Photo : Martin Sidorjak
As he often does, Denis Shapovalov looks to be finding his best form in the late stages of the season as the Canadian is through to the semi-finals of the Korea Open without having dropped a set.
He was clinical on Friday in his quarter-final clash with Radu Albot, never facing a break point on his way to a 6-2, 6-2 victory. It is the first time since February in Dubai that Shapovalov has won consecutive matches in straight sets, having beaten Jaime Munar in his opening match on Wednesday.
On Friday, Shapovalov was at his big-hitting best, ripping 26 winners as he blasted every ball in sight. He was also locked in on serve, winning 88 per cent of his first-serve points, only losing nine total points on serve, and never losing more than two points in a service game.
It was Shapovalov setting the tone for the match right from the start, breaking serve in the opening game to take an early lead.
Everything was going the Canadian’s way in the first set as Albot drove a backhand wide to give Shapovalov a double-break lead at 4-1. The only thing that did not go perfectly for Shapovalov was failing to convert two set points on his opponent’s serve at 5-1, but when called upon to serve out the set, he closed it out on his third set point with an ace.
Shapovalov continued to attack every ball with great success in the second set. When he was not hitting winners, he was drawing errors off the racket of Albot which allowed him to break for a 2-1 lead.
Once ahead, there was no let up from the fourth seed. His great hitting applied to everywhere in the court as some smart approaches to the net allowed him to add a second break to move ahead 5-2.
There were no hiccups getting over the finish line, as Shapovalov capped off his masterful performance with a hold to love.
The Canadian will meet American Jenson Brooksby, who got a free pass after second seed Cameron Norrie withdrew with an illness, in the semifinals.
Three of Shapovalov’s four career finals have come in the post-US Open period of the season.
