(Reuters) – A look at the key facts and records of American Sofia Kenin and Czech Republic’s Petra Kvitova before their French Open semi-final on Thursday (prefix number denotes seeding):
WTA ranking: 6 (Highest ranking: 4)
Grand Slam titles: 1 (Australian Open 2020)
Career WTA titles: 5
2019 French Open performance: Fourth round
Best French Open performance: Semi-finals (2020)
ROAD TO FINAL
First round: Ludmilla Samsonova (Russia) 6-4 3-6 6-3
Second round: Ana Bogdan (Romania) 3-6 6-3 6-2
Third round: Irina Bara (Romania) 6-2 6-0
Fourth round: Fiona Ferro (France) 2-6 6-2 6-1
Quarter-finals: Danielle Collins (U.S.) 6-4 4-6 6-0
The reigning Australian Open champion reached the French Open semi-finals for the first time in her career with a battling victory over American compatriot Danielle Collins.
Kenin has won four out of her five matches at Roland Garros in three sets and faces another potentially tough test against Kvitova, who has beaten her in their two previous meetings.
WTA ranking: 11 (Highest ranking: 2)
Grand Slam titles: 2 (Wimbledon 2011, 2014)
Career WTA titles: 27
2019 French Open performance: Did not play
Best French Open performance: Semi-finals (2020, 2012)
ROAD TO FINAL
First round: Oceane Dodin (France) 6-3 7-5
Second round: Jasmine Paolini (Italy) 6-3 6-3
Third round: Leylah Fernandez (Canada) 7-5 6-3
Fourth round: Zhang Shuai (China) 6-2 6-4
Quarter-finals: Laura Siegemund (Germany) 6-3 6-3
Kvitova is seeking a first crown at Roland Garros, where she made an emotional return to professional tennis three years ago after six months out following a burglar attack at home that left her with damaged nerves and tendons in her left hand.
She has looked in ominous form in Paris, reaching her second semi-final at the Grand Slam after a gap of eight years and is yet to lose a set at this year’s tournament.
HEAD-TO-HEAD: Kvitova leads 2-0
2019 Kvitova d Kenin 6-1 6-4 (Madrid Masters, clay)
2018 Kvitova d Kenin 3-6 6-2 6-4 (Miami Masters, hard)
(Compiled by Shrivathsa Sridhar in Bengaluru, editing by Ed Osmond)