No bigger week than this one in AFC East

I was away from the blog for much of Tuesday while traveling back from New Orleans, but I just got back to the new home office in suburban Buffalo and wanted to take a few moments to acknowledge the most significant seven days in the AFC East all year.

The stretch began Monday night, when the New England Patriots cracked the door for the rest of the AFC East by losing for the second time in three games. The Patriots remain the prohibitive favorite to win the division, but we'll learn gobs this week.

For the only time all year, all four AFC East teams will play each other.

The New York Jets will play the Buffalo Bills on Thursday night in Toronto. The Patriots will visit the Miami Dolphins on Sunday in Land Shark Stadium.

I will attend both games. Counting the Patriots' collapse in New Orleans, I'll trek 4,675 miles (205 miles in my car with Florida plates and still no ice scraper and 4,470 by air) to see every AFC East team in a concentrated period at a critical juncture.

The loser of the Jets-Bills game will be all but eliminated, and even if the Bills win they still will be two games below .500 with four to play. They'd both be 5-7, would have to run the table and get serious help along the way. In other words, the Jets need this one much more than the suddenly feisty Bills do.

Sunday's game is imperative to New England and Miami.

The Patriots, preparing on a short week for back-to-back away games, are 0-4 in true road games. They're desperate for a victory. So are the Dolphins, who could've nudged a game above .500 had they beaten the Bills on Sunday but imploded in the final minutes.

This is the week that could rip the AFC East wide open or bunch everybody together for a fascinating final month of jockeying.