Mikaela Shiffrin to enter women's downhill at Winter Olympics

Posted at 10:20 PM, Feb 13, 2022

Mikaela Shiffrin will enter the women’s downhill event at the 2022 Winter Olympics, she revealed following Monday's final training session.

“I’m planning to do it, yeah,” she told NBC’s Todd Lewis at the finish area. As recently as Saturday's first training session (the second session on Sunday was cancelled due to bad weather), the 26-year-old American was non-committal on her status in the downhill, an event in which she rarely competes. She said the decision would come down primarily to how well her training times stacked up in comparison to the rest of the field.

She went 1:34.58 in Monday's practice, the 17th-fastest time overall but just 1.40 behind the leading mark set by Switzerland's Joana Haehlen.

“My times are pretty good... speed’s pretty good," she said after the run. "I felt confident on my skis. I feel like I know where I am on the course, and those pieces are all pretty good."

"It's just important that I take any practice that I can get," she said.

It will mark the first time Shiffrin has competed in the downhill at the Winter Olympics. Just two of her 73 career World Cup victories – the vast majority having come in the more technical slalom and giant slalom disciplines – are in downhill.

The decision means that Shiffrin will almost certainly accomplish her pre-Olympics goal of competing in all five individual events at the Winter Games. She previously committed to the women's combined event, scheduled for February 17, in which she is the defending silver medalist.

Shiffrin also, according to multiple reports, plans to compete for the U.S. in the team parallel slalom event for the first time in her Olympic career, bringing her total Olympic schedule to six medal events.

The Vail, Colorado, native has struggled to find her footing thus far at her third Winter Olympics. She compiled a pair of shocking DNFs in her two best events, the slalom and giant slalom, then managed a ninth-place result in the super-G.