[obm-l] problema de geometria e link de IMO's

[Luís Lopes <qed_texte@xxxxxxxxxxx>]

Sauda,c~oes,

No email editado abaixo tem um problema de geometria,
sua fonte (um jornal de Hong Kong com o link) e uma
discussão de sua solução.

Se o Claudio (obrigado pelas demonstrações, muito claras)
não conhece, o jornal de HK traz muitos problemas tipo IMO.

[]'s
Luís



Dear all my friends:

The problem angles is from

Mathematical Excalibur
Vol 7 nº 3, problem 158

http://www.math.ust.hk/excalibur/v7_n3.pdf

Best regard

Ricardo


Vladimir Dubrovsky wrote: Dear Tuan, Ricardo, Kostas, Tarik, Francois and 
Nikolaos

Somehow I missed some posts and found another solution to Tuan's question.

It seems to be shorter, so I decided to add it to the collection.

So we have point P on AD such that Angle BPD=Angle BAC=2Angle DPC.

Extend AD to meet the circumcircle of ABC at E; let CE=d, BE=e.

Then

BD:DC=area(ABE) :area(ACE) =ce/bd. (*)

Triangle PBE is similar to ABC; hence PE=be/a. (This is, actually,

Francois's similarity.)

Triangle CPF is similar to A'AB, where AA' is the bisector of A (this is

another similarity, but a transformational argument escapes me; maybe

Francois can shed light on it).

Hence PE= cd/BA' =(b+c)d/a.

It remains to equate the two expressions for PE and substitute the resulting

e/d=(b+c)/b into (*).

Notice that if we take triangle EBC as the initial one (instead of ABC),

then P will be on the *extension* of ED. The original relation between the

angles is violated, but it will remain the same if we think of angles as

oriented angles between lines rather than rays. So, in a certain way, this

answers Tarik's question: BD/DC = e(e-d)/dd.

Best regards,

Vladimir

============ ==

[QTB]

>>This fact is still true for any triangle ABC. In

>>this case, D divide BC by one simple ratio depending

>>on sides a, b, c. What is this ratio and how is the

>>proof for this general case?

>

[TA]

>Dear kostas,

>if the point P is on the extension of the line AD

>will your proof work there?

>Moon Bangladesh

>

[ND]

>We construct a point D on BC such that

>BD/DC = k =c(b+c)/bb

>The parallel from D to AC meets AB at E.

>The circumcircle EBC meets AC at F.

>The circumcircle EBD meets AD at P.

>It is easy to prove that DF is parallel to

>AA' the A_bisector

>from CA'=ac/(b+c) DC=a/(k+1) AE = c/(k+1)

>AE.AB = AF.AC and CF/CA = CD/CA'

>Hence the ratio is k. For b=c we get k=2

>as in Barosso's problem.

>

>The contruction of D is as follows:

>We construct the parallelogram BCAC1.

>The parallel from C to AA' meets AB at C2

>The reflection of A in B is the point C3.

>The circumcircle of C1C2C3 meets the line

>BC1 at C4. The line AC4 meets BC at the

>point we want D.

>

>Best regards

>Nikos Dergiades

[FR]

>Here another proof where I have tried to minimize the number of auxiliary

>points.

>

>Let f be the direct similarity of center B sending A to P and E = f(C).

>By hypothesis, A, P, E are on a same line.

>We have the following equalities between angles of lines:

>(CA, CB) = (EP, EB) (invariance of angles by a direct similarity)

>(EP, EB) = (EA, EB) (for A, B, P are on a same line)

>So (EA, EB) = (CA, CB) and points A, B, C, E are cocyclic (i.e on a same

>circle)

>

>Now as A is on the perpendicular bisector of segment BC, line EA is a

>bisector of angle > The other bisector is through the point D" harmonic 
>conjugate of D wrt BC.

>But by assumption done on D (i.e BD = 2 DC), C is the middle of BD".

>

>Let F be the middle of segment BE. Then FC is parallel to ED"and hence

>perpendicular to line EDPA.

>Hence C and F are symmetric wrt this last line and we are done :

> >

>Un abrazo

>Francois

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