Alexander Zverev edged to victory over Jenson Brooksby in Acapulco early on Tuesday morning in the latest-ever finish to a professional tennis match.
The Olympic champion saved two match points before completing a 3-6, 7-6 (12/10), 6-2 win at 4:54 am local time (1054 GMT).
The first-round tie at the ATP event lasted three hours and 19 minutes, the third match of Monday's schedule to pass the three-hour mark.
The previous latest finish was a 2008 Australian Open match between Lleyton Hewitt and Marcos Baghdatis that concluded at 4:34 am.
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Zverev and Brooksby were still playing when a doubles match, scheduled for Tuesday, had already finished in Dubai, which is 10 hours ahead of Acapulco.
American Brooksby took the opening set but saw two match points come and go in the second-set tie-break.
Zverev eventually forced a decider on his fifth set point, winning a set that took 111 minutes.
Despite appearing to suffer from a cramp, the German world number three managed to find two breaks of serve in the third set.
Zverev will face his compatriot Peter Gojowczyk in the last 16.
Later on Tuesday, Rafael Nadal will play for the first time since his record-breaking Australian Open triumph when he takes on Denis Kudla.
Daniil Medvedev, chasing Novak Djokovic's world number one spot, faces Benoit Paire.
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