Australian Open: Daniil Medvedev and Stefanos Tsitsipas reach fourth round

Russian world number two Daniil Medvedev reached the Australian Open fourth round with a straight-set win over Dutchman Botic van de Zandschulp.

The 2021 finalist claimed a comfortable 6-4 6-4 6-2 win against the world number 57 inside Margaret Court Arena.

The US Open champion plays Australian wildcard Christopher O'Connell or American Maxime Cressy in the last 16.

Greek fourth seed Stefanos Tsitsipas also progressed with victory in four sets against Benoit Paire.

Tsitsipas, 23, recorded a 6-3 7-5 6-7 (2-7) 6-4 victory on Rod Laver Arena against Frenchman Paire, ranked 56th.

The two-time semi-finalist will face American 20th seed Taylor Fritz for a place in the quarter-finals.

Fritz overcame Spanish 15th seed Roberto Bautista Agut in five sets, battling back from losing consecutive sets to win 6-0 3-6 3-6 6-4 6-3.

After coming through an entertaining four-set match against Australian Nick Kyrgios, Medvedev encountered few problems against the only player to win a set against him on his run to the US Open title in September.

A break of serve in the third game of the opening two sets was enough to put the Russian, 25, in control of his third-round match — although the second seed did face a break point when leading 3-2 in the second.

But the impressive Medvedev, the highest ranked player in the draw, recorded a double break in the third set as he clinically closed out the match by winning five consecutive games.

Neither of his potential last-16 opponents O'Connell, 27, or Cressy, 24, have previously reached the fourth round of a Grand Slam.

Tsitsipas had won his previous three meetings with Paire and took his first break point of the match as he raced into a 3-0 lead on the way to claiming the first set.

The 23-year-old escaped two