MIAMI GARDENS, Fla. — Bianca Andreescu of Canada — the 2019 US Open champion — beat 2020 Wimbledon 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 one-time Grand Slam champions. Andreescu converted all three break opportunities she had.
In other matches Sunday, world No. 9 Belinda Bencic of Switzerland, the gold medalist at the Tokyo Olympics, outlasted Ekaterina Alexandrova of Russia 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 and No. 10 Taylor Fritz advanced in straight sets.
Paul beat Alejandro Davidovich Fokina of Spain 6-3, 7-5, helped by eight aces. Fokina had six double faults.
Fritz defeated Denis Shapovalov of Canada 6-4, 6-4. Shapovalov had seven double faults.
“It’s going to be a tough match I think for a round of 16,” Fritz said. “Myself playing Holger [Rune] is a very tough draw. I’m excited to play him. We’ve never played before. I’m not entirely I guess sure what to make of his game. Obviously he’s very good.”
Also, world No. 1 Carlos Alcaraz of Spain beat Dusan Lajovic of Serbia 6-0, 7-6 (5), 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.