Sofia Kenin named WTA Player of the Year
December 10, 2020
According to Tennis Now on Tuesday, Sofia Kenin of the United States has been named the Women’s Tennis Association Player of the Year. Kenin becomes the eighth American to win the award, following Tracy Austin (1980), Chris Evert (1981), Martina Navratilova (1982, 1983, 1984, 1985, 1986), Lindsay Davenport (1998, 1999), Venus Williams (2000), Jennifer Capriati (2001), and Serena Williams (2002, 2008, 2009, 2012, 2013, 2014, and 2015).
A major reason why Kenin was victorious was because of her stellar play at the 2020 Australian Open in Melbourne. Kenin won her first career grand slam title by beating Garbine Muguruza of Spain in the final, 4-6, 6-2, 6-2. Kenin was also the only women’s player this season to reach two Grand Slam singles finals. She reached the final of the French Open, before losing to Iga Swiatek of Poland, 6-4, 6-1.
Kenin won a second tournament on the WTA Tour in 2020. In March, she defeated Anna-Lena Friedsam of Germany in the final of the Lyon Open, 6-2, 4-6, 6-4.
The WTA also handed out four other notable awards. The WTA Doubles Team of the Year went to Kristina Mladenovic of France and Timea Babos of Hungary. Mladenovic and Babos both won the Australian Open and French Open together in 2020.
Swiatek won the WTA Most Improved Player of the Year Award. When she won the French Open, Swiatek was the lowest-ranked French Open women’s singles champion, at 54th in the world.
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The WTA Newcomer of the Year Award went to Argentina’s Nadia Podoroska. At the end of 2019, Podoroska was ranked 255th in the world. However, a shocking semifinal appearance at the French Open moved her up the rankings to 47th.
Finally, the WTA Comeback Player of the Year winner was Victoria Azarenka of Belarus. Azarenka reached her first grand slam final in seven years in 2020, as she reached the final of the 2020 U.S. Open. Azarenka was beaten by Naomi Osaka of Japan, 1-6, 6-3, 6-3. Azarenka had taken time off from the tour to give birth to her son, Leo.
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