Jordan Burroughs Becomes Most Decorated American Wrestler Ever

The New Jersey native made history on Friday at the world championships in Belgrade.

Over the last 11 years, American freestyle wrestler Jordan Burroughs has staked his claim as one of the greatest competitors in U.S. wrestling history.

On Friday, the 34-year-old New Jersey native further cemented that legacy by claiming his sixth world championship to become the most decorated American wrestler of all-time.

Using his signature double-leg takedown, Burroughs, a 2012 Olympic gold medalist in the 74 kg division, defeated Iran’s Mohammad Nokhodi, 4-2, to win the men’s freestyle 79 kg title at the World Championships in Belgrade. The victory secured the legendary grappler’s second 79 kg title and a record seventh combined Olympic/World title, breaking a tie with two-time Olympic gold medalist John Smith for the most ever by an American wrestler.

Since winning his first world title in 2011, Burroughs has gone 10-0 in medal matches between the Olympics and worlds; he also claimed gold in ’13, ’15, ’17 and ’21. Friday’s match also marked Burroughs’ fifth win in Belgrade in the last two days and served as a rematch of last year’s world final when Burroughs beat Nokhodi for the same title in Oslo. 

In addition to owning seven golds, Burroughs also boasts three bronze medals he earned competing in the 74kg weight class at the 2014, ’18 and ’19 world championships. His 10 total medals are the second-most in U.S. history behind Bruce Baumgartner‘s 13.

While Burroughs’s reign atop the sport has solidified his name among the greatest the country has ever seen, all signs indicate his career may soon be coming to a close. Burroughs, who has spent most of his career at 74 kg, has maintained in recent months that he plans to retire after the 2024 Olympics, regardless of whether or not he makes the team.