miquel ca t gic ton phnKt qu:Cho t gic ton phn . Khi y ng trn ngoi tip ca cc tam gic v ng quy. im ng quy c gi l im ca t gic ton phn.Ch dn chng minh:Gi s ng trn ngoi tip cc tam gic v giao nhau ti im khc Khi ta c:

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II.31)ng trn Miquel ca t gic ton phnKt qu:Cho t gic ton phn Khi im v tm ca cc ng trn ngoi tip tam gic v cng nm trn 1 ng trn - ng trn ca t gic ton phn.Ch dn chng minh:Gi v ln lt l tm ng trn ngoi tip cc tam gic v .Gi ln lt l chn cc ng vung gc k t ti v Do l trung im ca nn chng thng hng.Theo nh l o v ng thng Simson ta c cng nm trn 1 ng trn.Tng t ta suy ra dpcm.

(Thm phn t gic ton phn km theo, File BDF)

II.32)Hnh bnh hnh Varignon ca t gic .Kt qu:Cho t gic ABCD c M,N,P,Q ln lt l trung im ca AB,BC,CD,DA.Khi M,N,P,Q l bn nh ca mt hnh bnh hnh gi l hnh bnh hnh Varignon ca t gic ABCD. Ch dn chng minh:Chng minh kt qu ny kh n gin,d thy MN,PQ tng ng l ng trung bnh ca cc tam gic ABC v ACD th nn .Do vy MNPQ l mt hnh bnh hnh.

I.71)nh l Poncelet v bn knh ng trn ni tip,bng tip trong tam gic vung.nh l:Cho tam gic c ln lt l bn knh cc ng trn ni tip, bng tip gc .Chng minh rng: tam gic ABC vung ti A khi v ch khi .Chng minh::Ta c:tam gic vung ti (dpcm)

Nhng nh L Ton Hay(C Th Bn Cha Bit)

nh l Fuerbach:ng trn Eulerca mt tam gic lun tip xc trong vi ng trn ni tip v lun tip xc ngoi vi cc ng trn bng tip i vi mi cnh trong tam gic .ng thng Gauss:Trung im hai ng cho v trung im on thng ni giao im ca cc cnh i trong t gic l ba im thng hng.nh l Brianchon:Cc ng cho ca mt lc gic ngoi tip mt ng trn (hoc mt ng ellip) l ba ng thng ng qui.nh l Morley:Khi chia ba gc ca mt tam gic th giao im ca cc ng chia l ba nh ca mt hnh tam gic u.nh l khong cch Euler:Bnh phng khong cch t tm ca ng trn ngoi tip ti tm ca ng trn ni tip trong tam gic bng bnh phng bn knh ca ng trn ngoi tip tr cho hai ln tch gia bn knh ng trn ngoi tip v ni tip tam gic .ViRl bn knh ng trn ngoi tip vrl bn knh ng trn ni tip tam gic. Vy khong cch t tm ng trn ngoi tip v ni tip tam gic bng:

nh l ng thng Newton trong t gic ngoi tip:-iu 1:Nu mt t gic ngoi tip mt ng trn th tng cc cp cnh i bng nhau.-iu 2:Cc trung im hai ng cho trong t gic ngoi tip ng trn lun thng hng vi tm ca ng trn ni tip.nh l Casey:Nu cc ng trn tmkhng ct nhau v cng thuc min trong v ln lt tip xc trong vi ng trn tmth

-Trong :l tip tuyn ca cc ng trnv.Chui ng trn Steiner:Trong :-ng trn (vin) l ng trn nh ;-ng trn (vinxanh) l ng trn ln ;-ng trn (vinen) c gi l nhng ng trn tip xc xung quanh ;-ng trn (vincam) l ng trn ni cc im tip xc ngoi gia nhng ng trn tip xc xung quanh ;-ng trn (vinxanh l) l ng ellip ni tm cc ng trn tip xc xung quanh.Chui ng trn Pappus:Chui ng trn Pappusl trng hp c bit caChui ng trn Steiner.Trong , ng trn nh thuc min trong v tip xc trong vi ng trn ln.V tm nhng ng trn xung quanh lun nm trn cng mt ng trn.nh l Apollonius:Nu cho ba ng trn c chu vi khc nhau v mi ng trn cng ln lt tip xc vi cc ng trn cn li th lun lun tn ti mt ng trn tip xc vi c ba ng trn .nh l Brahmagupta:on thng ni giao im ca hai ng cho vung gc trong t gic ni tip ng trn vi trung im ca mt cnh bn th lun vung gc vi cnh bn i din.nh l Mbius (nh l Hnh lc gic Pascal Tng qut):Nu mt a gic 4n +2 cnh ni tip ng trn c cc cp cnh i khng song song th 2n + 1 giao im ca cc cp cnh i l cc im thng hng.nh l Euler:Nu cc s nguyn dng a v m nguyn t cng nhau th lun tn ti s t nhin k (k < m, k nguyn t cng nhau vi m) sao chochia ht cho m. Th k nhn mt trong hai gi tr:- Nu m l s nguyn t th k = m 1 ;- Nu m l hp s v c phn tch ra tha s nguyn t di dngth

nh l Euler - Fermat:Bt k s nguyn t no c dng 4n + 1 u l tng ca hai s bnh phng.

nh l Euler cho s hon chnh:S hon chnh chn ch c duy nht mt dngnh l Lagrange: Mi s t nhin u c th biu din di dng tng ca bn s bnh phng.nh l Gauss:Bt k mt a thc no trn trng s phc cng u phi c t nht mt nghim.Phng php dng hnh Tht thp gic u (Gauss):

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