ATP · Challenger 50

Phan Thiet 4 Challenger 2026 Tournament Results

In brief

Phan Thiet 4 Challenger 2026 is a Challenger 50 tournament on the ATP tour that started on September 14, 2026. The title went to Ilia Simakin, who beat Rodrigo Pacheco Mendez in the final (6-7(4) 6-4 6-3). The draw features 18 players from 6 countries. Best represented: Australia (6). 16 matches played, 3 tie-breaks and no retirements.

Bracket

Final

Ilia Simakin 2

Rodrigo Pacheco Mendez 1

Roll of honour

Year	Champion	Runner-up
2026	Ilia Simakin	Rodrigo Pacheco Mendez 6-7(4) 6-4 6-3

Every edition of Phan Thiet 4 Challenger →

The tournament in numbers

Matches played: 16
Tie-breaks: 3
Matches to a deciding set: 5
6-0 sets: 0
Retirements: 0

Most games in a match: Philip Sekulic-Kenta Miyoshi, 6-7(5) 7-6(5) 6-3 (35 games).

Countries in the draw

Country	Players	Best result
Australia	6	Quarterfinal
Japan	5	Round of 16
France	4	Semifinal
Russia	1	Title
Mexico	1	Final
Spain	1	R32

Points on offer

ATP points by round reached at this edition.

Round reached	Points
Title	50
Final	30
Semifinal	18
Quarterfinal	10
Round of 16	5
R32	1

R32

Winner	Loser	Score
Rodrigo Pacheco Mendez	Sergio Callejon Hernando	6-3 7-5
Kenta Miyoshi	Moerani Bouzige	6-4 6-4
Enzo Aguiard	Masamichi Imamura	6-3 6-2
Philip Sekulic	Yuta Shimizu	6-2 2-6 7-5
Arthur Weber	Koki Matsuda	6-1 6-2
Blake Ellis	Hiroki Moriya	6-4 5-7 6-3

Round of 16

Winner	Loser	Score
Rodrigo Pacheco Mendez	Arthur Weber	6-4 6-2
Philip Sekulic	Kenta Miyoshi	6-7(5) 7-6(5) 6-3
Matthew Dellavedova	Enzo Aguiard	6-3 6-1
Ilia Simakin	Blake Ellis	6-4 6-2
Cruz Hewitt	Constantin Bittoun Kouzmine	7-5 6-2

Quarterfinal

Winner	Loser	Score
Ilia Simakin	Cruz Hewitt	6-3 6-1
Moise Kouame	Florent Bax	6-4 6-2
Rodrigo Pacheco Mendez	Philip Sekulic	6-4 6-4

Semifinal

Winner	Loser	Score
Rodrigo Pacheco Mendez	Moise Kouame	2-6 6-3 6-4

Final

Winner	Loser	Score
Ilia Simakin	Rodrigo Pacheco Mendez	6-7(4) 6-4 6-3