Dear Luis,

If the line II_a meets BC at E then

since BI, BI_a are bisectors of <ABC the points A, E are

the harmonic conjugates relative to I, I_a.

Since AH_a is perpendicular to BC the circumcircle of AH_aE

is the Apollonius circle of triangle H_aII_a and hence

H_aE is internal bisector of triangle H_aII_a and

hence the line BC can be constructed.

Since we know that DI = DB = DC the points B, C can be

found as intersections of line BC with the circle (D, DI).

Finally A can be constructed as intersection of line II_a

with the perpendicular from H_a to BC.

Best regards

Nikos Dergiades

> Dear Hyacinthists,

>

> Problem 13 from the link below: TC given

> (H_a,I,I_a). I would like to have a hint to the

> TConstruction.

> Let D be the midpoint of the segment (I,I_a).

> I know that D belongs to the circuncircle and B,C

> belong to the circle (D,DI).

>

> Best regards,

> Luis

>

>

>

> To: Hyacinthos@yahoogroups.comFrom:

> alex_geom@...: Thu, 1 Jan 2009 10:37:22

> +0300Subject: [EMHL] Olimpiada

>

>

>

> Dear friends!Happy New Year!Here is a link for geometrical

> olimpiad in memory of

> I.F.Sharygin.http://www.geometry.ru/olimp.htmBest

> regards,Alexei Myakishev[Non-text portions of this message

> have been removed]

>

>

>

>

>

> _________________________________________________________________

> Cansado de espaço para só 50 fotos? Conheça o Spaces, o

> site de relacionamentos com até 6,000 fotos!

> http://www.amigosdomessenger.com.br

>

> [Non-text portions of this message have been removed]

>

>

> ------------------------------------

>

> Yahoo! Groups Links

>

>

>

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