Tennis player Botic van de Zandschulp has failed to reach the fourth round of the Australian Open. In the third round, for which he had qualified as the first Dutch man since 2011, he lost to Daniil Medvedev on Saturday. The Russian number two in the world won 6-4, 6-4 and 6-2.
Van de Zandschulp played more sloppily than Medvedev: he hit 32 balls out or into the net, while his opponent did it nineteen times. The Dutchman knew that he could afford few mistakes against Medvedev. At the end of last year, he already lost to him in the quarterfinals of the US Open, the tournament that Medvedev would eventually win. His service also went a lot better on Saturday: Medvedev hit ten aces, Van de Zandschulp three.
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Nevertheless, the scores in the first two sets were still fairly close to each other. The players kept all their own games, with one exception: Van de Zandschulp was broken once by Medvedev in both the first and second set. Because he got three break points himself but was unable to use any of them, both lost serves also meant an immediate loss of set. In the third set, Van de Zandschulp made more mistakes, and it was over quite quickly: Medvedev broke Van de Zandschulp twice and won on his first match point.
Number 57 in the world Van de Zandschulp was the first Dutchman since Robin Haase eleven years ago who survived two rounds in Melbourne. In the first round he dealt with the German Jan-Lennard Struff in three sets, after which he made it through the second round thanks to a withdrawal from the French former top ten player Richard Gasquet. Last year, Van de Zandschulp finally broke through into the world top thanks to his unexpected series of victories during the US Open. He is now the highest placed Dutchman in the world ranking.