Andrey Rublev wins 2020 St. Petersburg Open

Andrey Rublev of Moscow, Russia won the 2020 St. Petersburg Open on Sunday. With the win, Rublev moves from 10th to eighth in the ATP World Rankings.

In Sunday’s final, Rublev, who was the tournament’s third seed, defeated Borna Coric of Croatia, the tournament’s seventh seed, 7-6, 6-4. Rublev defeated Coric 7-5 in the first set tiebreak after being down 5-2, and then broke Coric’s serve in the fifth game of the second set to take complete control of the match.

One reason for Rublev’s success over Coric on Sunday came down to the percentage of points won on second serve. Rublev won 74% of his points on second serve, while Coric only won 48%.

In reaching Sunday’s final, Rublev defeated Vasek Pospisil of Canada 6-2, 6-4 in the first round, Ugo Humbert of France 4-6, 6-4, 7-5 in the second round, Cameron Norrie of Great Britain very easily 6-2, 6-1 in the quarterfinals, and second seed Denis Shapovalov of Canada 4-6, 6-3, 6-4 in the semifinals.

This was the first time that the St. Petersburg Open was part of the ATP Tour 500 series. Rublev became the fourth Russian to win the tournament. Yevgeny Kafelnikov won in 1995, Marat Safin won in back-to-back years in 2000 and 2001, and Daniil Medvedev won in 2019.

With the win, Rublev has now won four ATP Titles in 2020. That ties him for the most with world number one, Novak Djokovic of Serbia. Rublev won the Qatar Open and Adelaide International in back-to-back weeks in January, along with Hamburg European Open in Germany right before the French Open. Djokovic’s four titles in 2020 came at the Australian Open, the Dubai Tennis Championship, the Cincinnati Masters (played in New York due to coronavirus), and the Italian Open.

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