Alex de Minaur has a golden opportunity for his best-ever result at a grand slam after defeating Australian compatriot Jordan Thompson in the fourth round.
The 10th seed started the match on fire before weathering a comeback to win 6-0 3-6 6-3 7-5 in two hours, 59 minutes to reach the quarter-finals at Flushing Meadows for the second time.
De Minaur will now meet Britain left-hander Jack Draper in the last eight on Thursday morning (AEST), with both aiming for their first-ever major semi-final.
It was a stunning start for De Minaur, who blitzed Thompson 6-0 in the first set as the Aussie underdog struggled to find any rhythm.
But Thompson improved in the second set, hung in through the first few games, before breaking in the sixth game and eventually serving it out 6-3.
The third set was a mirror image of the second, except De Minaur secured the break and served it out - grabbing the two-sets-to-one advantage.
It was rollercoaster in the first with Thompson snaring an early break before De Minaur broke consecutively to gain the advantage.
Thompson's fighting spirit was on show as he broke back to make it 4-4 but he couldn't stop De Minaur, who broke and then served the match out 7-5.
Despite winning all three matches against his upcoming semi-final opponent Draper, it will be a tough match for the Australian with the Brit in career-best form.
Draper hasn't lost a set on the way to his first-ever major quarter-final, defeating solid opponents in Tomas Machac, Botic Van de Zandschulp and Zhizen Zhang.
The winner of the De Minaur-Draper quarter-final will meet either Daniil Medvedev or Jannik Sinner/Tommy Paul in the semi-final.
De Minaur has reached three consecutive quarter-finals at majors in 2024, at Wimbledon, French Open and now the US Open.
It also his second appearance in the last-eight at Flushing Meadows, after he was beaten by Andrey Rublev in the 2020 quarter-final.