Alex de Minaur upsets Nishikori in the 3rd round at the US Open

ATP
Friday, 30 August 2019 at 20:16
DeMinaur_Alex_Shanghai2018

Kei Nishikori advanced to the semi-final at the US Open last year, but will not repeat it this time as he was outplayed by Alex de Minaur in the third round 6-2, 6-4, 4-6, 6-4.

De Minaur did not ever make it to the last 16 on the Grand Slam. His Japanese opponent, on the other hand, played at least quarterfinals in the last five Grand Slams. The Australian wanted to change this statistic in the match played on Grandstand.
De Minaur came to Flushing Meadows in good form, having already won tournaments in Sydney and Atlanta this year. In the third match, de Minaur faced world number seven Nishikori with whom he had not met on the ATP Tour yet.
The match started much better for the young Australian, winning 6-2 in the first set and 6-4 in the second. In the third set, it looked like Nishikori could turn the match when he won it 6-2.
But the 4th set was again directed by de Minaur, who took the opponent's serve three times to win this set 6-3.
In the next round, Aussie will meet the winner of a match between Dimitrov or Majchrzak and will play for the first time among the last 16 on the Grand Slam.
[embed]https://twitter.com/usopen/status/1167499630369808384[/embed]

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