By Chris Oddo | @TheFanChild | Wednesday October 30, 2019
Alexander Zverev qualified for the 2019 Nitto ATP Finals, leaving just one spot to be claimed over the weekend in Paris.
Photo Source: Christopher Levy
Defending Nitto ATP Finals champion Alexander Zverev is heading back to London. The German officially qualified on Wednesday in Paris when Matteo Berrettini fell to Jo-Wilfried Tsonga in the last match of the evening.
Zverev will make his third consecutive trip to the season-ending championships, where he became the first man to defeat both Roger Federer and Novak Djokovic in the same ATP Finals en route to winning the title in 2018.
“I am really happy to qualify for the third consecutive year,” said Zverev. “Winning the title there last year has been the highlight of my career so far. I have such great memories of The O2.”
Berrettini’s loss also opens the door for a trio of challengers with designs on the eighth and final London slot.
Gael Monfils is closest at the moment. The Frenchman took out Benoit Paire to reach the third round and he can pass Berrettini (2660 points) if he reaches the semifinals in Paris. Monfils (2440 points) last reached the ATP Finals in 2016. He will face Radu Albot in the round of 16 at Paris on Thursday.
Stan Wawrinka (2000) points is still in the hunt, but he faces his nemesis Rafael Nadal on Thursday and would need to win the title to pass Berrettini. Alex de Minaur (1775 points) could potentially qualify, but he would need to win the title as well. He faces Stefanos Tsitsipas in third-round action on Thursdsay.
The draw for Paris shapes up as follow: