2005 BCR Open Romania – Doubles

Doubles
2005 BCR Open Romania
Champions José Acasuso
Sebastián Prieto
Runners-up Victor Hănescu
Andrei Pavel
Score6–3, 4–6, 6–3

Lucas Arnold Ker and Mariano Hood were the defending champions, but Arnold Ker did not compete this year. Hood teamed up with Martín García and lost in the quarterfinals to Juan Ignacio Chela and Luis Horna.

José Acasuso and Sebastián Prieto won the title by defeating Victor Hănescu and Andrei Pavel 6–3, 4–6, 6–3 in the final.[1]

Seeds

Draw

Key

Draw

First round Quarterfinals Semifinals Final
1 M García
M Hood
6 56 6
WC V Ioniță
R Sabău
1 68 2 1 M García
M Hood
3 3
JI Chela
L Horna
6 6 JI Chela
L Horna
6 6
T Behrend
A Waske
2 2 JI Chela
L Horna
3 4
3 J Acasuso
S Prieto
6 6 3 J Acasuso
S Prieto
6 6
J Coetzee
C Haggard
4 4 3 J Acasuso
S Prieto
6 6
F Mayer
R Wassen
6 67 F Mayer
R Wassen
1 2
PH Mathieu
R Rochus
4 51 3 J Acasuso
S Prieto
6 4 6
V Hănescu
A Pavel
6 67 V Hănescu
A Pavel
3 6 3
P Pála
D Škoch
4 53 V Hănescu
A Pavel
6 67
N Almagro
A Martín
2 2 4 M Fyrstenberg
M Matkowski
2 53
4 M Fyrstenberg
M Matkowski
6 6 V Hănescu
A Pavel
6 6
WC I Moldovan
G Moraru
6 6 WC I Moldovan
G Moraru
4 4
C Berlocq
M Puerta
1 2 WC I Moldovan
G Moraru
610 6
D Sanguinetti
T Vanhoudt
6 67 D Sanguinetti
T Vanhoudt
58 4
2 J Knowle
M Kohlmann
4 50

References

  1. ^ "Unseeded Serra Wins 1st Career Title at ATP Bucharest". Tennis-X. 19 September 2005. Retrieved 8 December 2020. In the doubles final, No. 3 seeds José Acasuso and Sebastián Prieto defeated unseeded Romanians Victor Hănescu and Andrei Pavel 6-3, 4-6, 6-3 for their second title in their fourth final of the year.