Phase shift through resistor in wilkinson power divider?

Is it common to make quarterwave sections a little shorter? Can this effect be neglected at 10ghz? If resistor dielectric have high Er phase shift can be significant. I did not found solution yet.

You mean that the SMD body (ceramic) above the line increases the effective dielectric constant of the line?

I can see your point, but most of the dielectric field will be between the line and ground, so whatever you place on top of the line will only change the fringing fields. I would expect only a small effect. What is the length of the line, and how much of that is covered by your SMD resistor?

For example i want make Wilkinson power divider in shape of ring. But it can be any shape, for example two 70 Ohm quarterwave lines with 100 Ohm resistor at the end. I can calculate this microstrip ring radius, ring consist of two quarterwave arms, so total line length is lambda/2. This bending is only for reduced size of the divider, and identical to simple lines arrangement.
Then microstrip ring radius is R=L/(2*Pi) = (lambda/2) / (2 *Pi) = lambda / Pi
There would be a gap for placing 100 Ohm resistor. Obviously this resistor must be part of 180 degree phase shift through two arms, so both arms must be shorter than quarterwave.
I searched for smd resistor structure images, and it seems that resistors ceramic substrate is below conducting layer. So as i understand it affects phase very much. And if resistor is something like 0805 or 0603, there about millimeter length with unknown phase shift.

Correct.

This is where it gets wrong. Each of the lines must be lambda/4 from coupler input to coupler output where the loads are connected. The resistor size isn't part of the line length calculation. Your ring length must be 2 * lambda/4 plus resistor length.

The resistor is not part of the 1:2 impedance tranformation in the coupler, and in normal operation the voltage at both resistor terminals is identical and there is no current in the resistor.