Miami Open: Andreescu, Paul advance to fourth round

Andreescu advanced to the fourth round of the Miami Open for the third time. She had seven aces to Kenin’s one and double-faulted only once.

Bianca Andreescu returns a shot to Iga Swiatek during the BNP Paribas Open at the Indian Wells Tennis Garden.. (FILE PHOTO) | Photo Credit: GETTY IMAGES

Bianca Andreescu of Canada — the 2019 US Open champion — beat 2020 Australian Open winner Sofia Kenin of the United States 6-4, 6-4 in the third round of the Miami Open on Sunday.

Andreescu advanced to the fourth round of the Miami Open for the third time. She had seven aces to Kenin’s one and double-faulted only once as she won the third straight matchup between the Grand Slam champions. Andreescu converted all three of her break opportunities.

In other matches, Ekaterina Alexandrova of Russia defeated world No. 9 Belinda Bencic of Switzerland, the gold medalist at the Tokyo Olympics, 7-6 (8), 6-3; Marketa Vondrousova ousted Karolina Pliskova 6-1, 6-2 in an all-Czech matchup; Sorana Cirstea of Romania beat Karolína Muchová of the Czech Republic 7-5, 6-1; and Varvara Gracheva of Russia defeated Magdalena Frech of Poland 6-1, 6-2.

On the men’s side, Americans Tommy Paul advanced in straight sets.

Paul beat Alejandro Davidovich Fokina of Spain 6-3, 7-5, helped by eight aces. Fokina had six double-faults.

Also, world No. 8 Rune of Denmark defeated Diego Schwartzman of Argentina 6-4, 6-2, and Andrey Rublev of Russia dispatched Miomir Kecmanović of Serbian 6-1, 6-2.

Later Sunday, Australian Open winner and world No. 2 Aryna Sabalenka of Belarus plays Marie Bouzkova of the Czech Republic, and world No. 4 Casper Ruud of Norway takes on Botic van de Zandschulp of the Netherlands.