Andrei Mlendea vs Teodor Bolanu — Romania F17 Round of 32
Romania F17 · ITF · Round of 32
ROUAndrei Mlendea
Last 10 · pre-match
LWLWWWWLWL
2 — 1
Sets
Monday, 22 August 2005 ·
clay
ROUTeodor Bolanu
Last 10 · pre-match
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Mlendea A.4-6 6-4 6-0
Bolanu T.
Mlendea A.
4662
Bolanu T.
6401
Andrei Mlendea defeated Teodor Bolanu 4-6 6-4 6-0 in the Romania F17 round of 32 on clay. The result followed the form book — Andrei Mlendea came in leading the head-to-head 2–0. Andrei Mlendea extended the head-to-head to 3–0.
Key context
- Match story: Andrei Mlendea dropped the opening set before beating Teodor Bolanu 4–6 6–4 6–0.
Head to head
2–0| Date | Tournament | Round | Surface | Winner | Score |
|---|---|---|---|---|---|
| 2006-08-21 | Romania F17 | QF | clay | Andrei Mlendea | 6-2 6-2 |
| 2005-08-29 | Romania F18 | R16 | clay | Andrei Mlendea | 7-5 6-4 |