Joe Burrow named AFC Offensive Player of the Week for the third time after record-tying win

CINCINNATI — Cincinnati Bengals Quarterback Joe Burrow was named AFC Offensive Player of the Week for the third time this season.

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Burrow was named AFC Offensive Player of the Week for Week 16 after his performance against the New England Patriots on Saturday, scoring a 22-18 win.

Burrow completed 40 of his 52 passes, at least two passes to eight different receivers, for a total of 375 yards and three touchdowns, rating him 99.4, a Cincinnati Bengals spokesperson stated.

In the first half of the game, Burrow completed 28 of 36 for 284 yards with three touchdowns, tying for the most completions in a single half this season.

The quarterback led all NFL players in Week 16 with his number of completions and passing yardage, the spokesperson continued. “His three passing scores were the most by an AFC quarterback.”

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The 40 completions in one game marked not only a career high, but also tied the historic single-game team record set by Ken Anderson on December 20, 1982, the spokesperson said.

“He now has seven career games with at least 350 passing years and three passing touchdowns,” the spokesperson informed.

The accomplishment ties the second-most record for such games within the player’s first three seasons in NFL history, the spokesperson stated.

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The recognition is Burrow’s third this season and fifth overall.

Bengals players combined have won this award six times this year.

The other Bengals players of the weeks were Trey Hendrickson, Evan McPherson, and Joe Mixon.

Cincinnati’s next game will be on January 2 when they host the Buffalo Bills at Paycor Stadium on Monday Night Football.

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