Bogdan Leonte vs Teodor Bolanu — Romania F16 Round of 32
Bogdan Leonte defeated Teodor Bolanu 6-3 5-7 6-2 in the Romania F16 round of 32 on clay. The upset overturned the form book — Teodor Bolanu came in leading the head-to-head 1–0, but Bogdan Leonte had been the form player, winning 3 of their last 9 on clay. Bogdan Leonte claimed a first win over Teodor Bolanu in two meetings, levelling the head-to-head at 1–1.
Key context
- Match story: Bogdan Leonte closed it out in three sets, 6–3 5–7 6–2.
How the match was decided
Auto-generatedLeonte B. came through in three sets.
Common opponents
0 shared · last 12 monthsNo common opponents in the last 12 months. Surface form and ELO carry more weight here.
Head-to-head
2 meetings · level 1–111
- Leonte B.August 2006Romania F16This matchClay6-3 5-7 6-2
- Bolanu T.July 2006Romania F12Clay6-3 6-4
Draw
Closing odds
Next up
Player context
Before the matchLeonte B.Bolanu T.1 503ELO1 4471 511Clay ELO1 4533–7Last 100–10
Cite this
Bogdan Leonte beat Teodor Bolanu 6-3 5-7 6-2. Umpiry, "Bogdan Leonte vs Teodor Bolanu — Romania F16 Round of 32", https://umpiry.com/match/2006-08-14/bolanu-vs-leonte-romania-f16-r32?utm_source=cite (as of 14 August 2006).