Kei Nishikori into fourth round at Australian Open

Associated PressJan 19, 2019, 1:37 AM EST

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MELBOURNE, Australia — Kei Nishikori got through the third round in straight sets against Joao Sousa, getting a bit of relief after his first two matches at Melbourne Park went to five sets.

The eighth-seeded Nishikori had a 7-6 (6), 6-1, 6-2 win on Margaret Court Arena right after Naomi Osaka’s three-set comeback win over Hsieh Su-wei.

Nishikori, the 2014 U.S. Open finalist, fended off 59 aces from 39-year-old Ivo Karlovic in the second round before winning the fifth-set tiebreaker 10-7.

Sousa had also advanced through two five-setters and needed a medical time out in the third set.

The last time Nishikori won consecutive matches in five sets at a major was at the U.S. Open in 2014, when he beat Milos Raonic and Stan Wawrinka in five on the way to his semifinal win over Novak Djokovic.

It was Nishikori’s third straight game on Margaret Court Arena, and he said he enjoyed the vibe – particularly after playing after fellow Japanese player Osaka, the U.S. Open champion.

“Forehand has been working very well. Second and third sets I was playing really aggressive, and everything was working well,” he said.

He has reached the fourth round in Australia for the seventh time and is on a roll after winning the season-opening tournament in Brisbane.

Alcaraz, Sabalenka advance to 4th round of Miami Open

Associated PressMar 27, 2023, 12:13 PM EDT

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MIAMI GARDENS, Fla. — Top-ranked Carlos Alcaraz of Spain beat Dusan Lajovic of Serbia 6-0, 7-6 (5) and will face American Tommy Paul.

World No. 2 Aryna Sabalenka of Belarus cruised past Marie Bouzkova of the Czech Republic 6-1, 6-2 on Sunday.

“I lost the first and only match that I played against Tommy” Alcaraz said. “I know that he’s a really talented and really tough player, so I have to play at my best. Let’s see what’s going to happen on Tuesday.

Paul and fellow American No. 10 Taylor Fritz advanced in straight sets.

Bianca Andreescu of Canada – the 2019 U.S. Open champion – beat 2020 Australian Open winner Sofia Kenin of the United States 6-4, 6-4.

Andreescu advanced to the tournament’s fourth round 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, Barbora Krejcikova of the Czech Republic outlasted American Madison Keys 7-6 (4), 6-3.

Also, 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, Botic van de Zandschulp of the Netherlands outlasted world No. 4 Casper Ruud of Norway 3-6, 6-4, 6-4.

Van de Zandschulp had 12 aces to Ruud’s six, but they each had three double-faults. Van de Zandschulp will face Emil Ruusuvuori of Finland in the next round.

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. Fritz will face world No. 8 Holger Rune of Denmark, who defeated Diego Schwartzman of Argentina 6-4, 6-2.

“It’s going to be a tough match I think for a round of 16,” Fritz said. “Myself playing Holger 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, Andrey Rublev of Russia dispatched Miomir Kecmanović of Serbia 6-1, 6-2.

Andreescu, Paul advance to 4th round at Miami Open

Associated PressMar 26, 2023, 6:06 PM EDT

Geoff Burke/USA TODAY Sports

MIAMI GARDENS, Fla. — Bianca Andreescu of Canada – the 2019 U.S. 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.

In other matches, Australian Open winner and world No. 2 Aryna Sabalenka of Belarus defeated Marie Bouzkova of the Czech Republic 6-1, 6-2; and Barbora Krejcikova of the Czech Republic outlasted American Madison Keys 7-6 (4), 6-3.

On the men’s side, Botic van de Zandschulp of the Netherlands outlasted world No. 4 Casper Ruud of Norway 3-6, 6-4, 6-4.

Van de Zandschulp had 12 aces to Ruud’s six, but they each had three double-faults. Van de Zandschulp will face Emil Ruusuvuori of Finland in the next round.

World No. 1 Carlos Alcaraz of Spain beat Dusan Lajovic of Serbia 6-0, 7-6 (5). Alcaraz faces American Tommy Paul on Tuesday.

“Yeah, today has been a great match, great level. I hope to play at this level on Tuesday,” Alcaraz said. “I lost the first and only match that I played against Tommy. I know that he’s a really talented and really tough player, so I have to play at my best. Let’s see what’s going to happen on Tuesday.”

Paul and fellow American 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. Fritz will face world No. 8 Holger Rune of Denmark, who defeated Diego Schwartzman of Argentina 6-4, 6-2.

Also, Andrey Rublev of Russia dispatched Miomir Kecmanović of Serbian 6-1, 6-2.