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Rublev wins epic 5-setter vs de Minaur; Djokovic, Sabalenka dominate

Melbourne, Australia, Jan 21 (EFE).- Russian Andrey Rublev fought back from two sets down on Sunday night to defeat Australia’s Alex de Minaur 6-4, 6-7 (5-7), 6-7 (4-7), 6-3, 6-0 in the fourth round of the Australian Open in Melbourne.

The fifth seed took four hours and 14 minutes to defeat his 10th-seeded opponent, whose defeat means the home crowd will have no more players left to root for in either the men’s or the women’s singles competitions.

No Australian male player has won the singles title in Melbourne since 1976.

Earlier on Sunday, reigning champion and top seed Novak Djokovic was dominant once again as he made light work of France’s 20th seed Adrian Mannarino.

The Serb, bidding to extend his Australian Open title haul to 11 and his all-time Grand Slam record to 25, sealed his spot in the quarter finals with a 6-0, 6-0, 6-3 victory.

Djokovic will face US 12th seed Taylor Fritz in the last eight, while Rublev will take on Italy’s fourth seed Jannik Sinner who defeated Russian Karen Khachanov in straight sets.

Aryna Sabalenka, the highest seed remaining in the women’s competition, continued her imperious form with a straight sets win over American Amanda Anisimova.

The reigning champion will face Czech Barbora Krejcikova in the quarter finals.

US fourth seed Coco Gauff also powered into the last eight with a 6-1, 6-2 win against Poland’s Magdalena Frech.

The American and reigning US Open champion, the second highest ranked player left in the draw, has yet to drop a set at the Australian Open this year.

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